Foundations of Single-Valued (T, I, N, F)-Neutrosophic Topological Spaces

1. Introduction

Fuzzy topology originates from the replacement of ordinary characteristic functions by membership functions with values in the unit interval. Zadeh’s fuzzy sets [20] and Chang’s fuzzy topological spaces [2] initiated this program, while Goguen’s L-fuzzy sets made explicit that the value domain may be a general ordered structure rather than only $[0,1]$ [7]. From a topological point of view, the relevant question is how generalized open sets should be ordered, intersected, and united so that the standard notions of interior, closure, subspace, product, and continuity remain available.

Atanassov’s intuitionistic fuzzy sets separate membership from non-membership [1], and Coker developed the corresponding intuitionistic fuzzy topology [4]. The relation between such two-coordinate systems and more general L-fuzzy structures was subsequently emphasized in the literature [8,19]. Neutrosophic theory enlarges the value domain further by treating truth, indeterminacy, and falsehood as separate components. Topological developments include the work of Lupiáñez on neutrosophic and interval neutrosophic topologies [10,11,12], the three-component neutrosophic topological framework of Salama and Alblowi [14], and later work on refined neutrosophic topologies [13,16,18].

Smarandache introduced the specific quadruple $(T,I,N,F)$ in 2017 [17]. In that source the established terminology is $(T,I,N,F)$ neutrosophic set; when all four components are numerical values in $[0,1]$, we shall speak more precisely of a single-valued $(T,I,N,F)$-neutrosophic set. For brevity only, after this first formal declaration we use the abbreviation TINF. Every element is then assigned a degree of truth T, pure indeterminacy I, neutrality N, and falsehood F, without a normalization constraint on their sum. The present paper takes this four-valued object as the membership scale of a topology. The semantic distinction between pure indeterminacy and neutrality is retained in the notation, but the aim here is not to develop an epistemic or applicative interpretation of the four coordinates. Instead, we ask what the corresponding topological spaces, maps, point constructions, and basic topological properties should be.

A topological study must also be positioned carefully with respect to other four-component extensions. Quadripartitioned neutrosophic sets use quadruples $(T,C,G,F)$ and already support a corresponding topology [3,5]. Turiyam sets use a fourth value in addition to truth, indeterminacy, and falsity [15], and recent Fuzzy Neutrosophic Turiyam work explicitly introduces open and closed sets, interior, and closure [6]. Consequently, the mathematical contribution of a TINF topology cannot be the mere presence of four coordinates. What is needed first is a topological development that makes explicit which definitions are genuinely part of the TINF setting and which are transported from closely related four-component frameworks.

The present paper provides such a foundation. We define topological spaces whose open sets are single-valued $(T,I,N,F)$-neutrosophic sets and develop the basic theory of open and closed TINF sets, interior, closure, boundary, subspaces, products, and continuous maps. The order and complement on the degree set are fixed only insofar as they are needed to define these operators and their dualities. Classical topological spaces are represented by crisp TINF characteristic sets, giving each TINF topology an ordinary crisp core. We also record the precise relation with quadripartitioned neutrosophic topology and with Turiyam min–max open-set operations, in order to delimit rather than inflate the novelty of the construction.

The main topological development is the systematic construction of the TINF point space and its induced ordinary topology, motivated by the neutrosophic singleton approach of Jafari, Nordo, and Thakur [9]. This construction yields an exact characterization of continuity and homeomorphisms, an exact support-Hausdorff correspondence, and corrected forms of connectedness and compactness. The corrections are necessary because unions of point-basic open sets do not always coincide with the point set associated with a join of TINF-valued open sets.

2. Single-Valued $(T,I,N,F)$-Neutrosophic Sets and Topological Operations

2.1. Single-Valued $(T,I,N,F)$-Neutrosophic Sets

Let X be a nonempty set.

Definition 1.

A single-valued $(T,I,N,F)$-neutrosophic set (briefly, a TINF set) on X is a map

$$A:X⟶{[0,1]}^4,A\left(x\right)=\left(T_A\left(x\right),I_A\left(x\right),N_A\left(x\right),F_A\left(x\right)\right).$$

No restriction such as $T_A\left(x\right)+I_A\left(x\right)+N_A\left(x\right)+F_A\left(x\right)=1$ is imposed. This agrees with the single-valued specialization of the $(T,I,N,F)$ framework [17].

For the present topological purpose, intersections and unions of TINF sets are required to behave as genuine set operations. We therefore use the coordinatewise min–max operations described below; no alternative t-norm or t-conorm operations are considered here.

2.2. Order, Complement, and Topological Set Operations

Definition 2.

The single-valued $(T,I,N,F)$-neutrosophic topological value scale, abbreviated TINF value scale, is

$${\mathbb{L}}_{\mathrm{TINF}}:=[0,1]\times {[0,1]}^{\mathrm{op}}\times [0,1]\times {[0,1]}^{\mathrm{op}}.$$

Here ${[0,1]}^{\mathrm{op}}$ denotes the unit interval endowed with the reverse of its usual order. Thus the truth and neutrality coordinates are increasing, whereas the pure indeterminacy and falsehood coordinates are decreasing. For

$$p=(t,i,n,f),q=(t^{\prime },i^{\prime },n^{\prime },f^{\prime }),$$

define

$$p{\le }_{\mathrm{TINF}}q\iff t\le t^{\prime },i\ge i^{\prime },n\le n^{\prime },f\ge f^{\prime }.$$

Its bottom and top elements are

$$\mathbf{0}=(0,1,0,1),\mathbf{1}=(1,0,1,0).$$

For a family $q_{\lambda }=(t_{\lambda },i_{\lambda },n_{\lambda },f_{\lambda })$, its TINF union and intersection are defined by

$$\begin{array}{cc}\hfill \underset{\lambda }{\bigvee }{q}_{\lambda }& =\left(\underset{\lambda }{sup}{t}_{\lambda },\underset{\lambda }{inf}{i}_{\lambda },\underset{\lambda }{sup}{n}_{\lambda },\underset{\lambda }{inf}{f}_{\lambda }\right),\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill \underset{\lambda }{\bigwedge }{q}_{\lambda }& =\left(\underset{\lambda }{inf}{t}_{\lambda },\underset{\lambda }{sup}{i}_{\lambda },\underset{\lambda }{inf}{n}_{\lambda },\underset{\lambda }{sup}{f}_{\lambda }\right).\hfill \end{array}$$

(2)

The union and intersection of an empty family are understood to be $\mathbf{0}$ and $\mathbf{1}$, respectively.

Definition 3.

Define

$$c:{\mathbb{L}}_{\mathrm{TINF}}⟶{\mathbb{L}}_{\mathrm{TINF}},c(t,i,n,f)=(f,n,i,t).$$

For a TINF set A, define its complement pointwise by

$$A^{\mathsf{c}}\left(x\right)=c\left(A\left(x\right)\right).$$

The complement is involutive, since $c\left(c\right(q\left)\right)=q$ for every $q\in {\mathbb{L}}_{\mathrm{TINF}}$.

All TINF set operations on ${\mathbb{L}}_{\mathrm{TINF}}^X$ are taken pointwise. In particular, for TINF sets $A,B$ on X, we write $A\le B$ when $A\left(x\right){\le }_{\mathrm{TINF}}B\left(x\right)$ for every $x\in X$. Their TINF intersection and union are

$$A\wedge B=\left(min(T_A,T_B),max(I_A,I_B),min(N_A,N_B),max(F_A,F_B)\right),$$

and

$$A\vee B=\left(max(T_A,T_B),min(I_A,I_B),max(N_A,N_B),min(F_A,F_B)\right).$$

We write ${\mathbf{0}}_X$ and ${\mathbf{1}}_X$ for the constant bottom and top TINF sets.

The preceding definitions imply that arbitrary TINF unions and intersections exist, finite intersections distribute over arbitrary unions, and complement interchanges unions and intersections. These are the only properties of the value scale used in the topological arguments below.

3. Single-valued $(T,I,N,F)$-neutrosophic topological spaces

Definition 4.

A single-valued $(T,I,N,F)$-neutrosophic topology (briefly, a TINF topology) on a nonempty set X is a family $\tau \subseteq {\mathbb{L}}_{\mathrm{TINF}}^X$ such that

(T1)

${\mathbf{0}}_X,{\mathbf{1}}_X\in \tau$;

(T2)

$U,V\in \tau$ implies $U\wedge V\in \tau$;

(T3)

if ${\left\{U_{\lambda }\right\}}_{\lambda \in \Lambda }\subseteq \tau$, then ${\bigvee }_{\lambda \in \Lambda }{U}_{\lambda }\in \tau$.

The pair $(X,\tau )$ is a single-valued $(T,I,N,F)$-neutrosophic topological space (briefly, a TINF topological space), and members of τ are$(T,I,N,F)$-neutrosophic open sets (briefly,TINF open sets).

This follows the Chang–Goguen tradition of topologies with generalized-valued open sets [2,7].

Example 1.

For every nonempty set X,

$${\tau }_{\mathrm{ind}}=\{{\mathbf{0}}_X,{\mathbf{1}}_X\}$$

is the indiscrete TINF topology and

$${\tau }_{\mathrm{dis}}={\mathbb{L}}_{\mathrm{TINF}}^X$$

is the discrete TINF topology.

3.1. Generated Topologies and Bases

The construction of subspaces and products will use generated topologies and bases in the same way as ordinary topology.

Proposition 1.

Let ${\left\{{\tau }_{\alpha }\right\}}_{\alpha \in A}$ be a nonempty family of TINF topologies on the same set X. Then ${\bigcap }_{\alpha \in A}{\tau }_{\alpha }$ is a TINF topology on X.

Proof. 

Every ${\tau }_{\alpha }$ contains ${\mathbf{0}}_X$ and ${\mathbf{1}}_X$, so both constant TINF sets belong to the intersection. If U and V belong to every ${\tau }_{\alpha }$, then so does $U\wedge V$. Likewise, if every member of a family $\left\{U_{\lambda }\right\}$ belongs to every ${\tau }_{\alpha }$, then ${\bigvee }_{\lambda }{U}_{\lambda }$ belongs to every ${\tau }_{\alpha }$. Hence the intersection satisfies the three axioms of a TINF topology. □

Definition 5.

