Bounded resolutions for spaces $C_{p}(X)$ and a characterization in terms of X

Abstract

An internal characterization of the Arkhangel’skiĭ-Calbrix main theorem from [4] is obtained by showing that the space $C_{p}(X)$ of continuous real-valued functions on a Tychonoff space X is K-analytic framed in $\mathbb {R}^{X}$ if and only if X admits a nice framing. This applies to show that a metrizable (or cosmic) space X is $\sigma$ -compact if and only if X has a nice framing. We analyse a few concepts which are useful while studying nice framings. For example, a class of Tychonoff spaces X containing strictly Lindelöf Čech-complete spaces is introduced for which a variant of Arkhangel’skiĭ-Calbrix theorem for $\sigma$-boundedness of X is shown.

1 Introduction

A classic result of Christensen asserts that for a metric and separable space X the space $C_{p}(X)$ is analytic (i. e., is a continuous image of the Polish space $\omega ^{\omega }$) if and only if X is $\sigma$-compact (see for example [23, Theorem 9.6] and references therein).

Calbrix proved [11] that the analyticity of $C_{p}(X)$ yields the $\sigma$-compactness of X for any Tychonoff space X. The converse fails in general. Nevertheless, as Okunev proved [27], if X is $\sigma$-bounded, $C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}$, i. e., there exists a K-analytic space Z such that $C_{p}(X)\subseteq Z\subseteq \mathbb {R}^{X}$. This latter result motivated paper [4], where Arkhangel’skiĭ and Calbrix characterized cosmic $\sigma$-compact spaces by showing

Theorem 1

A cosmic space X is $\sigma$-compact if and only if $C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}.$

The proof of Arkhangel’skiĭ-Calbrix theorem depends on a result of Christensen about fundamental compact resolutions in metric spaces and Okunev’s [28] about projectively $\sigma$-compact spaces X. They asked if the same holds when X is just a Lindelöf space. The answer to this question could already be found in Leiderman’s [24], and also recalled in [10, Remark 3.9]. We shall discuss again this example in a slightly stronger form. In [16] we extended the above mentioned Okunev’s theorem by showing the following useful

Theorem 2

$C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}$ if and only if it has a bounded resolution.

We will show that $C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}$ if and only if X admits a nice framing (Theorem 13) if and only if X has a fundamental resolution of functions (Theorem , Corollary 14). The latter concept will be directly used to construct an (usc) map from $\omega ^{\omega }$ into the compact sets of Z with $C_{p}(X)\subseteq Z\subseteq \mathbb {R}^{X}$ (showing K-analyticity of Z). Examples illustrating these results are presented in Sects. 3, 4 and 5 where we discuss some consequences of Theorem 13 for obtaining $\sigma$-compactness of X, and provide an alternative approach (independent of Arkhangel’skiĭ and Calbrix) to this fact. For example, we analyze situations when X admits a nice framing with a layer $\{U_{\alpha ,n}:n\in \omega \}$ giving a sequence of compact sets covering X, and we introduce a class of spaces containing Lindelöf Čech-complete spaces for which a variant of the Arkhangel’skiĭ-Calbrix theorem is obtained (Theorem 32). Recall that if $C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}$, the space X is projectively $\sigma$-compact, i. e. every continuous separable and metrizable image of X is $\sigma$-compact, [4, Theorem 2.3]. This motivated us to examine some versions of projectively $\sigma$-compactness and show (Theorem 6) that if X is projectively analytic (i. e., if every continuous metrizable and separable image of X is analytic) then every continuous metrizable image of X is separable. This yields that a metrizable (or cosmic) space X is $\sigma$-compact if and only if X has a nice framing. However, this fact fails if X is only separable and strictly dominated by a metric space.

2 Bounded resolutions for spaces $C_{p}(X)$

A covering $\mathcal {F}=\{A_{\alpha }:\alpha \in \omega ^{\omega }\}$ of a set X is called a resolution for X if $A_{\alpha }\subseteq A_{\beta }$ whenever $\alpha \le \beta$ coordinatewise. If X is a topological space and the sets $A_{\alpha }$ are compact, the family $\mathcal {F}$ is called a compact resolution. Recall that a subset B of a locally convex space (lcs) E is said to be bounded if for each neighborhood of the origin U in E there exists $\lambda >0$ such that $\lambda B\subseteq U$. A resolution $\{A_{\alpha }:\alpha \in \omega ^{\omega }\}$ for E consisting of bounded sets is called a bounded resolution. If additionally every bounded set in E is contained in some $A_{\alpha }$, we say that the family $\{A_{\alpha }:\alpha \in \omega ^{\omega }\}$ is a fundamental bounded resolution for E. For a Tychonoff space X, by $C_{p}(X)$ and $C_{k}(X)$ we denote the space of continuous real-valued functions on X with the pointwise $\tau _{p}$ and the compact-open topology $\tau _{k}$, respectively.

Proposition 3

If $C_{p}(X)$ is a continuous linear image of a metrizable lcs F, then $C_{p}(X)$ admits a bounded resolution.

Proof

Let $T:F\rightarrow C_{p}(X)$ be a continuous linear surjection; let $\left\{ U_{n}\right\} _{n=1}^{\infty }$ be a decreasing base of neighborhoods of zero in F. If for $\alpha =(\alpha (n))\in \omega ^{\omega }$ we set $A_{\alpha }=\bigcap _{n=1}^{\infty }\alpha (n)U_{\alpha (n)}$, the family $\{A_{\alpha }:\alpha \in \omega ^{\omega }\}$ is a fundamental bounded resolution for F. Hence, $\{T(A_{\alpha }):\alpha \in \omega ^{\omega }\}$ is a bounded resolution for $C_{p}(X)$. $\square$

We refer also the reader to [19, Proposition 22] for a sufficient condition for $C_{p}(X)$ to have a fundamental bounded resolution.

