Filter One-Point Sums and Lindelöf C_p-Spaces

1. Introduction

All spaces are assumed to be Tychonoff, and we use standard topological terminology as in [2]. For a space X, the symbol $C_p\left(X\right)$ denotes the space of all real-valued continuous functions on X with the topology of pointwise convergence. For general $C_p$-theory, see [1,13].

Hernández-Hernández, Ramírez-Chávez and Rojas-Hernández studied one-point extensions of countable free topological sums whose new point has neighbourhoods controlled by the Fréchet filter. If ${\left(X_n\right)}_{n<\omega }$ is a sequence of compact spaces and

$$X=\underset{n<\omega }{⨁}{X}_n\cup \{\infty \}$$

is endowed with the usual tail neighbourhoods at ∞, they proved ([4], Theorem 5.6) that

$$C_p\left(X\right)\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}⟺C_p\left(\underset{n<\omega }{⨁}{X}_n\right)\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}.$$

They asked whether an analogous statement holds when the Fréchet filter is replaced by an arbitrary filter on $\omega$ ([4], Problem 7.6). Related results on Lindelöf $C_p$-spaces and Lindelöf $\Sigma$-spaces include [6,9,10].

For analytic filters, the usual tail convergence from the Fréchet-filter case is replaced by convergence in $L_{\mathcal{F}}$, the space of real sequences tending to zero along $\mathcal{F}$. Lemma 2 gives the descriptive-set-theoretic reduction that, for filters on $\omega$, the space $L_{\mathcal{F}}$ is K-analytic if and only if $\mathcal{F}$ is analytic as a subspace of $2^{\omega }$. Under this hypothesis, the Lindelöfness of $C_p\left({⨁}_{n<\omega }{X}_n\right)$ implies the Lindelöfness of $C_p\left(X_{\mathcal{F}}\right)$. A complementary necessary condition is that, for each $A\in \mathcal{F}$, the sub-sum indexed by $\omega \setminus A$ must have Lindelöf $C_p$-space whenever $C_p\left(X_{\mathcal{F}}\right)$ is Lindelöf. Together with the Fréchet-filter case from ([4], Theorem 5.6), these two directions give a constant-fibre equivalence for analytic filters.

If the free sum has a countable network, no descriptive-set-theoretic assumption on the filter is needed. For compact fibres, any obstruction to removing analyticity must therefore come from non-metrizable compact fibres.

2. Preliminaries

Let $\mathcal{F}$ be a free filter on $\omega$. For a sequence ${\left(X_n\right)}_{n<\omega }$ of spaces, put

$$Y=\underset{n<\omega }{⨁}{X}_n.$$

The $\mathcal{F}$-one-point sum of the sequence ${\left(X_n\right)}_{n<\omega }$ is the space

$$X_{\mathcal{F}}=Y\cup \{\infty \}$$

in which every $X_n$ is clopen and carries its original topology, and a local base at ∞ consists of the sets

$$\{\infty \}\cup \bigcup _{n\in A}{X}_n,A\in \mathcal{F}.$$

For the Fréchet filter, this is the space considered in ([4], Theorem 5.6). If the spaces $X_n$ are Tychonoff, then so is $X_{\mathcal{F}}$. To see this, separate a point of $X_n$ from a closed set by complete regularity in $X_n$. The separating function is extended by the constant value 1 off $X_n$. Since $\mathcal{F}$ contains the Fréchet filter, the cofinite neighbourhood $\{\infty \}\cup {\bigcup }_{m\ne n}{X}_m$ of ∞ avoids $X_n$ and ensures continuity at ∞. If the point is ∞, choose a basic neighbourhood U of ∞ disjoint from the closed set. Then U is clopen, and the function that is 0 on U and 1 on $X_{\mathcal{F}}\setminus U$ gives the required separation.

Lemma 1.

Let $\mathcal{F}$ be a free filter on ω, and let ${\left(X_n\right)}_{n<\omega }$ be a sequence of spaces. A function $f:X_{\mathcal{F}}\to \mathbb{R}$ is continuous if and only if $f|X_n\in C\left(X_n\right)$ for each $n<\omega$, and for every $\epsilon >0$,

$$\left\{n<\omega :f\left[X_n\right]\subseteq (f(\infty )-\epsilon ,f(\infty )+\epsilon )\right\}\in \mathcal{F}.$$

Proof. 