Let $\mathcal{S}\subseteq {\mathbb{L}}_{\mathrm{TINF}}^X$. The TINF topology generated by $\mathcal{S}$, denoted by $\tau \left(\mathcal{S}\right)$, is the intersection of all TINF topologies on X that contain $\mathcal{S}$. This family is nonempty because the discrete TINF topology ${\mathbb{L}}_{\mathrm{TINF}}^X$ contains $\mathcal{S}$.

If τ is a TINF topology on X, a family $\mathcal{B}\subseteq \tau$ is a TINF base for τ if every $U\in \tau$ is the union of the base elements contained in U, that is,

$$U=\bigvee \{B\in \mathcal{B}:B\le U\}.$$

Proposition 2.

Suppose $\mathcal{B}\subseteq {\mathbb{L}}_{\mathrm{TINF}}^X$ contains ${\mathbf{1}}_X$ and is closed under finite intersections. Then

$$\tau \left(\mathcal{B}\right)=\left\{\bigvee \mathcal{E}:\mathcal{E}\subseteq \mathcal{B}\right\},$$

where the union of the empty family is ${\mathbf{0}}_X$. In particular, $\mathcal{B}$ is a TINF base for $\tau \left(\mathcal{B}\right)$.

Proof. 

Let ${\tau }_{\mathcal{B}}$ denote the family on the right. It contains ${\mathbf{0}}_X$ as the empty union and ${\mathbf{1}}_X$ because ${\mathbf{1}}_X\in \mathcal{B}$. Arbitrary unions of members of ${\tau }_{\mathcal{B}}$ are again unions of members of $\mathcal{B}$. If $U={\bigvee }_{i\in I}{B}_i$ and $V={\bigvee }_{j\in J}{C}_j$, distributivity gives

$$U\wedge V=\underset{(i,j)\in I\times J}{\bigvee }(B_i\wedge C_j).$$

Every $B_i\wedge C_j$ belongs to $\mathcal{B}$, so $U\wedge V\in {\tau }_{\mathcal{B}}$. Consequently, ${\tau }_{\mathcal{B}}$ is a TINF topology containing $\mathcal{B}$.

Any TINF topology containing $\mathcal{B}$ must contain every union of members of $\mathcal{B}$. Hence ${\tau }_{\mathcal{B}}$ is the smallest such topology and equals $\tau \left(\mathcal{B}\right)$. The displayed description also proves the asserted base property. □

3.2. Closed Sets, Interior, Closure, and Boundary

Definition 6.

A TINF set C on $(X,\tau )$ is closed if $C^{\mathsf{c}}\in \tau$. For $A\in {\mathbb{L}}_{\mathrm{TINF}}^X$, its interior and closure are, respectively,

$${Int}_{\tau }\left(A\right)=\bigvee \{U\in \tau :U\le A\}$$

and

$${Cl}_{\tau }\left(A\right)=\bigwedge \{C\in {\mathbb{L}}_{\mathrm{TINF}}^X:A\le C,C^{\mathsf{c}}\in \tau \}.$$

The TINF boundary of A is

$${Bd}_{\tau }\left(A\right):={Cl}_{\tau }\left(A\right)\wedge {Cl}_{\tau }\left(A^{\mathsf{c}}\right).$$

When the topology is clear, the subscript τ is omitted. If several spaces occur, we write ${Int}_X$, ${Cl}_X$, and ${Bd}_X$ for the operators determined by the topology on X.

Proposition 3.

The TINF closed sets contain ${\mathbf{0}}_X$ and ${\mathbf{1}}_X$, are closed under arbitrary intersections, and are closed under finite unions.

Proof. 

The complements of ${\mathbf{0}}_X$ and ${\mathbf{1}}_X$ are ${\mathbf{1}}_X$ and ${\mathbf{0}}_X$, respectively, and both are open. If every $C_{\lambda }$ is closed, then De Morgan’s rule gives ${\left({\bigwedge }_{\lambda }{C}_{\lambda }\right)}^{\mathsf{c}}={\bigvee }_{\lambda }{C}_{\lambda }^{\mathsf{c}}$, which is open; hence ${\bigwedge }_{\lambda }{C}_{\lambda }$ is closed. For two closed TINF sets C and D, ${(C\vee D)}^{\mathsf{c}}=C^{\mathsf{c}}\wedge D^{\mathsf{c}}$ is open, so $C\vee D$ is closed. □

Theorem 1.

For all TINF sets $A,B$ on X:

(i)

$Int\left(A\right)\le A\le Cl\left(A\right)$;

(ii)

$A\le B$ implies $Int\left(A\right)\le Int\left(B\right)$ and $Cl\left(A\right)\le Cl\left(B\right)$;

(iii)

$Int(Int(A\left)\right)=Int\left(A\right)$ and $Cl(Cl(A\left)\right)=Cl\left(A\right)$;

(iv)

$Int(A\wedge B)=Int\left(A\right)\wedge Int\left(B\right)$;

(v)

$Cl(A\vee B)=Cl\left(A\right)\vee Cl\left(B\right)$;

(vi)

$Cl\left(A\right)={(Int\left(A^{\mathsf{c}}\right))}^{\mathsf{c}},Int\left(A\right)={(Cl\left(A^{\mathsf{c}}\right))}^{\mathsf{c}};$

(vii)

$Bd\left(A\right)$ is closed and $Bd\left(A\right)\le Cl\left(A\right)$;

(viii)

$Bd\left(A^{\mathsf{c}}\right)=Bd\left(A\right)$.

Proof. 

By definition, $Int\left(A\right)$ is a union of open TINF sets lying below A; it is therefore open and satisfies $Int\left(A\right)\le A$. Similarly, $Cl\left(A\right)$ is the intersection of all closed TINF sets lying above A, so it is closed and $A\le Cl\left(A\right)$. This proves (i).

If $A\le B$, every open TINF set below A also lies below B. Taking their unions yields $Int\left(A\right)\le Int\left(B\right)$. Every closed TINF set above B is also above A; taking the corresponding intersections yields $Cl\left(A\right)\le Cl\left(B\right)$. Thus (ii) holds.

Since $Int\left(A\right)$ is open, it is one of the open TINF sets entering the definition of $Int(Int(A\left)\right)$. Consequently, $Int\left(A\right)\le Int(Int(A\left)\right)$. The reverse inequality follows from (i), applied to $Int\left(A\right)$. Hence $Int(Int(A\left)\right)=Int\left(A\right)$. The same argument with closed TINF sets gives $Cl(Cl(A\left)\right)=Cl\left(A\right)$, proving (iii).

For (iv), $Int\left(A\right)\wedge Int\left(B\right)$ is open and lies below $A\wedge B$, whence

$$Int\left(A\right)\wedge Int\left(B\right)\le Int(A\wedge B).$$

Conversely, $Int(A\wedge B)$ lies below both A and B. Since it is open, it lies below both $Int\left(A\right)$ and $Int\left(B\right)$, and therefore below their intersection. This proves (iv).

For (v), the closed TINF set $Cl\left(A\right)\vee Cl\left(B\right)$ contains $A\vee B$, so $Cl(A\vee B)\le Cl\left(A\right)\vee Cl\left(B\right)$. Conversely, monotonicity gives $Cl\left(A\right)\le Cl(A\vee B)$ and $Cl\left(B\right)\le Cl(A\vee B)$; hence their union also lies below $Cl(A\vee B)$.

To prove the first identity in (vi), note that ${(Int\left(A^{\mathsf{c}}\right))}^{\mathsf{c}}$ is closed and contains A, so

$$Cl\left(A\right)\le {(Int\left(A^{\mathsf{c}}\right))}^{\mathsf{c}}.$$

On the other hand, $Cl{\left(A\right)}^{\mathsf{c}}$ is open and lies below $A^{\mathsf{c}}$, which gives $Cl{\left(A\right)}^{\mathsf{c}}\le Int\left(A^{\mathsf{c}}\right)$. Taking complements reverses the inequality and proves the reverse inclusion. Replacing A by $A^{\mathsf{c}}$ gives the second identity.

Both $Cl\left(A\right)$ and $Cl\left(A^{\mathsf{c}}\right)$ are closed. Their intersection is therefore closed, and the definition of boundary gives $Bd\left(A\right)\le Cl\left(A\right)$. This proves (vii).

Finally, the definition of boundary and involutivity of complement give

$$Bd\left(A^{\mathsf{c}}\right)=Cl\left(A^{\mathsf{c}}\right)\wedge Cl\left(A\right)=Bd\left(A\right).$$

□

Theorem 2

(Boundary duality). For every TINF set A on $(X,\tau )$,

$${Bd}_{\tau }\left(A\right)={Cl}_{\tau }\left(A\right)\wedge {\left({Int}_{\tau }\left(A\right)\right)}^{\mathsf{c}}.$$

Thus the TINF boundary satisfies the usual closure–interior formula even though the value scale ${\mathbb{L}}_{\mathrm{TINF}}$ is not Boolean.

Proof. 

Applying Theorem 1(vi) to $A^{\mathsf{c}}$ and using involutivity gives

$${Cl}_{\tau }\left(A^{\mathsf{c}}\right)={\left({Int}_{\tau }\left({\left(A^{\mathsf{c}}\right)}^{\mathsf{c}}\right)\right)}^{\mathsf{c}}={\left({Int}_{\tau }\left(A\right)\right)}^{\mathsf{c}}.$$

Substitution in the definition of ${Bd}_{\tau }\left(A\right)$ proves the asserted identity.