A topological space X is pseudocompact if $f(X)\subseteq \mathbb {R}$ is bounded, $f\in C(X)$; X is pseudocompact if and only if $C_{p}(X)$ does not contain a complemented copy of $\mathbb {R}^{\omega }$ (see [1, Section 4]); X is called $\sigma$-bounded if $X=\bigcup _{n}X_{n}$ and every $X_{n}$ is functionally bounded, i. e., every $f\in C(X)$ is bounded on $X_{n}$. A special case of Theorem 2 is Proposition 4 (due to Uspenskiĭ, see [27, Theorem 3.1]). We provide a short alternative proof.

Proposition 4

A Tychonoff space X is pseudocompact if and only if there exists a $\sigma$-compact space K with $C_{p}(X)\subseteq K\subseteq \mathbb {R}^{X}$.

Proof

If X is pseudocompact and $S=\{f\in C(X):|f(x)|\le 1,x\in X\}$, the sequence $\left\{ nS\right\} _{n=1}^{\infty }$ covers $C_{p}(X)$. Hence, the closure of nS in $\mathbb {R}^{X}$ provides a sequence of compact sets in $\mathbb {R}^{X}$ whose union K contains $C\left( X\right)$. Conversely, if the conclusion holds, $C_{p}(X)$ is covered by a sequence of bounded sets. If X is not pseudocompact, $C_{p}(X)$ contains a complemented copy of $\mathbb {R}^{\omega }$, which is not covered by a sequence of functionally bounded sets. $\square$

One can ask whether a Tychonoff space X is $\sigma$-bounded if and only if $C_{p}(X)$ does not contain a copy of $\mathbb {R}^{Y}$ for some uncountable Y. The answer is negative: $C_{p}(\mathbb {R}^{\omega })$ does not contain a copy of $\mathbb {R}^{Y}$ for any set Y with $|Y|>\aleph _{0}$. Indeed, since the weak* dual of $C_{p}(\mathbb {R}^{\omega })$ is separable, $C_{p}(\mathbb {R}^{\omega })$ admits a weaker metrizable locally convex topology, but $\mathbb {R}^{Y}$ fails this property, whereas it is well known that $C_{p}(\omega )=\mathbb {R}^{\omega }$ is not $\sigma$-bounded. Nevertheless, the ‘only if’ part is true in general. In fact, if $\left\{ B_{n}\right\} _{n=1}^{\infty }$ is a sequence of functionally bounded sets covering X, the sets $A_{\alpha }=\{f\in C(X):\text {sup} _{x\in B_{n}}|f(x)|\le \alpha (n)\ \forall n\in \omega \}$ for $\alpha \in \omega ^{\omega }$ compose a bounded resolution for $C_{p}(X)$. If $C_{p}(X)$ contains a copy of $\mathbb {R}^{Y}$ then this latter Baire space also admits a bounded resolution. So, according to [23, Proposition 7.1], Y must be countable.

Recall that X is a $\mu$-space if every functionally bounded set in X is relatively compact.

Theorem 5

([22]) A Tychonoff space X is $\sigma$-compact if and only if X is a $\mu$-space and there exists a metrizable locally convex topology $\xi$ on C(X) such that $\tau _{p}\le \xi \le \tau _{k}$.

Recall that X is projectively analytic if each continuous metrizable and separable image of X is analytic. The space X is said to have the Discrete Countable Chain Condition (DCCC) if every discrete family of open sets is countable, which is equivalent to require that each continuous metrizable image of X is separable.

Theorem 6

If an infinite Tychonoff space X is projectively analytic, then it has the DCCC.

Proof

Assume there exists a continuous surjective map $h:X\rightarrow Z$ and Z is metrizable but not separable. Choose a closed discrete set D in Z with $|D|=\aleph _{1}$. Such set exists since $d(Z)=w(Z)=e(Z)$, where e(Z) means the extent of Z, see [13]. Then there exists a continuous one-to-one map $f:D\rightarrow Y$ onto a metrizable and separable space Y, which is not analytic. Indeed, such Y can be obtained as follows. Under (CH) we know that $\mathbb {R}$ contains $2^{\mathfrak {c}}$ subsets of the cardinality continuum, but only a continuum number of analytic subsets. So, one of those $2^{\mathfrak {c}}$ subsets Y is not analytic. Under $(\lnot CH)$, take a subset $Y\subseteq \mathbb {R}$ of cardinality $\aleph _{1}$. Then it is not analytic. Indeed, every uncountable analytic subset of $\mathbb {R}$ contains a copy of the Cantor set and hence has cardinality $\mathfrak {c}.$

The map f admits a (canonical) extension $Pf:PD\rightarrow PY$ to spaces of finitely supported maps, where PY is the space of finitely supported probability measures endowed with the weak* topology determined by the subspace $C^{b}\left( X\right)$ of C(Y) consisting of bounded functions. It turns out that PY is a separable and metrizable convex set by Prokhorov-Wasserman-Kantorovich metric, see [7, Lemma 4.3]. As follows from the proof of [2, 0.5.9 Proposition], the $y\mapsto \delta _{y}$ copy of Y in L(Y) (the dual of $C_{p}(Y)$) is closed in $L\left( Y\right)$ when the latter linear space is provided with the weak topology of the dual pair $\left\langle L\left( Y\right) ,C^{b}\left( Y\right) \right\rangle$. Hence Y is closed in PY. Since PY is a convex metrizable subset of a lcs, $f:D\rightarrow Y\subseteq PY$ admits a continuous extension ${\bar{f}}:Z\rightarrow PY$ by Dugundji theorem [14, page 185]. ${\bar{f}}(Z)$ in PY is not analytic since it contains a closed subset Y which is not analytic. Then ${\bar{f}}\circ h$ has a non analytic (separable) metrizable image, a contradiction. $\square$

A Tychonoff space X is called strongly projectively $\sigma$-compact if every continuous metrizable image of X is $\sigma$-compact.