It remains only to check continuity at ∞. Suppose first that f is continuous. Let $\epsilon >0$. Since

$$\left(f\right(\infty )-\epsilon ,f(\infty )+\epsilon )$$

is a neighbourhood of $f(\infty )$, there is $F\in \mathcal{F}$ such that

$$f\left[\{\infty \}\cup \bigcup _{n\in F}{X}_n\right]\subseteq (f(\infty )-\epsilon ,f(\infty )+\epsilon ).$$

The required set contains F, hence belongs to $\mathcal{F}$.

Conversely, assume that the stated condition holds. Let V be a neighbourhood of $f(\infty )$ in $\mathbb{R}$. Choose $\epsilon >0$ such that

$$\left(f\right(\infty )-\epsilon ,f(\infty )+\epsilon )\subseteq V.$$

By assumption,

$$F=\left\{n<\omega :f\left[X_n\right]\subseteq (f(\infty )-\epsilon ,f(\infty )+\epsilon )\right\}\in \mathcal{F}.$$

Then

$$\{\infty \}\cup \bigcup _{n\in F}{X}_n$$

is mapped into V. Thus f is continuous at ∞. Since each $X_n$ is clopen, the proof is complete. □

Let

$$\xi \left(\mathcal{F}\right)=\omega \cup \{\infty \}$$

be the space in which every point of $\omega$ is isolated and a local base at ∞ consists of the sets

$$\{\infty \}\cup F,F\in \mathcal{F}.$$

We use this notation for the one-point space associated with $\mathcal{F}$. Closely related countable one-point spaces occur in the study of function spaces over countable spaces with one non-isolated point [3]. Related descriptive-set-theoretic questions for analytic or coanalytic $C_p$-spaces and for spaces of filter convergence were considered in [7,11]. Put

$$C_p^0\left(\xi \left(\mathcal{F}\right)\right)=\{g\in C_p\left(\xi \left(\mathcal{F}\right)\right):g(\infty )=0\}.$$

The next lemma relates $\mathcal{F}$ to the following subspace of ${\mathbb{R}}^{\omega }$:

$$L_{\mathcal{F}}=\left\{a\in {\mathbb{R}}^{\omega }:\forall k<\omega ,\{n<\omega :|a\left(n\right)|<2^{-k}\}\in \mathcal{F}\right\}.$$

We use the standard compact-valued description of K-analytic spaces [12]. A space Z is K-analytic if there is an upper semicontinuous compact-valued map

$$\phi :{\omega }^{\omega }⟶\mathcal{K}\left(Z\right)$$

such that $Z={\bigcup }_{\alpha \in {\omega }^{\omega }}\phi \left(\alpha \right)$. Here $\mathcal{K}\left(Z\right)$ denotes the family of non-empty compact subsets of Z, and upper semicontinuity means that $\{\alpha :\phi (\alpha )\subseteq U\}$ is open whenever U is open in Z.

Lemma 2.

For a free filter $\mathcal{F}$ on ω, the following are equivalent.

1.

$\mathcal{F}$ is analytic as a subspace of $2^{\omega }$.

2.

$L_{\mathcal{F}}$ is analytic as a subspace of ${\mathbb{R}}^{\omega }$.

3.

$L_{\mathcal{F}}$ is K-analytic.

Proof. 

Assume first that $\mathcal{F}$ is analytic. For each $k<\omega$, define

$$A_k\left(a\right)=\{n<\omega :|a\left(n\right)|<2^{-k}\}$$

for $a\in {\mathbb{R}}^{\omega }$. The map

$$T:{\mathbb{R}}^{\omega }⟶{\left(2^{\omega }\right)}^{\omega },T\left(a\right)={\left(A_k\left(a\right)\right)}_{k<\omega },$$

is Borel, because each coordinate condition $n\in A_k\left(a\right)$ is given by the open inequality $\left|a\left(n\right)\right|<2^{-k}$. By the standard closure properties of analytic sets ([5], §14), the product ${\mathcal{F}}^{\omega }$ is analytic in ${\left(2^{\omega }\right)}^{\omega }$. Therefore

$$L_{\mathcal{F}}=T^{-1}\left({\mathcal{F}}^{\omega }\right)$$

is analytic in ${\mathbb{R}}^{\omega }$, because analytic sets are closed under Borel inverse images between standard Borel spaces. This proves that (1) implies (2). As an analytic subspace of a Polish space, $L_{\mathcal{F}}$ is K-analytic; hence (2) implies (3).