For an explicit coordinate verification, fix $x\in X$ and write

$${Cl}_{\tau }\left(A\right)\left(x\right)=(c_T,c_I,c_N,c_F),{Int}_{\tau }\left(A\right)\left(x\right)=(u_T,u_I,u_N,u_F).$$

The preceding duality and the definition of complement yield

$${Cl}_{\tau }\left(A^{\mathsf{c}}\right)\left(x\right)=(u_F,u_N,u_I,u_T).$$

Since meet in ${\mathbb{L}}_{\mathrm{TINF}}$ is coordinatewise minimum in the truth and neutrality coordinates and coordinatewise maximum in the pure-indeterminacy and falsehood coordinates, both sides of the claimed identity have value

$$\left(min\{c_T,u_F\},max\{c_I,u_N\},min\{c_N,u_I\},max\{c_F,u_T\}\right)$$

at x. Because x was arbitrary, the two TINF sets are equal. The argument uses the De Morgan duality and involutivity of the TINF complement, but no Boolean identity such as $A\vee A^{\mathsf{c}}={\mathbf{1}}_X$. □

Corollary 1.

A TINF set A is open if and only if $Int\left(A\right)=A$, and it is closed if and only if $Cl\left(A\right)=A$.

3.3. Subspaces

Definition 7.

Let $(X,\tau )$ be a TINF topological space and let $⌀\ne Y\subseteq X$. For every TINF set $A:X\to {\mathbb{L}}_{\mathrm{TINF}}$, we write ${A|}_Y:Y\to {\mathbb{L}}_{\mathrm{TINF}}$ for the restriction of A to Y, defined by $\left(A{|}_Y\right)\left(y\right)=A\left(y\right)$ for every $y\in Y$. The subspace TINF topology on Y is

$${\tau |}_Y:=\{U{|}_Y:U\in \tau \}.$$

The pair $(Y,\tau {|}_Y)$ is called the TINF subspace of $(X,\tau )$ determined by Y.

Proposition 4.

${\tau |}_Y$ is a TINF topology on Y.

Proof. 

The restrictions of ${\mathbf{0}}_X$ and ${\mathbf{1}}_X$ are ${\mathbf{0}}_Y$ and ${\mathbf{1}}_Y$. If ${U|}_Y,V{{|}_Y\in \tau |}_Y$, then

$${U|}_Y{\wedge V|}_Y=(U\wedge V){{|}_Y\in \tau |}_Y.$$

For a family $\left\{U_{\lambda }{|}_Y\right\}$ of members of ${\tau |}_Y$,

$$\underset{\lambda }{\bigvee }{U}_{\lambda }{|}_Y=\left(\underset{\lambda }{\bigvee }{U}_{\lambda }\right){{|}_Y\in \tau |}_Y.$$

Thus ${\tau |}_Y$ satisfies all three TINF topology axioms. □

For a TINF set A on Y, define its bottom extension $A^{0,X}$ to X by

$$A^{0,X}\left(x\right)=\left\{\begin{array}{cc}A\left(x\right),\hfill & x\in Y,\hfill \\ \mathbf{0},\hfill & x\in X\setminus Y.\hfill \end{array}\right.$$

Proposition 5

(Closure in a subspace). For every TINF set A on Y,

$${Cl}_Y\left(A\right)={Cl}_X\left(A^{0,X}\right){|}_Y.$$

Proof. 

A TINF set C is closed in Y exactly when ${C=D|}_Y$ for some TINF closed set D on X. Indeed, if C is closed in Y, then $C^{\mathsf{c}}{=U|}_Y$ for some $U\in \tau$, and hence $C=U^{\mathsf{c}}{|}_Y$; the converse follows by taking complements and restricting.

For a TINF closed set D on X, the inequality ${A\le D|}_Y$ is equivalent to $A^{0,X}\le D$: on Y the two inequalities are the same, and outside Y the latter holds because $A^{0,X}$ has the bottom value. Therefore the closed TINF sets used to compute ${Cl}_Y\left(A\right)$ are precisely the restrictions of those used to compute ${Cl}_X\left(A^{0,X}\right)$. Since restriction preserves arbitrary intersections, their intersections satisfy the displayed identity. □

4. Continuous Maps and Products

4.1. Inverse Images and Continuity

Let $f:X\to Y$ be a map. For a TINF set B on Y, its inverse image under f is the TINF set on X defined by

$$f^{\leftarrow }\left(B\right):=B\circ f.$$

Lemma 1.

For every family $\left\{B_{\lambda }\right\}$ of TINF sets on Y and all TINF sets $B,C$ on Y,

$$\begin{array}{cccc}\hfill f^{\leftarrow }\left(\underset{\lambda }{\bigvee }{B}_{\lambda }\right)& =\underset{\lambda }{\bigvee }{f}^{\leftarrow }\left(B_{\lambda }\right),\hfill & \hfill f^{\leftarrow }\left(\underset{\lambda }{\bigwedge }{B}_{\lambda }\right)& =\underset{\lambda }{\bigwedge }{f}^{\leftarrow }\left(B_{\lambda }\right),\hfill \\ \hfill f^{\leftarrow }\left(B^{\mathsf{c}}\right)& ={\left(f^{\leftarrow }\left(B\right)\right)}^{\mathsf{c}},\hfill & \hfill f^{\leftarrow }\left({\mathbf{0}}_Y\right)& ={\mathbf{0}}_X,f^{\leftarrow }\left({\mathbf{1}}_Y\right)={\mathbf{1}}_X.\hfill \end{array}$$

Proof. 

Fix $x\in X$. Evaluating the first identity at x gives

$$f^{\leftarrow }\left(\underset{\lambda }{\bigvee }{B}_{\lambda }\right)\left(x\right)=\left(\underset{\lambda }{\bigvee }{B}_{\lambda }\right)\left(f\left(x\right)\right)=\underset{\lambda }{\bigvee }{B}_{\lambda }\left(f\left(x\right)\right)=\left(\underset{\lambda }{\bigvee }{f}^{\leftarrow }\left(B_{\lambda }\right)\right)\left(x\right).$$

The same calculation with intersections proves the second identity. Since complement and the two constant TINF sets are also defined pointwise, the remaining identities follow by evaluating both sides at an arbitrary $x\in X$. □

Definition 8.

A map $f:(X,{\tau }_X)\to (Y,{\tau }_Y)$ is TINF continuous if $f^{\leftarrow }\left(U\right)\in {\tau }_X$ for every $U\in {\tau }_Y$.

Theorem 3

(Equivalent continuity criteria). For a map $f:(X,{\tau }_X)\to (Y,{\tau }_Y)$, the following are equivalent:

(i)

f is TINF continuous;

(ii)

the inverse image of every TINF closed set is TINF closed;

(iii)

for every $B\in {\mathbb{L}}_{\mathrm{TINF}}^Y$, $f^{\leftarrow }\left({Int}_Y\left(B\right)\right)\le {Int}_X\left(f^{\leftarrow }\left(B\right)\right);$

(iv)

for every $B\in {\mathbb{L}}_{\mathrm{TINF}}^Y$, ${Cl}_X\left(f^{\leftarrow }\left(B\right)\right)\le f^{\leftarrow }\left({Cl}_Y\left(B\right)\right)$.

Proof. 

Assume (i), and let C be TINF closed in Y. Since $C^{\mathsf{c}}$ is open, $f^{\leftarrow }\left(C^{\mathsf{c}}\right)$ is open in X. By Lemma 1, this TINF set is the complement of $f^{\leftarrow }\left(C\right)$, so $f^{\leftarrow }\left(C\right)$ is closed. This proves (ii). Conversely, if inverse images preserve closed TINF sets and U is open in Y, then $U^{\mathsf{c}}$ is closed. Hence $f^{\leftarrow }\left(U^{\mathsf{c}}\right)$ is closed, and its complement $f^{\leftarrow }\left(U\right)$ is open. Thus (ii) implies (i).

Assume (i). The TINF set $f^{\leftarrow }\left({Int}_Y\left(B\right)\right)$ is open and lies below $f^{\leftarrow }\left(B\right)$. By the defining maximality of the interior, it lies below ${Int}_X\left(f^{\leftarrow }\left(B\right)\right)$, proving (iii). Conversely, suppose (iii) holds and let B be open in Y. Then $B={Int}_Y\left(B\right)$, and therefore

$$f^{\leftarrow }\left(B\right)\le {Int}_X\left(f^{\leftarrow }\left(B\right)\right)\le f^{\leftarrow }\left(B\right).$$

Equality follows, so $f^{\leftarrow }\left(B\right)$ is open. Hence (iii) implies (i).

Assume (ii). The TINF set $f^{\leftarrow }\left({Cl}_Y\left(B\right)\right)$ is closed and contains $f^{\leftarrow }\left(B\right)$. By the defining minimality of closure, ${Cl}_X\left(f^{\leftarrow }\left(B\right)\right)\le f^{\leftarrow }\left({Cl}_Y\left(B\right)\right)$, which is (iv). Finally, suppose (iv) holds and let B be closed in Y. Since ${Cl}_Y\left(B\right)=B$, we obtain

$${Cl}_X\left(f^{\leftarrow }\left(B\right)\right)\le f^{\leftarrow }\left(B\right)\le {Cl}_X\left(f^{\leftarrow }\left(B\right)\right).$$

Thus $f^{\leftarrow }\left(B\right)$ equals its closure and is closed. This proves (ii), and completes the cycle of equivalences. □

4.2. Products

Let $(X,{\tau }_X)$ and $(Y,{\tau }_Y)$ be TINF spaces, and let $p_X:X\times Y\to X$ and $p_Y:X\times Y\to Y$ be the coordinate projections.

Definition 9.

For $U\in {\tau }_X$ and $V\in {\tau }_Y$, the TINF set

$$p_X^{\leftarrow }\left(U\right)\wedge p_Y^{\leftarrow }\left(V\right)$$

is called an open TINF rectangle. The product TINF topology is

$${\tau }_X\otimes {\tau }_Y:=\tau \left(\mathcal{B}\right),$$

where

$$\mathcal{B}=\left\{p_X^{\leftarrow }\left(U\right)\wedge p_Y^{\leftarrow }\left(V\right):U\in {\tau }_X,V\in {\tau }_Y\right\}.$$

The family $\mathcal{B}$ contains ${\mathbf{1}}_{X\times Y}$ and is closed under finite intersections, since

$$\left(p_X^{\leftarrow }\left(U_1\right)\wedge p_Y^{\leftarrow }\left(V_1\right)\right)\wedge \left(p_X^{\leftarrow }\left(U_2\right)\wedge p_Y^{\leftarrow }\left(V_2\right)\right)=p_X^{\leftarrow }(U_1\wedge U_2)\wedge p_Y^{\leftarrow }(V_1\wedge V_2).$$

By Proposition 2, the open TINF rectangles form a TINF base for ${\tau }_X\otimes {\tau }_Y$.