Corollary 7

Let X be an infinite Tychonoff space. Then X is projectively $\sigma$-compact if and only if X is strongly projectively $\sigma$-compact.

Theorem 6 and Okunev’s [27, Theorem 1.3] yield the following

Corollary 8

A metrizable space X is analytic if and only if every continuous metrizable and separable image of X is analytic.

Corollary 9

A paracompact Čech-complete space X is $\sigma$-compact if and only if $C_{p}(X)$ has a bounded resolution.

Indeed, if $C_p(X)$ has a bounded resolution, X is strongly projectively $\sigma$-compact by [4, Theorem 2.3] and Corollary 7. Since X is mapped onto a completely metrizable space Y by a perfect map T, see [13, 5.5.9(a)], the space Y is $\sigma$-compact. Hence X is $\sigma$-compact (since T is perfect). The converse implication is clear.

From [4, p. 5200] the one-point Lindeöfication of an uncountable discrete space X is projectively $\sigma$-compact but is not $\sigma$-bounded. Even more can be shown.

Example 10

$C_{p}(X)$ is not K-analytic framed in $\mathbb {R} ^{X}$ for the one-point Lindeöfication X of an uncountable discrete space.

Indeed, X is an $\omega$-space, i. e. every continuous metrizable separable image of X is countable, see [3] or [2]. So, X is projectively $\sigma$-compact. Since X is a P-space, $C_{p}(X)$ is a Baire lcs. So, if $C_{p}(X)$ admits a bounded resolution, [23, Proposition 7.1] ensures that the space $C_{p}(X)$ is metrizable. This X must be countable, a contradiction.

3 A characterization in terms of X

This section deals with the following

Problem 11

Characterize Tychonoff spaces X such that $C_{p}(X)$ has a bounded resolution.

According to [15] a family $\{U_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ of closed subsets of X is called a framing if (i) for each $\alpha \in \omega ^{\omega }$ the layer $\{U_{\alpha ,n}:n\in \omega \}$ is an increasing covering of X, and (ii) for every $n\in \omega$ one has that $U_{\beta ,n}\subseteq U_{\alpha ,n}$, $\alpha \le \beta$.

Lemma 12

([15, Lemma 104]) A set $A\subseteq C_p\left( X\right)$ is bounded if and only if there is an increasing covering $\left\{ V_{n}:n\in \omega \right\}$ of X by closed sets such that $\sup _{f\in A}\left| f\left( x\right) \right| \le n$, $x\in V_{n}$.

Theorem 13

The space $C_{p}(X)$ has a bounded resolution if and only if there exists a framing $\{U_{\alpha ,n}:(\alpha ,n)\in \omega ^{\omega }\times \omega \}$ in X enjoying the property that if $f\in C(X)$ there exists $\gamma \in \omega ^{\omega }$ such that $| f(x)| \le n$ for each $x\in U_{\gamma ,n}$ and $n\in \omega$.

Proof

If there is a framing $\{U_{\alpha ,n}:(\alpha ,n)\in \omega ^{\omega }\times \omega \}$ of the aforementioned characteristics, the sets

$$\begin{aligned} A_{\alpha }:=\left\{ f\in C(X):\sup _{x\in U_{\alpha ,n}}|f(x)|\le n\ \forall n\in \omega \right\} \end{aligned}$$

compose a bounded resolution for $C\left( X\right)$. Indeed, each set $A_{\alpha }$ is pointwise bounded by virtue of Lemma 12, since $\left\{ U_{\alpha ,n}:n\in \omega \right\}$ is an increasing covering of X by closed sets such that $\sup \nolimits _{f\in A_{\alpha }}\left| f\left( x\right) \right| \le n$ for all $x\in U_{\alpha ,n}$. Moreover, $A_{\alpha }\subseteq A_{\beta }$ if $\alpha \le \beta$ since $U_{\beta ,n}\subseteq U_{\alpha ,n}$. If $f\in C\left( X\right)$, by the statement of the theorem there exists $\gamma \in \omega ^{\omega }$ such that $\left| f\left( x\right) \right| \le n$ for each $x\in U_{\gamma ,n}$ and all $n\in \omega$ . Hence $f\in A_{\gamma }$, so $\{A_{\alpha }:\alpha \in \omega ^{\omega }\}$ covers $C\left( X\right)$. Conversely, assume $C_{p}\left( X\right)$ has a bounded resolution $\{B_{\alpha }:\alpha \in \omega ^{\omega }\}$. If $V_{\alpha ,n}=\left\{ x\in X:\sup \nolimits _{f\in B_{\alpha }}\left| f\left( x\right) \right| \le n\right\} ,$ then $\left\{ V_{\alpha ,n}:n\in \omega \right\}$ is an increasing covering of X by closed sets for each $\alpha \in \omega ^{\omega }$ with $V_{\beta ,n}\subseteq V_{\alpha ,n}$ whenever $\alpha \le \beta ,$ $n\in \omega$. If $f\in C\left( X\right)$ there is $\delta \in \omega ^{\omega }$ such that $f\in B_{\delta }$. Hence $\left| f\left( x\right) \right| \le n$ for each $x\in V_{\delta ,n}$ $n\in \omega$, so $\{V_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ is a framing satisfying the required property. $\square$

We say that X has a nice framing if X admits a framing as stated in Theorem 13.

The following concept can also be used when studying the role of framings, see also [10], where a similar concept was fixed for uniform spaces X with uniformly continuous functions $f_{\alpha }$. A Tychonoff space X admits a fundamental resolution of functions if there exists on X a family of nonnegative real-valued functions $\{f_{\alpha }:\alpha \in \omega ^{\omega }\}$ such that $f_{\alpha }\le f_{\beta }$ for $\alpha \le \beta$ and for each $f\in C(X)$ there exists $\alpha \in \omega ^{\omega }$ with $|f|\le f_{\alpha }$.