It remains to prove that (3) implies (1). We recall the short argument that, in this metrizable case, K-analyticity gives analyticity. Let $\phi :{\omega }^{\omega }\to \mathcal{K}\left(L_{\mathcal{F}}\right)$ be an upper semicontinuous compact-valued map whose union is $L_{\mathcal{F}}$. Its graph

$$G=\{(\alpha ,x)\in {\omega }^{\omega }\times {\mathbb{R}}^{\omega }:x\in \phi \left(\alpha \right)\}$$

is closed. Indeed, if $(\alpha ,x)\notin G$, then $x\notin \phi \left(\alpha \right)$. Since ${\mathbb{R}}^{\omega }$ is metrizable and $\phi \left(\alpha \right)$ is compact, choose an open set U in ${\mathbb{R}}^{\omega }$ with $\phi \left(\alpha \right)\subseteq U$ and $x\notin \overline{U}$. Applying upper semicontinuity to the open set $U\cap L_{\mathcal{F}}$ of $L_{\mathcal{F}}$, there is a neighbourhood V of $\alpha$ such that $\phi \left(\beta \right)\subseteq U$ for all $\beta \in V$. Then $V\times ({\mathbb{R}}^{\omega }\setminus \overline{U})$ misses G. Hence G is closed. Thus $L_{\mathcal{F}}$ is analytic in ${\mathbb{R}}^{\omega }$, being the projection of the closed set G from the Polish product ${\omega }^{\omega }\times {\mathbb{R}}^{\omega }$.

Now define a continuous map $\theta :2^{\omega }\to {\mathbb{R}}^{\omega }$ by

$$\theta \left(A\right)\left(n\right)=\left\{\begin{array}{cc}0,\hfill & n\in A,\hfill \\ 1,\hfill & n\notin A.\hfill \end{array}\right.$$

For every $A\subseteq \omega$, we have $\theta \left(A\right)\in L_{\mathcal{F}}$ if and only if $A\in \mathcal{F}$. Thus

$$\mathcal{F}={\theta }^{-1}\left(L_{\mathcal{F}}\right).$$

Since $\theta$ is continuous and $L_{\mathcal{F}}$ is analytic in the Polish space ${\mathbb{R}}^{\omega }$, the inverse image ${\theta }^{-1}\left(L_{\mathcal{F}}\right)$ is analytic in $2^{\omega }$. Hence $\mathcal{F}$ is analytic. This proves (3) implies (1). □

Lemma 3

([4], Proposition 5.1). Let P be a space, and let M be a K-analytic space. If $P\times {\omega }^{\omega }$ is Lindelöf, then $P\times M$ is Lindelöf.

3. Main Results

3.1. Sufficient and Necessary Conditions

Theorem 1.

Let $\mathcal{F}$ be an analytic free filter on ω, and let ${\left(X_n\right)}_{n<\omega }$ be a sequence of non-empty compact spaces. Let $Y={⨁}_{n<\omega }{X}_n$, and let $X_{\mathcal{F}}$ be the corresponding $\mathcal{F}$-one-point sum. If $C_p\left(Y\right)$ is Lindelöf, then $C_p\left(X_{\mathcal{F}}\right)$ is Lindelöf.

Proof. 

The proof adapts the argument used for the Fréchet-filter case ([4], Theorem 5.6). Lemma 2 supplies the needed K-analyticity of $L_{\mathcal{F}}$. Let $I=[-1,1]$ and put

$$W=\prod _{n<\omega }{C}_p(X_n,I).$$

Since

$$C_p\left(Y\right)\cong \prod _{n<\omega }{C}_p\left(X_n\right),$$

the space W is a closed subspace of $C_p\left(Y\right)$.

The space Y is not pseudocompact. Indeed, the function $h:Y\to \mathbb{R}$ defined by $h|X_n=n$ for each $n<\omega$ is continuous and unbounded. By ([10], Proposition 1.1), the space

$$C_p\left(Y\right)\times {\omega }^{\omega }$$

is homeomorphic to a closed subspace of $C_p\left(Y\right)$. Thus $C_p\left(Y\right)\times {\omega }^{\omega }$ is Lindelöf. Since W is closed in $C_p\left(Y\right)$, the subspace

$$W\times {\omega }^{\omega }$$

is closed in $C_p\left(Y\right)\times {\omega }^{\omega }$, and hence is Lindelöf.