Theorem 4

(Product continuity criterion). The projections $p_X$ and $p_Y$ are TINF continuous. A map

$$h:(Z,{\tau }_Z)⟶(X\times Y,{\tau }_X\otimes {\tau }_Y)$$

is TINF continuous if and only if $p_X\circ h$ and $p_Y\circ h$ are TINF continuous.

Proof. 

For $U\in {\tau }_X$,

$$p_X^{\leftarrow }\left(U\right)=p_X^{\leftarrow }\left(U\right)\wedge p_Y^{\leftarrow }\left({\mathbf{1}}_Y\right)$$

is an open TINF rectangle; hence $p_X$ is continuous. The same argument proves continuity of $p_Y$.

If h is continuous, then for every $U\in {\tau }_X$,

$${(p_X\circ h)}^{\leftarrow }\left(U\right)=h^{\leftarrow }\left(p_X^{\leftarrow }\left(U\right)\right)$$

is open in Z. Thus $p_X\circ h$ is continuous, and the proof for $p_Y\circ h$ is the same.

Conversely, suppose both coordinate maps are continuous. For every open TINF rectangle,

$$h^{\leftarrow }\left(p_X^{\leftarrow }\left(U\right)\wedge p_Y^{\leftarrow }\left(V\right)\right)={(p_X\circ h)}^{\leftarrow }\left(U\right)\wedge {(p_Y\circ h)}^{\leftarrow }\left(V\right).$$

The right-hand side is open in Z. Every member of the product topology is a union of open TINF rectangles, and inverse images preserve arbitrary unions by Lemma 1. Therefore the inverse image under h of every product-open TINF set is open, and h is continuous. □

5. Classical Topologies and the Crisp Core

Definition 10.

For $S\subseteq X$, define the crisp TINF characteristic set

$${\chi }_S^{\mathrm{TINF}}\left(x\right)=\left\{\begin{array}{cc}\mathbf{1},\hfill & x\in S,\hfill \\ \mathbf{0},\hfill & x\notin S.\hfill \end{array}\right.$$

Proposition 6.

For any family $\left\{S_{\lambda }\right\}$ of ordinary subsets of X,

$${\chi }_{{\bigcup }_{\lambda }{S}_{\lambda }}^{\mathrm{TINF}}=\underset{\lambda }{\bigvee }{\chi }_{S_{\lambda }}^{\mathrm{TINF}},{\chi }_{{\bigcap }_{\lambda }{S}_{\lambda }}^{\mathrm{TINF}}=\underset{\lambda }{\bigwedge }{\chi }_{S_{\lambda }}^{\mathrm{TINF}},$$

and

$${\left({\chi }_S^{\mathrm{TINF}}\right)}^{\mathsf{c}}={\chi }_{X\setminus S}^{\mathrm{TINF}}.$$

Consequently, every ordinary topology $\mathcal{T}$ on X determines a TINF topology

$${\mathcal{T}}^{\mathrm{TINF}}=\{{\chi }_U^{\mathrm{TINF}}:U\in \mathcal{T}\},$$

and every TINF topology τ determines an ordinary topology

$$\mathcal{C}\left(\tau \right):=\{U\subseteq X:{\chi }_U^{\mathrm{TINF}}\in \tau \},$$

called the crisp core of τ.

A TINF topology of the form ${\mathcal{T}}^{\mathrm{TINF}}$ for an ordinary topology $\mathcal{T}$ is called a crisp TINF topology.

Proof. 

At a point $x\in X$, the value of ${\chi }_{{\bigcup }_{\lambda }{S}_{\lambda }}^{\mathrm{TINF}}$ is $\mathbf{1}$ exactly when x belongs to at least one $S_{\lambda }$; this is exactly the pointwise union of the ${\chi }_{S_{\lambda }}^{\mathrm{TINF}}$. The intersection identity is proved in the same way. Moreover, complement exchanges the two values $\mathbf{0}$ and $\mathbf{1}$, so ${\left({\chi }_S^{\mathrm{TINF}}\right)}^{\mathsf{c}}={\chi }_{X\setminus S}^{\mathrm{TINF}}$.

It follows that ${\mathcal{T}}^{\mathrm{TINF}}$ contains the two constant TINF sets and is closed under arbitrary TINF unions and finite TINF intersections. Hence it is a TINF topology. In the opposite direction, $⌀,X\in \mathcal{C}\left(\tau \right)$ because their TINF characteristic sets are ${\mathbf{0}}_X$ and ${\mathbf{1}}_X$. The two displayed identities show that $\mathcal{C}\left(\tau \right)$ is closed under ordinary arbitrary unions and finite intersections. Therefore $\mathcal{C}\left(\tau \right)$ is an ordinary topology on X. □

Proposition 7.

If

$$f:(X,{\tau }_X)⟶(Y,{\tau }_Y)$$

is TINF continuous, then

$$f:(X,\mathcal{C}\left({\tau }_X\right))⟶(Y,\mathcal{C}\left({\tau }_Y\right))$$

is continuous in the ordinary sense.

Proof. 

Let U be open in $\mathcal{C}\left({\tau }_Y\right)$. By definition, ${\chi }_U^{\mathrm{TINF}}\in {\tau }_Y$. TINF continuity of f gives

$$f^{\leftarrow }\left({\chi }_U^{\mathrm{TINF}}\right)={\chi }_{f^{-1}\left(U\right)}^{\mathrm{TINF}}\in {\tau }_X.$$

The equality follows by evaluating both sides at an arbitrary $x\in X$. Hence $f^{-1}\left(U\right)\in \mathcal{C}\left({\tau }_X\right)$, which proves ordinary continuity of f between the crisp cores. □

6. Single-valued $(T,I,N,F)$-neutrosophic points and the induced point topology

Neutrosophic singleton spaces provide a useful bridge between generalized-valued topology and ordinary topology [9]. The TINF setting allows a parallel construction, but arbitrary unions require special care.

Definition 11.

Let $x\in X$ and $q\in {\mathbb{L}}_{\mathrm{TINF}}\setminus \left\{\mathbf{0}\right\}$. The single-valued $(T,I,N,F)$-neutrosophic point (briefly, TINF point) $x_q$ is the TINF set

$$x_q\left(y\right)=\left\{\begin{array}{cc}q,\hfill & y=x,\hfill \\ \mathbf{0},\hfill & y\ne x.\hfill \end{array}\right.$$

The point x is the support of $x_q$. Let

$${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)=\{x_q:x\in X,q\in {\mathbb{L}}_{\mathrm{TINF}}\setminus \left\{\mathbf{0}\right\}\}.$$

We write

$$x_q⊑A$$

if $q\le A\left(x\right)$, and define

$$V_A:=\{x_q\in {\mathcal{P}}_{\mathrm{TINF}}\left(X\right):x_q⊑A\}.$$

Lemma 2

(Point-set representation). For TINF sets $A,B$ on X and an arbitrary family $\left\{A_{\lambda }\right\}$:

(i)

$A\le B$ if and only if $V_A\subseteq V_B$;

(ii)

$V_{A\wedge B}=V_A\cap V_B$;

(iii)

${\bigcup }_{\lambda }{V}_{A_{\lambda }}\subseteq V_{{\bigvee }_{\lambda }{A}_{\lambda }}$, and the inclusion may be strict;

(iv)

$V_A=⌀$ if and only if $A={\mathbf{0}}_X$;

(v)

if $V_A={\bigcup }_{\lambda }{V}_{A_{\lambda }}$, then $A={\bigvee }_{\lambda }{A}_{\lambda }$.

Proof. 

For (i), suppose $A\le B$. If $x_q\in V_A$, then $q\le A\left(x\right)\le B\left(x\right)$, and hence $x_q\in V_B$. Thus $V_A\subseteq V_B$. Conversely, assume $V_A\subseteq V_B$ and fix $x\in X$. If $A\left(x\right)\ne \mathbf{0}$, the TINF point $x_{A\left(x\right)}$ belongs to $V_A$ and therefore to $V_B$, which means $A\left(x\right)\le B\left(x\right)$. If $A\left(x\right)=\mathbf{0}$, the same inequality holds because $\mathbf{0}$ is the bottom degree. Hence $A\le B$.

For (ii), a TINF point $x_q$ belongs to $V_{A\wedge B}$ exactly when

$$q\le A\left(x\right)\wedge B\left(x\right)\iff q\le A\left(x\right)\mathrm{and}q\le B\left(x\right).$$

This is equivalent to $x_q\in V_A\cap V_B$.

For (iii), $A_{\lambda }\le {\bigvee }_{\mu }{A}_{\mu }$ for every $\lambda$. Part (i) then gives $V_{A_{\lambda }}\subseteq V_{{\bigvee }_{\mu }{A}_{\mu }}$, and taking the ordinary union over $\lambda$ proves the stated inclusion. The strictness assertion is established in Example 2 below.

For (iv), $A={\mathbf{0}}_X$ clearly gives $V_A=⌀$. Conversely, if $A\ne {\mathbf{0}}_X$, there is $x\in X$ for which $A\left(x\right)\ne \mathbf{0}$; then $x_{A\left(x\right)}\in V_A$, so $V_A$ is nonempty.