Corollary 14

A Tychonoff space X has a fundamental resolution of functions if and only if $C_{p}(X)$ has a bounded resolution, if and only if X has a nice framing.

Indeed, if $\{A_{\alpha }:\alpha \in \omega ^{\omega }\}$ is a bounded resolution on $C_{p}(X)$, then $f_{\alpha }(x)=\sup \{|f(x)|:f\in A_{\alpha }\}$ form a fundamental resolution of functions, and if $(f_{\alpha })$ is a fundamental resolution of functions, sets $A_{\alpha }=\{f\in C_{p}(X):|f|\le f_{\alpha }\}$ form a bounded resolution on $C_{p}(X)$. The last statement follows from Theorem 13.

To keep the paper self-contained we apply this concept to present a short proof of Theorem 1 if X is metrizable (or cosmic), although the main idea remains similar (see also [6, Proof of Theorem 2.2] for a similar argument). Nevertheless, theorem fails if X is only separable with a stronger metric topology, Example 37.

Theorem 15

A metrizable space X is $\sigma$-compact if and only if X admits a nice framing. The same statement holds if X is cosmic.

Proof

Assume first that X is metrizable and separable, with a nice framing. Let $\{f_{\alpha }:\alpha \in \omega ^{\omega }\}$ be a fundamental resolution of functions on X (we apply Corollary 14). Let ${\overline{X}}$ be a metric compactification (see [20] for details). For $\alpha \in \omega ^{\omega }$ set $K_{\alpha }=\bigcap _{y\in X}({\overline{X}} \setminus B(y,\exp (-f_{\alpha }(y))$, where B(y, r) is the open ball at y and radius r. Clearly each $K_{\alpha }$ is a compact subset of $\overline{ X}\setminus X$ and $K_{\alpha }\subseteq K_{\beta }$, if $\alpha ,\beta \in \omega ^{\omega }$ with $\alpha \le \beta$. Let $K\subseteq {\overline{X}} \setminus X$ be compact. For $h(y)=|\ln d(K,y)|$, $y\in X$, there exists $\sigma \in \omega ^{\omega }$ with $h\le f_{\sigma }$. Hence $d(K,y)\ge \exp (-f_{\sigma }(y)),$ and then $K\subseteq {\overline{X}}\setminus B(y,\exp (-f_{\sigma }(y)))$ for every $y\in X;$ so $K\subseteq K_{\sigma }$. Thus $\{K_{\alpha }:\alpha \in \omega ^{\omega }\}$ is a fundamental compact resolution for the metrizable and separable space ${\overline{X}}\setminus X$. By Christensen’s [23, Theorem 6.1] ${\overline{X}}\setminus X$ is Polish, so X is $\sigma$-compact.

Next, assume that X is metrizable and contains a nice framing. By Corollary 14 the space $C_{p}(X)$ has a bounded resolution. Assume that X is continuously mapped on a metrizable and separable space Y. Since $C_{p}(Y)$ is isomorphic to a subspace of $C_{p}(X)$, the space $C_{p}(Y)$ has a bounded resolution; consequently the metrizable and separable space Y admits a nice framing. By the first case we derive that Y is $\sigma$-compact. Now, Corollary 7 applies to get that X is $\sigma$-compact. The converse follows from the fact, mentioned earlier, that if $\left\{ B_{n}\right\} _{n=1}^{\infty }$ is a sequence of functionally bounded sets covering X, the sets $A_{\alpha }=\{f\in C(X):\sup _{x\in B_{n}}|f(x)|\le \alpha (n)\ \forall n\in \omega \}$ for $\alpha \in \omega ^{\omega }$ compose a bounded resolution for $C_{p}(X)$.

Finally, assume that X is cosmic with a nice framing, and let Y be a continuous metrizable and separable image of X. By the previous argument Y is $\sigma$-compact. So, according to [27, Theorem 1.5], the space X is $\sigma$-compact. The converse is clear. $\square$

A regular space X is angelic if every relatively countably compact subset A of X is relatively compact and for every $x\in {\overline{A}}$ there exists a sequence in A which converges to x. The concept of a fundamental resolution of functions will be directly used to define an (usc) map F from $\omega ^{\omega }$ into compact subsets of some space Z where $C_{p}(X)\subseteq Z\subseteq \mathbb {R}^{X}$.

Theorem 16

If X has a nice framing, $C_{p}\left( X\right)$ is K-analytic-framed in $\mathbb {R}^{X}$ and angelic.

We provide two proofs of Theorem 16. For the first one we need the following two simple technical lemmas (which might be already known).

Lemma 17

Each increasing function $\varphi : \omega ^{\omega } \rightarrow [0, \infty )$ is bounded on some non-empty open subset of $\omega ^{\omega }$.