By Lemma 2, the space $L_{\mathcal{F}}$ is K-analytic. Since $W\times {\omega }^{\omega }$ is Lindelöf, Lemma 3 implies that $W\times L_{\mathcal{F}}$ is Lindelöf.

Put $C_p^0\left(X_{\mathcal{F}}\right)=\{f\in C_p\left(X_{\mathcal{F}}\right):f(\infty )=0\}$, and define $\Phi :W\times L_{\mathcal{F}}⟶C_p^0\left(X_{\mathcal{F}}\right)$ by

$$\Phi ({\left(u_n\right)}_{n<\omega },a)(\infty )=0,\Phi ({\left(u_n\right)}_{n<\omega },a)\left(x\right)=a\left(n\right)u_n\left(x\right)(x\in X_n).$$

The map $\Phi$ is well defined. Let $\epsilon >0$, and choose $k<\omega$ with $2^{-k}<\epsilon$. Since

$$\{n<\omega :|a\left(n\right)|<2^{-k}\}\in \mathcal{F}$$

and $|u_n\left(x\right)|\le 1$ for all $x\in X_n$, Lemma 1 applies to $\Phi (\left(u_n\right),a)$. Continuity on each $X_n$ follows at once. The map $\Phi$ is continuous for the topology of pointwise convergence, because each coordinate map is a composition of evaluation maps and multiplication in $\mathbb{R}$.

It remains to prove that $\Phi$ is onto. Let

$$f\in C_p^0\left(X_{\mathcal{F}}\right).$$

For each $n<\omega$, set

$$s\left(n\right)=max\left\{\right|f\left(x\right)|:x\in X_n\}.$$

The maximum exists because $X_n$ is compact. By Lemma 1,

$$s={\left(s\left(n\right)\right)}_{n<\omega }\in L_{\mathcal{F}}.$$

For each $n<\omega$, define

$$u_n=\left\{\begin{array}{cc}s{\left(n\right)}^{-1}{f|}_{X_n},\hfill & s\left(n\right)>0,\hfill \\ 0,\hfill & s\left(n\right)=0.\hfill \end{array}\right.$$

Then $u_n\in C_p(X_n,I)$ and

$$\Phi ({\left(u_n\right)}_{n<\omega },s)=f.$$

Therefore $C_p^0\left(X_{\mathcal{F}}\right)$ is a continuous image of the Lindelöf space $W\times L_{\mathcal{F}}$. Hence $C_p^0\left(X_{\mathcal{F}}\right)$ is Lindelöf.

Finally,

$$C_p\left(X_{\mathcal{F}}\right)\cong C_p^0\left(X_{\mathcal{F}}\right)\times \mathbb{R}$$

under the map

$$f\mapsto (f-f(\infty ),f(\infty \left)\right).$$

Since $\mathbb{R}$ is $\sigma$-compact, $C_p\left(X_{\mathcal{F}}\right)$ is Lindelöf. □

Proposition 1.

Let $\mathcal{F}$ be a free filter on ω, and let ${\left(X_n\right)}_{n<\omega }$ be a sequence of Tychonoff spaces. Let $Y={⨁}_{n<\omega }{X}_n$, and let $X_{\mathcal{F}}$ be the corresponding $\mathcal{F}$-one-point sum. If $C_p\left(X_{\mathcal{F}}\right)$ is Lindelöf, then for every $A\in \mathcal{F}$,

$$C_p\left(\underset{n\in \omega \setminus A}{⨁}{X}_n\right)$$

is Lindelöf.

Proof. 

Fix $A\in \mathcal{F}$, and put

$$Z_A=\underset{n\in \omega \setminus A}{⨁}{X}_n.$$

Consider the subspace

$$E_A=\{f\in C_p\left(X_{\mathcal{F}}\right):f(\infty )=0\mathrm{and}f|\bigcup _{n\in A}{X}_n=0\}.$$