For (v), the equality of point sets gives $V_{A_{\lambda }}\subseteq V_A$ for every $\lambda$. Part (i) yields $A_{\lambda }\le A$, and therefore ${\bigvee }_{\lambda }{A}_{\lambda }\le A$. For the reverse inequality, fix $x\in X$ with $A\left(x\right)\ne \mathbf{0}$. The point $x_{A\left(x\right)}$ lies in $V_A$ and hence in some $V_{A_{\lambda }}$. Consequently,

$$A\left(x\right)\le A_{\lambda }\left(x\right)\le \underset{\mu }{\bigvee }{A}_{\mu }\left(x\right).$$

If $A\left(x\right)=\mathbf{0}$, the same inequality follows from the fact that $\mathbf{0}$ is the bottom degree. Thus $A\le {\bigvee }_{\lambda }{A}_{\lambda }$, proving equality. □

Example 2

(A strict union inclusion). Let $X=\left\{x\right\}$ and

$$A\left(x\right)=(0.7,0.1,0.1,0.1),B\left(x\right)=(0.1,0.1,0.7,0.1).$$

Then

$$A\left(x\right)\vee B\left(x\right)=(0.7,0.1,0.7,0.1).$$

The point $x_{A\left(x\right)\vee B\left(x\right)}$ lies in $V_{A\vee B}$, but in neither $V_A$ nor $V_B$. Therefore

$$V_A\cup V_B⊊V_{A\vee B}.$$

Theorem 5

(Induced point topology). For a TINF space $(X,\tau )$, the family

$${\mathcal{B}}_{\tau }:=\{V_U:U\in \tau \}$$

is a basis for an ordinary topology ${\sigma }_{\tau }$ on ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$.

Proof. 

Since $V_{{\mathbf{1}}_X}={\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$, the family covers the point space. Moreover,

$$V_U\cap V_V=V_{U\wedge V}$$

by Lemma 2, and $U\wedge V\in \tau$. Thus the intersection of any two members of ${\mathcal{B}}_{\tau }$ is again a member of ${\mathcal{B}}_{\tau }$. The two standard basis conditions are satisfied, so ${\mathcal{B}}_{\tau }$ determines an ordinary topology on ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$. □

Every ordinary map $f:X\to Y$ induces

$$\widehat{f}:{\mathcal{P}}_{\mathrm{TINF}}\left(X\right)⟶{\mathcal{P}}_{\mathrm{TINF}}\left(Y\right),\widehat{f}\left(x_q\right)=f{\left(x\right)}_q.$$

Lemma 3.

For every TINF set B on Y,

$${\widehat{f}}^{-1}\left(V_B\right)=V_{f^{\leftarrow }\left(B\right)}.$$

Proof. 

For $x_q\in {\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$,

$$\widehat{f}\left(x_q\right)\in V_B\iff q\le B\left(f\left(x\right)\right)\iff q\le \left(f^{\leftarrow }\left(B\right)\right)\left(x\right)\iff x_q\in V_{f^{\leftarrow }\left(B\right)}.$$

□

Theorem 6

(Continuity represented on TINF points). A map

$$f:(X,{\tau }_X)⟶(Y,{\tau }_Y)$$

is TINF continuous if and only if

$$\widehat{f}:({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{{\tau }_X})⟶({\mathcal{P}}_{\mathrm{TINF}}\left(Y\right),{\sigma }_{{\tau }_Y})$$

is continuous in the ordinary sense.

Proof. 

Suppose first that f is TINF continuous. If $V_U$ is basic open in the target point space, then Lemma 3 gives ${\widehat{f}}^{-1}\left(V_U\right)=V_{f^{\leftarrow }\left(U\right)}$. Since $f^{\leftarrow }\left(U\right)\in {\tau }_X$, this is a basic open set in the source point space. Hence $\widehat{f}$ is ordinarily continuous.

Conversely, suppose $\widehat{f}$ is continuous and let $U\in {\tau }_Y$. Then

$$V_{f^{\leftarrow }\left(U\right)}={\widehat{f}}^{-1}\left(V_U\right)$$

is open in ${\sigma }_{{\tau }_X}$, hence

$$V_{f^{\leftarrow }\left(U\right)}=\bigcup _{\lambda }{V}_{A_{\lambda }}$$

for some $A_{\lambda }\in {\tau }_X$. By Lemma 2(v),

$$f^{\leftarrow }\left(U\right)=\underset{\lambda }{\bigvee }{A}_{\lambda }\in {\tau }_X.$$

Thus f is TINF continuous. □

Definition 12.

A bijection

$$f:(X,{\tau }_X)\to (Y,{\tau }_Y)$$

is a TINF homeomorphism if both f and $f^{-1}$ are TINF continuous.

Theorem 7

(Homeomorphisms represented on point spaces). A bijection

$$f:(X,{\tau }_X)⟶(Y,{\tau }_Y)$$

is a TINF homeomorphism if and only if the induced map

$$\widehat{f}:({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{{\tau }_X})⟶({\mathcal{P}}_{\mathrm{TINF}}\left(Y\right),{\sigma }_{{\tau }_Y})$$

is an ordinary homeomorphism.

Proof. 

If f is a TINF homeomorphism, Theorem 6 shows that both $\widehat{f}$ and $\widehat{f^{-1}}$ are ordinarily continuous. Directly from their definitions,

$$\widehat{f^{-1}}={\left(\widehat{f}\right)}^{-1}.$$

Therefore $\widehat{f}$ is an ordinary homeomorphism. Conversely, if $\widehat{f}$ is a homeomorphism, its continuity gives TINF continuity of f by Theorem 6. Continuity of ${\left(\widehat{f}\right)}^{-1}=\widehat{f^{-1}}$ gives TINF continuity of $f^{-1}$ by the same theorem. Hence f is a TINF homeomorphism. □

7. Separation, Connectedness, and Compactness

The three-component singleton construction of Jafari, Nordo, and Thakur motivates the following questions [9]. In the TINF setting, however, Lemma 2(iii) shows that TINF unions and unions of point-basic opens need not coincide. This distinction changes the correct formulation of connectedness and compactness.

Separation in the induced point space has two components: separation of distinct supports and separation of distinct degrees over one support. To isolate the first component, put

$$M_X:=\{x_{\mathbf{1}}:x\in X\}\subseteq {\mathcal{P}}_{\mathrm{TINF}}\left(X\right).$$

We call $M_X$ the maximal-point copy of X in ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$. The qualifications “support-pseudo” below refer to separation on this copy, except in the Hausdorff case, where an equivalent stronger formulation is available for all degrees on distinct supports.

7.1. TINF-$T_0$ and support-pseudo-$T_0$ separation

Definition 13.

A TINF space $(X,\tau )$ is TINF-$T_0$ if, for every pair of distinct points $x,y\in X$, there exists $U\in \tau$ such that either $U\left(x\right)=\mathbf{1}$ and $U\left(y\right)\ne \mathbf{1}$, or $U\left(y\right)=\mathbf{1}$ and $U\left(x\right)\ne \mathbf{1}$.

Definition 14.

The induced point space $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-$T_0$ if, for every pair of distinct supports $x,y\in X$, there exists $\mathcal{O}\in {\sigma }_{\tau }$ such that either $x_{\mathbf{1}}\in \mathcal{O}$ and $y_{\mathbf{1}}\notin \mathcal{O}$, or $y_{\mathbf{1}}\in \mathcal{O}$ and $x_{\mathbf{1}}\notin \mathcal{O}$. Equivalently, the subspace $M_X$ is an ordinary $T_0$ space.

Theorem 8

($T_0$ equivalence). A TINF space $(X,\tau )$ is TINF-$T_0$ if and only if $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-$T_0$.

Proof. 

Suppose first that $(X,\tau )$ is TINF-$T_0$, and let $x,y\in X$ be distinct. After interchanging x and y if necessary, choose $U\in \tau$ such that $U\left(x\right)=\mathbf{1}$ and $U\left(y\right)\ne \mathbf{1}$. By the definition of point-belonging, $x_{\mathbf{1}}\in V_U$. On the other hand, $y_{\mathbf{1}}\in V_U$ would mean $\mathbf{1}\le U\left(y\right)$ and hence $U\left(y\right)=\mathbf{1}$, contrary to the choice of U. Thus the ordinary open set $V_U$ contains $x_{\mathbf{1}}$ and does not contain $y_{\mathbf{1}}$. Hence the induced point space is support-pseudo-$T_0$.

Conversely, suppose that $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-$T_0$, and take distinct $x,y\in X$. There is an ordinary open set $\mathcal{O}$ that contains one of the maximal points $x_{\mathbf{1}},y_{\mathbf{1}}$ and not the other. Assume, without loss of generality, that $x_{\mathbf{1}}\in \mathcal{O}$ and $y_{\mathbf{1}}\notin \mathcal{O}$. Since ${\mathcal{B}}_{\tau }$ is a basis for ${\sigma }_{\tau }$, there exists $U\in \tau$ such that

$$x_{\mathbf{1}}\in V_U\subseteq \mathcal{O}.$$

The first inclusion gives $\mathbf{1}\le U\left(x\right)$, so $U\left(x\right)=\mathbf{1}$. Since $y_{\mathbf{1}}\notin \mathcal{O}$, it does not belong to $V_U$; therefore $U\left(y\right)\ne \mathbf{1}$. This is precisely the TINF-$T_0$ condition. □

The use of maximal representatives in Definition 14 is essential. From $U\left(y\right)\ne \mathbf{1}$ one obtains $y_{\mathbf{1}}\notin V_U$, but an arbitrary point $y_r$ may still belong to $V_U$ when $r\le U\left(y\right)$. Requiring the ordinary $T_0$ condition for every pair $x_q,y_r$ with $x\ne y$ would therefore be strictly stronger and would not be equivalent to Definition 13.

7.2. TINF-$T_1$ and support-pseudo-$T_1$ separation

Definition 15.

A TINF space $(X,\tau )$ is TINF-$T_1$ if, for every pair of distinct points $x,y\in X$, there exist $U,V\in \tau$ such that $U\left(x\right)=\mathbf{1}$, $U\left(y\right)\ne \mathbf{1}$, $V\left(y\right)=\mathbf{1}$, and $V\left(x\right)\ne \mathbf{1}$.

Definition 16.