Proof

Suppose, by contrary, that $\varphi$ is unbounded on every non-empty open subset of $\omega ^{\omega }$. Let $\beta ^{1}=(\beta _{n}^{1})\in \omega ^{\omega }$ with $\varphi (\beta ^{1})\ge 1.$ Let $\gamma ^{1}\in \{\beta _{1}^{1}\}\times \omega ^{\omega }$ with $\varphi (\gamma ^{1})\ge 2.$ Put $\beta ^{2}=(\beta _{n}^{2})=\max \{\beta ^{1},\gamma ^{1}\};$ then $\varphi (\beta ^{2})\ge 2,\beta ^{2}\ge \beta ^{1}$ and $\beta _{1}^{2}=\beta _{1}^{1}.$ Let $\gamma ^{2}\in \{(\beta _{1}^{1},\beta _{2}^{2})\}\times \omega ^{\omega }$ with $\varphi (\gamma ^{2})\ge 3.$ Put $\beta ^{3}=(\beta _{n}^{3})=\max \{\beta ^{2},\gamma ^{2}\};$ then $\varphi (\beta ^{3})\ge 3,\beta ^{3}\ge \beta ^{2}$ and $\beta _{1}^{3}=\beta _{1}^{1},\beta _{2}^{3}=\beta _{2}^{2}$. Following this procedure we get an element $\beta =(\beta _{n}^{n})\in \omega ^{\omega }$ and an increasing sequence $(\beta ^{k})\subseteq \omega ^{\omega }$ such that $\varphi (\beta ^{k})\ge k$ and $\beta _{i}^{k}=\beta _{i}^{i}$ for all $1\le i\le k,k\in \omega .$ Then $\beta ^{k}\le \beta$ and $k\le \varphi (\beta ^{k})\le \varphi (\beta )<+\infty$ for any $k\in \omega$, a contradiction. $\square$

Lemma 18

(1) Each increasing function $\varphi :\omega ^{\omega }\rightarrow [0,\infty )$ is locally bounded, i. e. each point $x\in \omega ^{\omega }$ has an open neighborhood U such that $\varphi (U)$ is bounded. (2) For every locally bounded function $\varphi :\omega ^{\omega }\rightarrow [0,\infty )$ there exists a locally constant function $g:\omega ^{\omega }\rightarrow [0,\infty )$ with $g\ge \varphi ;$ in particular, g is continuous.

Proof

(1) Assume the claim fails. Then there exists $\alpha \in \omega ^{\omega }$ such that $\varphi$ is unbounded on $\{(\alpha _1, \ldots , \alpha _m)\} \times \omega ^{\omega }$ for every $m\in \omega$. Hence for every $\beta \ge \alpha$ the function $\varphi$ is unbounded on $\{(\beta _1, \ldots , \beta _m)\} \times \omega ^{\omega }$ for every $m\in \omega$.

Set $\psi : \omega ^{\omega } \rightarrow [0, +\infty ), \psi ((\beta _n))=\varphi ((\beta _n+\alpha _n))$. Then $\psi$ is increasing and unbounded on any non-empty open subset of $\omega ^{\omega }.$ Indeed, let $\beta =(\beta _n) \in \omega ^{\omega }$ and $m\in \omega .$ Let $\gamma _i=\beta _i+\alpha _i$ for $1\le i \le m$ and $A=\{(\lambda _i)\in \omega ^{\omega }: \lambda _i>\alpha _{i+m}, i\in \omega \}.$ Then $\psi ( \{(\beta _1, \ldots , \beta _m)\}\times N^N)=\varphi (\{(\gamma _1, \ldots , \gamma _m)\}\times A)$ and for any $(\lambda ^{\prime }_i)\in \omega ^{\omega }$ we have $\varphi ((\gamma _1, \ldots , \gamma _m, \lambda ^{\prime }_1, \lambda ^{\prime }_2, \ldots ))\le \varphi ((\gamma _1, \ldots , \gamma _m, \lambda ^{\prime }_1+\alpha _{m+1}, \lambda ^{\prime }_2 +\alpha _{m+1}, \ldots ))$ and $(\gamma _1, \ldots , \gamma _m, \lambda ^{\prime }_1+\alpha _{m+1}, \lambda ^{\prime }_2 +\alpha _{m+1}, \ldots ) \in \{(\gamma _1, \ldots , \gamma _m)\}\times A$. Thus $\psi ( \{(\beta _1, \ldots , \beta _m)\}\times \omega ^{\omega })= \varphi (\{(\gamma _1, \ldots , \gamma _m)\}\times A)$ is unbounded, since $\varphi ( \{(\gamma _1, \ldots , \gamma _m)\}\times \omega ^{\omega })$ is unbounded. It follows that $\psi$ is unbounded on any non-empty open subset of $\omega ^{\omega }$, a contradiction with Lemma 17.

(2) For $\alpha \in \omega ^{\omega }$ let $m(\alpha )$ be the least integer such that $\varphi$ is bounded on $\{(\alpha _{1},\ldots ,\alpha _{m(\alpha )})\}\times \omega ^{\omega }$. Put $V_{\alpha }=\{(\alpha _{1},\ldots ,\alpha _{m(\alpha )})\}\times \omega ^{\omega }$ for any $\alpha \in \omega ^{\omega }.$ Clearly, $\bigcup \{V_{\alpha }:\alpha \in \omega ^{\omega }\}=\omega ^{\omega }.$ For all $\alpha ,\beta \in \omega ^{\omega }$ we have $V_{\alpha }=V_{\beta }$ or $V_{\alpha }\cap V_{\beta }=\emptyset .$ Indeed, if $m(\alpha )=m(\beta )$ and $\alpha _{i}=\beta _{i}$ for $1\le i\le m(\alpha )$, then $V_{\alpha }=V_{\beta }$; if $m(\alpha )=m(\beta )$ and $\alpha _{i}\ne \beta _{i}$ for some $1\le i\le m(\alpha )$, then $V_{\alpha }\cap V_{\beta }=\emptyset$; if $m(\alpha )\ne m(\beta )$, then $\alpha _{i}\ne \beta _{i}$ for some $1\le i\le \min \{m(\alpha ),m(\beta )\}$ and $V_{\alpha }\cap V_{\beta }=\emptyset$. Thus for some $W\subseteq \omega ^{\omega }$ the family $\{V_{\alpha }:\alpha \in W\}$ is a partition of $\omega ^{\omega }$ on non-empty clopen subsets such that $\varphi$ is bounded on $V_{\alpha }$ for every $\alpha \in W$. Let $t_{\alpha }=\sup \varphi (V_{\alpha })$ for $\alpha \in W$. Let $g:\omega ^{\omega }\rightarrow [0,+\infty )$ be the function such that $g(\beta )=t_{\alpha }$ for any $\beta \in V_{\alpha },\alpha \in W.$ Then $g\ge \varphi$ and g is locally constant, so it is continuous. $\square$