The space $E_A$ is closed in $C_p\left(X_{\mathcal{F}}\right)$. The restriction map

$$E_A⟶C_p\left(Z_A\right)$$

is a homeomorphism. Injectivity is immediate from the definition of $E_A$. For surjectivity, take $g\in C_p\left(Z_A\right)$ and define $\tilde{g}$ on $X_{\mathcal{F}}$ by putting $\tilde{g}=g$ on $Z_A$ and $\tilde{g}=0$ on

$$\{\infty \}\cup \bigcup _{n\in A}{X}_n.$$

The function $\tilde{g}$ is continuous on each summand. It is continuous at ∞, because the neighbourhood

$$\{\infty \}\cup \bigcup _{n\in A}{X}_n$$

is mapped to 0. Thus $\tilde{g}\in E_A$, and the restriction map is onto. Both the restriction map and its inverse are continuous for the topology of pointwise convergence. Hence $E_A\cong C_p\left(Z_A\right)$. Since closed subspaces of Lindelöf spaces are Lindelöf, $C_p\left(Z_A\right)$ is Lindelöf. □

Theorem 2.

Let K be a non-empty compact space, and let $\mathcal{F}$ be a non-Fréchet analytic free filter on ω. Define

$$X_{\mathcal{F}}\left(K\right)=(\omega \times K)\cup \{\infty \},$$

where ω is discrete and the neighbourhoods of ∞ are given by $\mathcal{F}$. Then

$$C_p\left(X_{\mathcal{F}}\left(K\right)\right)\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}$$

if and only if

$$C_p{\left(K\right)}^{\omega }\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}.$$

Proof. 

Assume first that $C_p{\left(K\right)}^{\omega }$ is Lindelöf. Since

$$C_p(\omega \times K)\cong C_p{\left(K\right)}^{\omega },$$

Theorem 1 gives that $C_p\left(X_{\mathcal{F}}\left(K\right)\right)$ is Lindelöf.

Conversely, assume that $C_p\left(X_{\mathcal{F}}\left(K\right)\right)$ is Lindelöf. Since every free filter on $\omega$ contains the Fréchet filter and $\mathcal{F}$ is not the Fréchet filter, there is $A\in \mathcal{F}$ such that

$$B=\omega \setminus A$$

is infinite. By Proposition 1,

$$C_p(B\times K)$$

is Lindelöf. Since B is countably infinite, with the discrete topology,

$$C_p(B\times K)\cong C_p{\left(K\right)}^B\cong C_p{\left(K\right)}^{\omega },$$

it follows that $C_p{\left(K\right)}^{\omega }$ is Lindelöf. □

Combining Theorem 2 with the Fréchet-filter case ([4], Theorem 5.6), the same constant-fibre equivalence holds for all analytic free filters.

3.2. A Countable-Network Case

Analyticity of the filter is not needed when the free sum has a countable network. The proof is a standard network argument. For $C_p$-spaces and network arguments, see [1,2,8,13].

Proposition 2.

Let $\mathcal{F}$ be any free filter on ω, and let ${\left(X_n\right)}_{n<\omega }$ be a sequence of Tychonoff spaces. Let $Y={⨁}_{n<\omega }{X}_n$, and let $X_{\mathcal{F}}$ be the corresponding $\mathcal{F}$-one-point sum. If Y has a countable network, then $C_p\left(X_{\mathcal{F}}\right)$ has a countable network. In particular, $C_p\left(X_{\mathcal{F}}\right)$ is hereditarily Lindelöf.

Proof. 

Let $\mathcal{N}$ be a countable network for Y. We claim that the following countable family is a network for $C_p\left(Y\right)$. For finite $\mathcal{A}\subseteq \mathcal{N}$ and rational open intervals $I_N$, $N\in \mathcal{A}$, put

$$[\mathcal{A},{\left(I_N\right)}_{N\in \mathcal{A}}]=\{f\in C_p\left(Y\right):f\left[N\right]\subseteq I_N\mathrm{for}\mathrm{every}N\in \mathcal{A}\}.$$

The family of all such sets is countable. Let $f\in C_p\left(Y\right)$, and let

$$O=\{g\in C_p\left(Y\right):g\left(y_i\right)\in U_i\mathrm{for}i<m\}$$

be a basic neighbourhood of f. For each $i<m$, choose a rational open interval $J_i$ such that

$$f\left(y_i\right)\in J_i\subseteq U_i.$$

Since $f^{-1}\left(J_i\right)$ is open and $\mathcal{N}$ is a network, choose $N_i\in \mathcal{N}$ such that

$$y_i\in N_i\subseteq f^{-1}\left(J_i\right).$$

Let

$$\mathcal{A}=\{N_i:i<m\}.$$

For each $N\in \mathcal{A}$, put

$$I_N=\bigcap \{J_i:i<m\mathrm{and}{N}_i=N\}.$$

This is a non-empty open interval and satisfies

$$f\left[N\right]\subseteq I_N\subseteq \bigcap \{U_i:i<m\mathrm{and}{N}_i=N\}.$$

Then

$$[\mathcal{A},{\left(I_N\right)}_{N\in \mathcal{A}}]$$

contains f and is contained in O. Thus $C_p\left(Y\right)$ has a countable network.