The induced point space $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-$T_1$ if, for every pair of distinct supports $x,y\in X$, there exist ordinary open sets ${\mathcal{O}}_1,{\mathcal{O}}_2\in {\sigma }_{\tau }$ such that $x_{\mathbf{1}}\in {\mathcal{O}}_1$, $y_{\mathbf{1}}\notin {\mathcal{O}}_1$, $y_{\mathbf{1}}\in {\mathcal{O}}_2$, and $x_{\mathbf{1}}\notin {\mathcal{O}}_2$. Equivalently, the subspace $M_X$ is an ordinary $T_1$ space.

Theorem 9

($T_1$ equivalence). A TINF space $(X,\tau )$ is TINF-$T_1$ if and only if $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-$T_1$.

Proof. 

Assume that $(X,\tau )$ is TINF-$T_1$, and let $x,y\in X$ be distinct. Choose $U,V\in \tau$ as in Definition 15. Since $U\left(x\right)=\mathbf{1}$ and $V\left(y\right)=\mathbf{1}$, we have $x_{\mathbf{1}}\in V_U$ and $y_{\mathbf{1}}\in V_V$. The inequalities $U\left(y\right)\ne \mathbf{1}$ and $V\left(x\right)\ne \mathbf{1}$ give, respectively, $y_{\mathbf{1}}\notin V_U$ and $x_{\mathbf{1}}\notin V_V$. Thus $V_U$ and $V_V$ provide the two ordinary open sets required for support-pseudo-$T_1$ separation.

Conversely, assume that the induced point space is support-pseudo-$T_1$, and fix distinct $x,y\in X$. Choose ${\mathcal{O}}_1,{\mathcal{O}}_2\in {\sigma }_{\tau }$ such that $x_{\mathbf{1}}\in {\mathcal{O}}_1$, $y_{\mathbf{1}}\notin {\mathcal{O}}_1$, $y_{\mathbf{1}}\in {\mathcal{O}}_2$, and $x_{\mathbf{1}}\notin {\mathcal{O}}_2$. By the basis property, there are $U,V\in \tau$ such that

$$x_{\mathbf{1}}\in V_U\subseteq {\mathcal{O}}_1,y_{\mathbf{1}}\in V_V\subseteq {\mathcal{O}}_2.$$

The two memberships imply $U\left(x\right)=\mathbf{1}$ and $V\left(y\right)=\mathbf{1}$. Moreover, $y_{\mathbf{1}}\notin {\mathcal{O}}_1$ implies $y_{\mathbf{1}}\notin V_U$, so $U\left(y\right)\ne \mathbf{1}$; similarly, $x_{\mathbf{1}}\notin {\mathcal{O}}_2$ implies $V\left(x\right)\ne \mathbf{1}$. Hence $(X,\tau )$ is TINF-$T_1$. □

7.3. TINF-Hausdorffness ($T_2$)

Definition 17

(TINF Hausdorff and support-pseudo-Hausdorff). A TINF space $(X,\tau )$ is TINF Hausdorff, or TINF-$T_2$, if for every pair of distinct points $x,y\in X$ there exist $U,V\in \tau$ such that

$$U\left(x\right)=\mathbf{1},V\left(y\right)=\mathbf{1},U\wedge V={\mathbf{0}}_X.$$

The induced point space $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-Hausdorff if any two TINF points $x_q,y_r$ with distinct supports $x\ne y$ have disjoint ordinary open neighborhoods. In particular, the subspace $M_X$ is an ordinary Hausdorff space.

Theorem 10

($T_2$ equivalence). A TINF space $(X,\tau )$ is TINF Hausdorff if and only if $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is support-pseudo-Hausdorff.

Proof. 

Assume that $(X,\tau )$ is TINF Hausdorff, and take $x_q,y_r\in {\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$ with $x\ne y$. Choose $U,V\in \tau$ as in Definition 17. Since $q,r\le \mathbf{1}$,

$$x_q\in V_U,y_r\in V_V.$$

By Lemma 2(ii),

$$V_U\cap V_V=V_{U\wedge V}=V_{{\mathbf{0}}_X}=⌀.$$

Thus $V_U$ and $V_V$ are disjoint ordinary open neighborhoods of $x_q$ and $y_r$.

Conversely, let $x,y\in X$ be distinct. Support-pseudo-Hausdorffness applied to $x_{\mathbf{1}}$ and $y_{\mathbf{1}}$ gives disjoint ordinary open neighborhoods ${\mathcal{O}}_x$ and ${\mathcal{O}}_y$. Since ${\mathcal{B}}_{\tau }$ is a basis, choose $U,V\in \tau$ such that

$$x_{\mathbf{1}}\in V_U\subseteq {\mathcal{O}}_x,y_{\mathbf{1}}\in V_V\subseteq {\mathcal{O}}_y.$$

Then $U\left(x\right)=\mathbf{1}$ and $V\left(y\right)=\mathbf{1}$. Furthermore,

$$V_{U\wedge V}=V_U\cap V_V=⌀.$$

By Lemma 2(iv), $U\wedge V={\mathbf{0}}_X$. Hence $(X,\tau )$ is TINF Hausdorff. □

Proposition 8

(Co-supported points and the fiber order). Fix $x\in X$ and distinct degrees $q,r\in {\mathbb{L}}_{\mathrm{TINF}}\setminus \left\{\mathbf{0}\right\}$.

(i)

If $q\le r$, then every ${\sigma }_{\tau }$-open set containing $x_r$ also contains $x_q$.

(ii)

The points $x_q$ and $x_r$ are $T_0$-separated in $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ if and only if there exists $U\in \tau$ such that either $q\le U\left(x\right)$ and $r\nleqq U\left(x\right)$, or $r\le U\left(x\right)$ and $q\nleqq U\left(x\right)$.

(iii)

If $q\nleqq r$ and the TINF point $x_r$, regarded as a TINF set on X, belongs to τ, then $V_{x_r}$ contains $x_r$ and does not contain $x_q$. Similarly, if $r\nleqq q$ and $x_q\in \tau$, then $V_{x_q}$ contains $x_q$ and does not contain $x_r$. In particular, for the discrete TINF topology, any two distinct co-supported points are $T_0$-separated.

(iv)

If $q<r$, no ordinary open set can contain $x_r$ while excluding $x_q$. Consequently, for every nonempty X and every TINF topology τ on X, the full induced point space $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is neither ordinary $T_1$ nor ordinary Hausdorff.

Proof. 

For (i), let $\mathcal{O}\in {\sigma }_{\tau }$ contain $x_r$. By the basis property, there is $U\in \tau$ such that $x_r\in V_U\subseteq \mathcal{O}$. Thus $r\le U\left(x\right)$. If $q\le r$, transitivity gives $q\le U\left(x\right)$, and hence $x_q\in V_U\subseteq \mathcal{O}$.

For (ii), suppose first that an ordinary open set $\mathcal{O}$ contains $x_q$ and not $x_r$. Choose $U\in \tau$ with $x_q\in V_U\subseteq \mathcal{O}$. Then $q\le U\left(x\right)$, whereas $x_r\notin V_U$ gives $r\nleqq U\left(x\right)$. The case in which $\mathcal{O}$ contains $x_r$ and not $x_q$ is symmetric. Conversely, either pair of inequalities makes the corresponding basic open set $V_U$ contain exactly one of $x_q,x_r$.

For (iii), recall that the TINF point $x_r$ has value r at x and $\mathbf{0}$ elsewhere. If $x_r\in \tau$, then $V_{x_r}$ is a basic open set. Since $r\le x_r\left(x\right)=r$, it contains $x_r$; since $q\nleqq r$, it does not contain $x_q$. The other assertion is obtained by interchanging q and r. If $\tau ={\mathbb{L}}_{\mathrm{TINF}}^X$, both TINF points are open TINF sets, and antisymmetry ensures that at least one of $q\nleqq r$ and $r\nleqq q$ holds whenever $q\ne r$.

For (iv), the first assertion is exactly part (i) for $q<r$. To prove the consequence, choose any $x\in X$ and, for example, degrees

$$q=\left({\displaystyle \frac{1}{2}},1,0,1\right),r=\mathbf{1}.$$

They satisfy $\mathbf{0}<q<r$. Every open neighborhood of $x_r$ contains $x_q$, so the two points do not satisfy the ordinary $T_1$ axiom. Hence the full point space is not $T_1$; since every Hausdorff space is $T_1$, it is not Hausdorff either. □

Proposition 8 explains the role of the support-pseudo conditions. The obstruction to ordinary $T_1$ and Hausdorff separation is intrinsic to the degree order inside each fiber $\{x_q:q\ne \mathbf{0}\}$. The support-pseudo axioms remove only this unavoidable co-supported obstruction and measure instead how the topology separates distinct carrier points. For $T_0$ and $T_1$, maximal representatives are necessary for exact equivalence; for $T_2$, disjointness at the maximal representatives automatically separates all lower degrees on the two supports.

Corollary 2

(Separation hierarchy). Every TINF Hausdorff space is TINF-$T_1$, and every TINF-$T_1$ space is TINF-$T_0$. Correspondingly, support-pseudo-Hausdorffness implies support-pseudo-$T_1$, which implies support-pseudo-$T_0$.

Proof. 

Let $(X,\tau )$ be TINF Hausdorff and let $x\ne y$. Choose $U,V\in \tau$ such that $U\left(x\right)=\mathbf{1}$, $V\left(y\right)=\mathbf{1}$, and $U\wedge V={\mathbf{0}}_X$. If $U\left(y\right)=\mathbf{1}$, then $(U\wedge V)\left(y\right)=\mathbf{1}$, a contradiction; hence $U\left(y\right)\ne \mathbf{1}$. Similarly, $V\left(x\right)\ne \mathbf{1}$, so the space is TINF-$T_1$. The implication from TINF-$T_1$ to TINF-$T_0$ follows immediately by retaining either one of the two separating open TINF sets. The support-pseudo implications follow from Theorems 8–10. □

Both implications in Corollary 2 are strict. Indeed, by Proposition 6, an ordinary topology $\mathcal{T}$ gives the crisp TINF topology ${\mathcal{T}}^{\mathrm{TINF}}$, and the three TINF separation axioms reduce respectively to the ordinary $T_0$, $T_1$, and Hausdorff axioms of $\mathcal{T}$. The Sierpiński topology is $T_0$ but not $T_1$, while the cofinite topology on an infinite set is $T_1$ but not Hausdorff.