First proof of Theorem 16

By Corollary 14 fix a fundamental resolution of functions $\{f_{\alpha }:\alpha \in \omega ^{\omega }\}$ for X. Let $x\in X.$ Then $\varphi _{x}:\omega ^{\omega }\rightarrow [0,+\infty ),\alpha \rightarrow f_{\alpha }(x)$ is increasing. By Lemma 18 there exists a locally constant function $g_{x}:\omega ^{\omega }\rightarrow [0,+\infty )$ with $g_{x}\ge \varphi _{x}.$ Let $g:\omega ^{\omega }\times X\rightarrow [0,+\infty ),g(\alpha ,x)=g_{x}(\alpha ).$ Clearly, for any $x\in X$ the function $\omega ^{\omega }\rightarrow [0,+\infty ),\alpha \rightarrow g(\alpha ,x)$ is locally constant. Moreover for any function $f\in C_{p}(X)$ there is an $\alpha \in \omega ^{\omega }$ with $|f(x)|\le f_{\alpha }(x)=\varphi _{x}(\alpha )\le g_{x}(\alpha )=g(\alpha ,x)$ for every $x\in X.$ For any $\alpha \in \omega ^{\omega }$ the set $F_{\alpha }=\prod _{x\in X}[-g(\alpha ,x),g(\alpha ,x)]$ in $\mathbb { R}^{X}$ is compact. Put $Z=\bigcup \{F_{\alpha }:\alpha \in \omega ^{\omega }\}$. Then $C_{p}(X)\subseteq Z\subseteq \mathbb {R}^{X}$. Using the continuity of g with respect to the first variable it is easy to see that $F:\alpha \mapsto F_{\alpha }$ is an upper semi-continuous (usc) set-valued map from $\omega ^{\omega }$ with compact values in Z. Thus $C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}$. $\square$

We propose another proof of Theorem 16, which uses an idea included in the proof of [2, Proposition IV 9.3]. First we prove the following

Lemma 19

If X has a nice framing, there exists a countable nice framing $\{W_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ for X.

Proof

Let $\{U_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ be a nice framing for X. For each $\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega$, define the closed set

$$\begin{aligned} W_{\alpha ,n}=\bigcap \left\{ U_{\beta ,n}:\beta \in \omega ^{\omega },\beta \left( i\right) =\alpha \left( i\right) ,1\le i\le n\right\} . \end{aligned}$$

Observe that $W_{\alpha ,n}\subseteq W_{\alpha ,n+1}$ for each $\alpha \in \omega ^{\omega }$ and $W_{\beta ,n}\subseteq W_{\alpha ,n}$ for each $n\in \omega$ whenever $\alpha \le \beta$. We claim that $\bigcup _{n\in \omega }W_{\alpha ,n}=X$ for each $\alpha \in \omega ^{\omega }$. Indeed, suppose otherwise that there exists $x\not \in \bigcup _{n\in \omega }W_{\alpha ,n}$ for some $\alpha \in \omega ^{\omega }$. For every $n\in \omega$ choose $\beta _{n}\in \omega ^{\omega }$ with $\beta _{n}\left( i\right) =\alpha \left( i\right)$ for $1\le i\le n$ such that $x\not \in U_{\beta _{n},n}$. Put $\gamma :=\sup \{\beta _{n}:n\in \omega \}$. Then, for every $n\in \omega$, $\beta _{n}\le \gamma$ and hence $x\not \in U_{\gamma ,n}$ since $U_{\gamma ,n}\subseteq U_{\beta _{n},n}$ by the definition of framing. Hence $x\not \in \bigcup _{n\in \omega }U_{\gamma ,n}=X$, a contradiction. All this means that $\{W_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ is a framing for X. Note that the family $\{W_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ is countable since $W_{\alpha ,n}$ depends only on $\alpha (1),\dots ,\alpha (n)$. Finally, if $f\in C\left( X\right)$, by Theorem 13 there is $\gamma \in \omega ^{\omega }$ such that $\left| f\left( x\right) \right| \le n$ for every $x\in U_{\gamma ,n}$ and all $n\in \omega$. So $\left| f\left( x\right) \right| \le n$, $x\in W_{\gamma ,n}$, $n\in \omega$ $\square$

Second proof of Theorem 16

By Lemma 19 let $\mathcal {F}=\left\{ U_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \right\}$ be a countable nice framing for X. First we prove that $C_{p}\left( X\right)$ is Lindelöf $\Sigma$-framed in $\mathbb {R}^{X}$. Let us say that a function $f\in \mathbb {R}^{X}$ is $\mathcal {F}$-bounded if for each $x\in X$ there exists $\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega$ such that $x\in U_{\alpha ,n}$ and $f(U_{\alpha ,n})\subseteq \left[ -n,n\right]$ . Let us denote by Z the subset of $\mathbb {R}^{X}$ consisting of all $\mathcal {F}$-bounded functions on X. We claim that $C\left( X\right) \subseteq Z$. Indeed, if $f\in C\left( X\right)$ there exists $\delta \in \omega ^{\omega }$ such that $f(U_{\delta ,n})\subseteq \left[ -n,n\right]$ for every $n\in \omega$. Since $\{U_{\delta ,n}:n\in \omega ^{\omega }\}$ covers X, given $x\in X$ there exists $m\in \mathbb {N}^{\mathbb {N}}$ with $x\in U_{\delta ,m}$ and $f(U_{\delta ,m})\subseteq \left[ -m,m\right]$, which shows that $f\in Z$. Thus $C\left( X\right) \subseteq Z$, as stated. Now we prove that Z is a Lindelöf $\Sigma$-space. If $\overline{ \mathbb {R}}$ designates the usual two points compactification of $\mathbb {R}$ , then $\overline{\mathbb {R}}^{X}$ is a compactification of Z. For $\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega$ define

$$\begin{aligned} L_{\alpha ,n}=\{f\in \overline{\mathbb {R}}^{X}:f(U_{\alpha ,n})\subseteq [-n,n]\,\}. \end{aligned}$$