The restriction map

$$R:C_p\left(X_{\mathcal{F}}\right)⟶C_p\left(Y\right)\times \mathbb{R},R\left(f\right)=\left(f\right|Y,f(\infty )),$$

is a topological embedding. Indeed, the topology of pointwise convergence on $C_p\left(X_{\mathcal{F}}\right)$ is determined by the coordinates in Y together with the coordinate ∞. The product of two spaces with countable networks again has a countable network, so $C_p\left(Y\right)\times \mathbb{R}$ has a countable network. Since the property of having a countable network is hereditary, $C_p\left(X_{\mathcal{F}}\right)$ has a countable network. Finally, every space with a countable network is hereditarily Lindelöf. □

Corollary 1.

Let $\mathcal{F}$ be any free filter on ω, and let ${\left(X_n\right)}_{n<\omega }$ be a sequence of compact metrizable spaces. Then $C_p\left(X_{\mathcal{F}}\right)$ is hereditarily Lindelöf. In particular, if K is compact metrizable, then $C_p\left(X_{\mathcal{F}}\left(K\right)\right)$ is hereditarily Lindelöf for every free filter $\mathcal{F}$ on ω.

Proof. 

For each $n<\omega$, choose a countable base ${\mathcal{B}}_n$ for $X_n$. Then

$$\mathcal{B}=\{B:B\in {\mathcal{B}}_n\mathrm{for}\mathrm{some}n<\omega \}$$

is countable. Since each $X_n$ is clopen in the free sum, $\mathcal{B}$ is a countable network for ${⨁}_{n<\omega }{X}_n$. The result follows from Proposition 2. □

3.3. Examples and an Open Question

Countably generated free filters provide a basic source of filters covered by Lemma 2. Let ${\left(F_m\right)}_{m<\omega }$ be a decreasing sequence of infinite subsets of $\omega$ such that

$$\bigcap _{m<\omega }{F}_m=\varnothing .$$

The filter

$$\mathcal{F}=\{B\subseteq \omega :F_m\subseteq B\mathrm{for}\mathrm{some}m<\omega \}$$

is free and is an $F_{\sigma }$ subset of $2^{\omega }$, since

$$\mathcal{F}=\bigcup _{m<\omega }\{B\subseteq \omega :F_m\subseteq B\}.$$

Thus $\mathcal{F}$ is analytic. If, for some m, the complement $\omega \setminus F_m$ is infinite, then $\mathcal{F}$ is not the Fréchet filter. For instance, if $A\subseteq \omega$ is infinite and coinfinite and

$${\mathcal{F}}_A=\{B\subseteq \omega :A\setminus B\mathrm{is}\mathrm{finite}\},$$

then ${\mathcal{F}}_A$ is a non-Fréchet analytic free filter. Hence the results above apply to $X_{{\mathcal{F}}_A}\left(K\right)$ for every non-empty compact space K.

Corollary 1 shows that non-analytic filters do not, by themselves, prevent the conclusion. For compact metrizable fibres, the Lindelöf conclusion holds for every free filter. Therefore any obstruction to removing analyticity from Theorem 1, or from the constant-fibre equivalence obtained from Theorem 2 together with ([4], Theorem 5.6), must use non-metrizable compact fibres.

The arbitrary-filter case remains open. By Lemma 2, the descriptive-set-theoretic hypothesis used in Theorem 1 is analyticity of the filter, while Proposition 1 gives only a necessary condition. Thus any obstruction to the arbitrary-filter form must involve filters beyond the analytic class together with compact fibres without a countable network, rather than the elementary coding of filter convergence.

Can one characterize, in purely filter-theoretic terms, those free filters $\mathcal{F}$ on $\omega$ for which the implication

$$C_p\left(\underset{n<\omega }{⨁}{X}_n\right)\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}⟹C_p\left(X_{\mathcal{F}}\right)\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}$$

holds for every sequence ${\left(X_n\right)}_{n<\omega }$ of non-empty compact spaces?