7.4. Connectedness

Definition 18.

A pair $U,V\in \tau$ is a TINF separation if $U\ne {\mathbf{0}}_X$, $V\ne {\mathbf{0}}_X$, $U\wedge V={\mathbf{0}}_X$, and $U\vee V={\mathbf{1}}_X$. A TINF space is TINF connected if it admits no TINF separation.

A pair $U,V\in \tau$ is a strong TINF separation if $U\ne {\mathbf{0}}_X$, $V\ne {\mathbf{0}}_X$, $U\wedge V={\mathbf{0}}_X$, and, for every $x\in X$, either $U\left(x\right)=\mathbf{1}$ or $V\left(x\right)=\mathbf{1}$. A TINF space is strongly TINF connected if it has no strong TINF separation.

The pointwise condition in a strong TINF separation implies $U\vee V={\mathbf{1}}_X$, so every strong TINF separation is a TINF separation. It is also equivalent to

$$V_U\cup V_V={\mathcal{P}}_{\mathrm{TINF}}\left(X\right).$$

Indeed, the forward implication follows because every degree lies below $\mathbf{1}$. Conversely, if the two point sets cover ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$, then the top-valued point $x_{\mathbf{1}}$ belongs to one of them for each $x\in X$, forcing $U\left(x\right)=\mathbf{1}$ or $V\left(x\right)=\mathbf{1}$.

Theorem 11

(Connectedness representation). $(X,\tau )$ is strongly TINF connected if and only if $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is connected.

Proof. 

If $U,V$ form a strong TINF separation, then $V_U$ and $V_V$ are nonempty disjoint open sets whose union is ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$. They therefore form an ordinary separation of the induced point space.

Conversely, suppose

$${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)=\mathcal{O}\cup \mathcal{W},\mathcal{O}\cap \mathcal{W}=⌀,$$

with $\mathcal{O},\mathcal{W}$ nonempty open. Write

$$\mathcal{O}=\bigcup _{a\in A}{V}_{U_a},\mathcal{W}=\bigcup _{b\in B}{V}_{V_b},$$

where $U_a,V_b\in \tau$, and put

$$U=\underset{a\in A}{\bigvee }{U}_a,V=\underset{b\in B}{\bigvee }{V}_b.$$

Because $\mathcal{O}$ and $\mathcal{W}$ are nonempty, at least one $U_a$ and at least one $V_b$ are nonbottom; hence $U\ne {\mathbf{0}}_X$ and $V\ne {\mathbf{0}}_X$. Since $V_{U_a}\cap V_{V_b}=⌀$, Lemma 2(ii),(iv) gives $U_a\wedge V_b={\mathbf{0}}_X$ for every $a,b$. Applying twice the finite-meet distributivity over arbitrary unions recorded in Section 2.2, we obtain

$$U\wedge V=\underset{a,b}{\bigvee }(U_a\wedge V_b)={\mathbf{0}}_X.$$

For every $x\in X$, the top-valued point $x_{\mathbf{1}}$ lies in $\mathcal{O}$ or $\mathcal{W}$. In the first case it lies in some $V_{U_a}$, forcing $U_a\left(x\right)=\mathbf{1}$ and hence $U\left(x\right)=\mathbf{1}$. In the second case it lies in some $V_{V_b}$, forcing $V_b\left(x\right)=\mathbf{1}$ and hence $V\left(x\right)=\mathbf{1}$. Thus $U,V$ form a strong TINF separation. □

Corollary 3.

If $(X,\tau )$ is TINF connected, then its induced point space is connected.

Proof. 

Every strong TINF separation is, in particular, a TINF separation. □

The converse need not hold.

Example 3

(A standard separation invisible in the point space). Let $X=\left\{x\right\}$ and define

$$A\left(x\right)=(1,1,0,0),B\left(x\right)=(0,0,1,1).$$

Then

$$A\wedge B={\mathbf{0}}_X,A\vee B={\mathbf{1}}_X,$$

so

$$\tau =\{{\mathbf{0}}_X,A,B,{\mathbf{1}}_X\}$$

is not TINF connected. However,

$$V_A\cup V_B⊊{\mathcal{P}}_{\mathrm{TINF}}\left(X\right),$$

because $x_{\mathbf{1}}$ belongs to neither $V_A$ nor $V_B$. In fact, the only basic open containing $x_{\mathbf{1}}$ is ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$ itself, so the induced point space is connected.

7.5. Compactness

Definition 19.

A family ${\left\{U_i\right\}}_{i\in I}\subseteq \tau$ is a TINF open cover if ${\bigvee }_{i\in I}{U}_i={\mathbf{1}}_X$. A subfamily indexed by $J\subseteq I$ is a subcover if ${\bigvee }_{j\in J}{U}_j={\mathbf{1}}_X$. The space is TINF compact if every TINF open cover has a finite subcover.

The family ${\left\{U_i\right\}}_{i\in I}$ is a strong TINF open cover if, for every $x\in X$, there exists $i\in I$ such that $U_i\left(x\right)=\mathbf{1}$. A strong subcover is a subfamily having the same pointwise property. A TINF space is strongly TINF compact if every strong TINF open cover has a finite strong subcover.

Equivalently, $\left\{U_i\right\}$ is a strong TINF open cover if and only if $\left\{V_{U_i}\right\}$ is an ordinary open cover of ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$.

Theorem 12

(Compactness representation). $(X,\tau )$ is strongly TINF compact if and only if $({\mathcal{P}}_{\mathrm{TINF}}\left(X\right),{\sigma }_{\tau })$ is compact.

Proof. 

Assume $(X,\tau )$ is strongly TINF compact and let $\mathcal{U}$ be an ordinary open cover of ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$. For each TINF point $x_q$, choose a member ${\mathcal{O}}_{x_q}\in \mathcal{U}$ that contains it. Since ${\mathcal{B}}_{\tau }$ is a basis, choose $U_{x_q}\in \tau$ such that

$$x_q\in V_{U_{x_q}}\subseteq {\mathcal{O}}_{x_q}.$$

The family $\{V_{U_{x_q}}:x_q\in {\mathcal{P}}_{\mathrm{TINF}}\left(X\right)\}$ covers the point space. Indeed, for each $x\in X$, the maximal point $x_{\mathbf{1}}$ belongs to some $V_{U_{y_r}}$; hence $\mathbf{1}\le U_{y_r}\left(x\right)$, which forces $U_{y_r}\left(x\right)=\mathbf{1}$. Thus $\{U_{x_q}:x_q\in {\mathcal{P}}_{\mathrm{TINF}}\left(X\right)\}$ is a strong TINF open cover and has a finite strong subcover. The finitely many corresponding sets ${\mathcal{O}}_{x_q}$ then cover ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$, proving that the point space is compact.

Conversely, if the point space is compact and $\left\{U_i\right\}$ is a strong TINF open cover, then $\left\{V_{U_i}\right\}$ is an ordinary open cover and therefore admits a finite subcover. The corresponding finitely many $U_i$ form a strong TINF subcover. □

Remark 1.

The TINF open-cover condition

$$\underset{i}{\bigvee }{U}_i={\mathbf{1}}_X$$

does not imply that $\left\{V_{U_i}\right\}$ covers ${\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$. The pair $A,B$ in Example 3 already shows this: $A\vee B={\mathbf{1}}_X$ while $V_A\cup V_B\ne {\mathcal{P}}_{\mathrm{TINF}}\left(X\right)$. Thus the two cover conditions are different in general and should not be conflated. Every strong TINF open cover is a TINF open cover. However, TINF compactness yields only a finite subfamily whose join is ${\mathbf{1}}_X$, and that subfamily need not retain the pointwise top-value condition required of a strong subcover. Thus the inclusion of cover classes, by itself, gives no implication between the two compactness notions. The two notions coincide on crisp TINF topologies, because a crisp family has pointwise union $\mathbf{1}$ at x exactly when at least one member has value $\mathbf{1}$ at x.

8. Structural Comparison with Related Four-Component Topologies

This section records which parts of the preceding topological construction coincide with already existing four-component frameworks. No semantic identification between different formalisms is assumed.

8.1. Quadripartitioned Neutrosophic Topology

The quadripartitioned neutrosophic set formalism was introduced in [3], while its topological development was initiated in [5].

A quadripartitioned neutrosophic (QNS) degree is written $(T,C,G,F)$ [3]. The order used in quadripartitioned neutrosophic topology [5] is

$$(t,c,g,f){\le }_{\mathrm{QNS}}(t^{\prime },c^{\prime },g^{\prime },f^{\prime })\iff t\le t^{\prime },c\le c^{\prime },g\ge g^{\prime },f\ge f^{\prime }.$$

Its bottom, top, and complement are

$$(0,0,1,1),(1,1,0,0),{(t,c,g,f)}^{\mathsf{c}}=(f,g,c,t).$$

A quadripartitioned neutrosophic set on X is a map from X into this degree scale. In the min–max topological convention, a quadripartitioned neutrosophic topology is a family of such maps containing the two constant extreme sets and closed under arbitrary pointwise unions and finite pointwise intersections determined by the displayed order.

Theorem 13

(QNS coordinate correspondence). Define

$$\mathsf{\Phi }:{\mathbb{L}}_{\mathrm{TINF}}⟶[0,1]\times [0,1]\times {[0,1]}^{\mathrm{op}}\times {[0,1]}^{\mathrm{op}},\mathsf{\Phi }(t,i,n,f)=(t,n,i,f).$$

Then Φ is a bijection from the TINF value scale onto the quadripartitioned neutrosophic value scale. It preserves inclusion, the bottom and top degrees, arbitrary unions, finite intersections, and complement.