The sets $L_{\alpha ,n}$ are compact since they are closed in $\overline{ \mathbb {R}}^{X}$, and compose a countably family because the framing $\mathcal {F}$ is countable. Choose $f\in Z$ and $g\in \overline{\mathbb {R}} ^{X}\setminus Z$. As $g\in \overline{\mathbb {R}}^{X}\setminus Z$, there exists $y\in X$ such that $g\left( U_{\alpha ,n}\right) \not \subseteq [-n,n]$ for each $\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega$ for which $y\in U_{\alpha ,n}$. Due to $f\in Z$ there is $\left( \gamma ,m\right) \in \omega ^{\omega }\times \omega$ with $y\in U_{\gamma ,m}$ and $f(U_{\gamma ,m})\subseteq \left[ -m,m\right]$, so $f\in L_{\gamma ,m}$. On the other hand $g\notin L_{\gamma ,m}$ since $g\left( U_{\gamma ,m}\right) \not \subseteq [-m,m]$ because $y\in U_{\gamma ,m}$. Since $\overline{\mathbb {R}}^{X}$ is a compactification of Z, [2, 4.9.2 Proposition] applies to get that Z is a Lindelöf $\Sigma$-space.

Next we show that $C_{p}\left( X\right)$ if K-analytic-framed in $\mathbb { R}^{X}$. Indeed, for each $\alpha \in \omega ^{\omega }$ we set $A_{\alpha }:=\{f\in C(X):\sup \nolimits _{x\in U_{\alpha ,n}}|f(x)|\le n\ \forall n\in \omega \}$ and put $B_{\alpha }=\overline{A_{\alpha }}$, where the closure is in $\mathbb {R}^{X}$. Note that $B_{\alpha }$ is a compact set in $\mathbb { R}^{X}$. We claim that $B_{\alpha }\subseteq Z$. Indeed, if $f\in B_{\alpha }$ there is a net $\left\{ f_{d}:d\in D\right\}$ in $A_{\alpha }$ such that $f_{d}\left( x\right) \rightarrow f\left( x\right)$ for every $x\in X$ . So, given $n\in \omega$, one has in particular $f_{d}\left( x\right) \rightarrow f\left( x\right)$ for every $x\in U_{\alpha ,n}$, which implies that $\sup \nolimits _{x\in U_{\alpha ,n}}|f(x)|\le n$. Hence $f(U_{\alpha ,n})\subseteq [-n,n]\,$, so that $f\in Z$, and $B_{\alpha }\subseteq Z$.

Define $Y=\bigcup \{B_{\alpha }:\alpha \in \omega ^{\omega }\}$ and note that, as a consequence of the previous claim, $Z\subseteq Y$. Since Y quasi-Suslin [23, Proposition 3.11], there is a set-valued map $T:\omega ^{\omega }\rightarrow 2^{Y}$ with $\bigcup \{T\left( \alpha \right) :\alpha \in \omega ^{\omega }\}=Y$ and if $\alpha _{n}\rightarrow \alpha$ in $\omega ^{\omega }$ and $x_{n}\in T\left( \alpha _{n}\right)$ for all $n\in \omega$ the sequence $\left\{ x_{n}\right\} _{n=1}^{\infty }$ has a cluster point $x\in T\left( \alpha \right)$. By a result of Cascales, we may assume $T\left( \alpha \right) \subseteq T\left( \beta \right)$ whenever $\alpha \le \beta$ (see [23, Theorem 3.1]). Define $S:\omega ^{\omega }\rightarrow 2^{Y}$ by $S\left( \alpha \right) =\overline{ T\left( \alpha \right) }$, closure in Z, and put $\Omega =\bigcup \{S\left( \alpha \right) :\alpha \in \omega ^{\omega }\}$. Then, the fact that Y is quasi-Suslin implies that $T\left( \alpha \right)$ is countably compact, hence functionally bounded in Y, so $S\left( \alpha \right)$ is functionally bounded in Z. Since Z is Lindelöf, $S\left( \alpha \right)$ compact. So, the map S is compactly-valued. If $\alpha _{n}\rightarrow \alpha$ in $\omega ^{\omega }$ and $z_{n}\in S\left( \alpha _{n}\right)$, we may proceed as in the proof of [15, Theorem 57] to show that $\left\{ z_{n}\right\} _{n=1}^{\infty }$ has a cluster point $z\in S\left( \alpha \right)$. This proves that $\Omega$ is K-analytic. Since $C_{p}\left( X\right) \subseteq \Omega \subseteq \mathbb {R}^{X}$, the space $C_{p}\left( X\right)$ is K-analytic-framed in $\mathbb {R}^{X}$. Proof that $C_{p}(X)$ is angelic: By Okunev’s [27, Theorem 3.5] the space $\upsilon X$ is a Lindelöf $\Sigma$-space, and then by Orihuela’s angelic theorem [23, Theorem 4.5] the space $C_{p}(\upsilon X)$ is angelic, and the same holds also for $C_{p}(X)$, see [23, Lemma 9.2]. $\square$

Corollary 20

([16, Theorem 1]) $C_{p}\left( X\right)$ has a bounded resolution if and only if $C_{p}\left( X\right)$ is K-analytic-framed in $\mathbb {R}^{X}$.

Corollary 21

If $C_{p}(X)$ and $C_{p}(Y)$ are linearly homeomorphic, X has a nice framing if and only if Y has a nice framing.