Proof. 

The inverse of $\mathsf{\Phi }$ is the same exchange of the second and third coordinates, so $\mathsf{\Phi }$ is bijective. Under this exchange, the increasing TINF coordinates $T,N$ become the increasing QNS coordinates $T,C$, while the decreasing coordinates $I,F$ become the decreasing QNS coordinates $G,F$. Hence $p{\le }_{\mathrm{TINF}}q$ holds exactly when $\mathsf{\Phi }\left(p\right){\le }_{\mathrm{QNS}}\mathsf{\Phi }\left(q\right)$. Applying the coordinatewise suprema and infima in the two scales shows that $\mathsf{\Phi }$ carries every TINF union and intersection to the corresponding QNS operation. Finally,

$$\mathsf{\Phi }(f,n,i,t)=(f,i,n,t)$$

is the QNS complement of $\mathsf{\Phi }(t,i,n,f)=(t,n,i,f)$. Thus complement is also preserved. □

For $A\in {\mathbb{L}}_{\mathrm{TINF}}^X$, set ${\mathsf{\Phi }}_X\left(A\right)\left(x\right)=\mathsf{\Phi }\left(A\left(x\right)\right)$.

Corollary 4.

A family $\tau \subseteq {\mathbb{L}}_{\mathrm{TINF}}^X$ is a TINF topology if and only if ${\mathsf{\Phi }}_X\left(\tau \right)$ is a quadripartitioned neutrosophic topology. Moreover, ${\mathsf{\Phi }}_X$ commutes with interior, closure, boundary, restrictions, and inverse images.

Proof. 

Suppose first that $\tau$ is a TINF topology. By Theorem 13, ${\mathsf{\Phi }}_X\left(\tau \right)$ contains the two constant QNS sets and is closed under arbitrary QNS unions and finite QNS intersections. It is therefore a quadripartitioned neutrosophic topology. Applying the inverse coordinate permutation proves the converse.

Interior is the union of all open sets below a given set, and closure is the intersection of all closed sets above it. Since ${\mathsf{\Phi }}_X$ preserves these operations and complement, it commutes with interior, closure, and boundary. Coordinate permutation also commutes with restriction to a subset and with evaluation after a map, which proves the assertions about restrictions and inverse images. □

Remark 2

(Topological meaning of the QNS correspondence). Theorem 13 and Corollary 4 show that the TINF and quadripartitioned neutrosophic frameworks are indistinguishable at the level of the ordered complemented structures used to define their topologies. Consequently, every result formulated solely in terms of inclusion, arbitrary unions, finite intersections, complement, open and closed sets, interior, closure, boundary, subspaces, and inverse images is transported by the coordinate permutation Φ.

This structural equivalence does not identify the semantic roles of the four coordinates. Rather, it shows that the specifically topological contribution of the present work does not arise from a new four-coordinate min–max calculus. It lies in the systematic development of the TINF point space and in the resulting representations of continuity, separation, connectedness, and compactness. These results are therefore structural results for the common topological pattern represented by the two formalisms, while their interpretation remains theory-dependent.

8.2. Turiyam Open-Set Operations

The Turiyam set formalism was introduced in [15]. Its recent Fuzzy Neutrosophic Turiyam extension includes open and closed sets, interior, closure, and a specific complement operation [6].

Turiyam sets employ four degrees $(t,i,f,l)$ and min–max operations in which $t,l$ are increasing coordinates and $i,f$ are decreasing coordinates [15]. The corresponding ordered value domain is

$${\mathbb{L}}_{\mathrm{Tur}}=[0,1]\times {[0,1]}^{\mathrm{op}}\times {[0,1]}^{\mathrm{op}}\times [0,1].$$

Thus

$$(t,i,f,l){\le }_{\mathrm{Tur}}(t^{\prime },i^{\prime },f^{\prime },l^{\prime })\iff t\le t^{\prime },i\ge i^{\prime },f\ge f^{\prime },l\le l^{\prime }.$$

A Turiyam set on X is a map from X into ${\mathbb{L}}_{\mathrm{Tur}}$. In this subsection, a Turiyam open-set topology means a family containing the constant bottom and top sets and closed under the pointwise min–max arbitrary unions and finite intersections determined by ${\mathbb{L}}_{\mathrm{Tur}}$.

Theorem 14

(TINF–Turiyam open-set correspondence). The map

$$\mathsf{\Psi }:{\mathbb{L}}_{\mathrm{TINF}}⟶{\mathbb{L}}_{\mathrm{Tur}},\mathsf{\Psi }(t,i,n,f)=(t,i,f,n),$$

is a bijection that preserves inclusion, the bottom and top degrees, arbitrary unions, and finite intersections. Consequently, τ is a TINF topology if and only if ${\mathsf{\Psi }}_X\left(\tau \right):=\{\mathsf{\Psi }\circ U:U\in \tau \}$ is a Turiyam open-set topology.

Proof. 

The inverse map is ${\mathsf{\Psi }}^{-1}(t,i,f,l)=(t,i,l,f)$, so $\mathsf{\Psi }$ is bijective. It sends the increasing coordinates $(T,N)$ to the increasing Turiyam coordinates $(t,l)$ and the decreasing coordinates $(I,F)$ to the decreasing coordinates $(i,f)$. Thus the four order inequalities defining $p{\le }_{\mathrm{TINF}}q$ are precisely those defining $\mathsf{\Psi }\left(p\right){\le }_{\mathrm{Tur}}\mathsf{\Psi }\left(q\right)$. The coordinate formulas for arbitrary unions and finite intersections are therefore preserved, as are the two constant extreme degrees. Applying these identities pointwise proves the final assertion about topologies. □

Recent Fuzzy Neutrosophic Turiyam (FNT) topology defines the complement

$$c_{\mathrm{FNT}}(t,i,f,l)=(f,i,t,1-l)$$

and uses it to define closed sets, interior, and closure [6]. This complement is not the transport of the TINF involution.

Proposition 9

(Complement incompatibility). For $q=(t,i,n,f)\in {\mathbb{L}}_{\mathrm{TINF}}$,

$$\mathsf{\Psi }\left(c\right(q\left)\right)=(f,n,t,i),$$

whereas

$$c_{\mathrm{FNT}}\left(\mathsf{\Psi }\left(q\right)\right)=(f,i,t,1-n).$$

Therefore

$$\mathsf{\Psi }\left(c\left(q\right)\right)=c_{\mathrm{FNT}}\left(\mathsf{\Psi }\left(q\right)\right)⟺i=n={\displaystyle \frac{1}{2}}.$$

Proof. 

Applying the two complements gives the displayed values directly. Equality $\mathsf{\Psi }\left(c\left(q\right)\right)=c_{\mathrm{FNT}}\left(\mathsf{\Psi }\left(q\right)\right)$ requires equality in every coordinate. The first and third coordinates already agree, whereas the second and fourth give $n=i$ and $i=1-n$. Therefore $i=n=1/2$. Conversely, these two equalities make all four coordinates agree, so the complements coincide exactly in the stated case. □

Remark 3.

The TINF and Turiyam systems therefore have the same min–max open-set calculus after this coordinate permutation, but their published complemented structures are not isomorphic through Ψ. Open-set statements depending only on bottom, top, finite intersections, and arbitrary unions can be transported, whereas statements involving closed sets or complement duality require separate verification.

9. Structural Perspective

The preceding construction separates three topological layers. First, a TINF topology assigns generalized open sets to a carrier and supports the usual operators of general topology: interior, closure, boundary, subspaces, products, and continuous maps. Second, the crisp core extracts an ordinary topology on the carrier. Third, the TINF point construction produces a generally much larger ordinary topological space whose points remember both support and degree. The exact continuity and homeomorphism theorems show that this induced point topology is not merely auxiliary: it captures continuity and homeomorphism in ordinary topological terms.

The comparison with quadripartitioned neutrosophic topology shows that many basic open-set and closed-set constructions are representation invariant under a change of coordinates. The Turiyam comparison identifies a complementary limitation: open-set operations may coincide even when complement and closed-set structures differ. The separation hierarchy adds a specifically point-topological distinction. Separation of carrier points is represented by support-pseudo axioms, whereas comparable degrees over one support form nested neighborhood systems and prevent the full point space from being $T_1$.

10. Conclusions

We have developed a focused topological foundation for single-valued $(T,I,N,F)$-neutrosophic topological spaces. The underlying value scale is

$${\mathbb{L}}_{\mathrm{TINF}}=[0,1]\times {[0,1]}^{\mathrm{op}}\times [0,1]\times {[0,1]}^{\mathrm{op}},$$

with bottom $(0,1,0,1)$, top $(1,0,1,0)$, and complement ${(t,i,n,f)}^{\mathsf{c}}=(f,n,i,t)$. On this scale, the coordinatewise min–max operations give the TINF unions, intersections, and complements used throughout the paper.

Within this framework we established the standard theory of open and closed sets, interior, closure, boundary, subspaces, products, and continuous maps, and we identified the ordinary crisp core. The comparison with quadripartitioned neutrosophic topology shows that the corresponding TINF topological operators are transported by a coordinate permutation, while the comparison with Turiyam sets shows that the min–max open-set operations are likewise transported after a different permutation. The latter correspondence does not preserve the Fuzzy Neutrosophic Turiyam complement, so the associated closed-set structures remain distinct.

The induced TINF point topology provides the main topological refinement developed here. TINF continuity and homeomorphism are exactly represented by ordinary continuity and homeomorphism of the induced point maps. The TINF separation hierarchy from $T_0$ to Hausdorffness is represented by the corresponding support-pseudo axioms, while the ordered degree fibers explain why the full point space is never ordinary $T_1$. The strict inclusion

$$\bigcup _{\lambda }{V}_{A_{\lambda }}\subseteq V_{{\bigvee }_{\lambda }{A}_{\lambda }}$$

explains why connectedness and compactness cannot be transferred naively from earlier singleton constructions; the strengthened notions introduced here restore exact correspondences with ordinary connectedness and compactness.