Corollary 22

([5]) Let X be $\sigma$-bounded and Y metric, and assume that there exists a continuous linear surjection from $C_{p}(X)$ onto $C_{p}(Y)$. Then Y is $\sigma$-compact.

Remark 23

A countable infinite product X of metrizable non-compact spaces $X_{n}$ each with a nice framing does not have a nice framing, since each $X_{n}$ is $\sigma$-compact but X is not $\sigma$-compact (as X contains a closed copy of $\omega ^{\omega }$).

4 Strong framings, $\sigma$-compactess

One may expect that each nice framing for a separable and metrizable X should contain a layer consisting of compact sets, so providing a $\sigma$-compact cover of X. We prove however the following

Theorem 24

Let $\mathfrak {M}$ be the class of metrizable and separable spaces with a nice framing.

• (1) If $X\in \mathfrak {M}$, then X admits a nice framing such that for each $\alpha \in \omega ^{\omega }$ the layer $\{U_{\alpha ,n}:n\in \omega \}$ consists of compact sets.

• (2) If $X\in \mathfrak {M}$ is non-Polish, then X admits also a nice framing such that for each $\alpha \in \omega ^{\omega }$ there exists $n\in \omega$ such that $U_{\alpha ,n}$ is not compact.

• (3) There exists a countable Polish space $\Gamma \in \mathfrak {M}$ with the conclusion like in item (2).

First we show some auxiliary results. The first one, when dealing with $X\in \mathfrak {M}$, asserts that X admits a nice framing each layer $\{U_{\alpha ,n}:n\in \omega \}$ consists of compact sets. We need also the following concept. For $\alpha ,\beta \in \omega ^{\omega }$ we write $\alpha \preceq \beta$, if there exists $m\in \omega$ such that $\alpha _{n}\le \beta _{n}$ for every $n\ge m.$ A nice framing is said to be a strong framing, if for all $\alpha ,\beta \in \omega ^{\omega }$ with $\alpha \preceq \beta$ there exists $p\in \omega$ such that $U_{\beta ,n}\subseteq U_{\alpha ,n}$ for every $n\ge p.$

Proposition 25

For a topological space X the following statements are equivalent:

• (1) X is $\sigma$-bounded.

• (2) X admits a strong framing such that for each $\alpha \in \omega ^{ \omega }$ the layer $\{U_{\alpha ,n}:n\in \omega \}$ consists of functionally bounded sets.

• (3) X admits a strong framing such that there exists $\alpha \in \omega ^{\omega }$ for which $\{U_{\alpha ,n}:n\in \omega \}$ consists of functionally bounded sets.

• (4) X admits a nice framing such that for each $\alpha \in \omega ^{\omega }$ the layer $\{U_{\alpha ,n}:n\in \omega \}$ consists of functionally bounded sets.

Proof

Only $(1)\Rightarrow (2)$ needs to be shown. Let X be a $\sigma$-bounded space with an increasing cover $(X_{n})_{n\ge 0}$ of functionally bounded (closed) sets, $X_{0}=\emptyset .$ Then X has a strong framing $\left\{ U_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \right\}$ such that $\{U_{\alpha ,n}:n\in \omega \}=\{X_{n}:n\ge 0\}$ for every $\alpha \in \omega ^{\omega }.$

Indeed, let $\alpha =(\alpha _n)\in \omega ^\omega .$ Let $\alpha _0=0$ and $\hat{ \alpha }_n= n+ \max \{k\alpha _k: 0\le k\le n\}$ for $n\ge 0.$ Clearly ${\hat{\alpha }}_0=0$ and ${\hat{\alpha }}_n<{\hat{\alpha }}_{n+1}$ for every $n\ge 0.$ Let ${\hat{\alpha }}=({\hat{\alpha }}_1, {\hat{\alpha }}_2, {\hat{\alpha }}_3, \ldots ).$ Clearly, ${\hat{\alpha }}\in \omega ^\omega .$

Let $\alpha ,\beta \in \omega ^{\omega }.$ If $\alpha \le \beta$, then ${\hat{\alpha }}\le {\hat{\beta }}.$ Moreover, if $\alpha \preceq \beta$, then ${\hat{\alpha }}\preceq {\hat{\beta }}.$ In fact, there exists $m\in \omega$ such that $\alpha _{n}\le \beta _{n}$ for every $n\ge m$. Put $A=\max \{k\alpha _{k}:0\le k\le m\},B=\max \{k\beta _{k}:0\le k\le m\}$ and $C=\max \{A,B\}$. Let $n\in \omega$ with $n\ge C.$ Then ${\hat{\alpha }}_{n}=n+\max (\{A\}\cup \{k\alpha _{k}:m<k\le n\})=n+\max \{k\alpha _{k}:m<k\le n\}\le n+\max \{k\beta _{k}:m<k\le n\}=n+\max (\{B\}\cup \{k\beta _{k}:m<k\le n\})={\hat{\beta }}_{n}.$ Thus ${\hat{\alpha }}_{n}\le {\hat{\beta }}_{n}$ for every $n\ge C,$ so ${\hat{\alpha }}\preceq {\hat{\beta }}.$ Let $n\in \omega .$ Then there exists $m\ge 0$ such that ${\hat{\alpha }}_{m}<n\le {\hat{\alpha }}_{m+1}.$ Put $U_{\alpha ,n}=X_{m}.$ Then $U_{\alpha ,n}\subseteq U_{\alpha ,n+1}$ for all $n\in \omega$ and $\bigcup _{n=1}^{\infty }U_{\alpha ,n}=\bigcup _{n=0}^{\infty }X_{n}=X.$

Let $f\in C_p(X).$ Then there exists $\alpha = (\alpha _n) \in \omega ^\omega$ such that $\Vert f|X_k\Vert _{\infty } \le \alpha _k$ for every $k\in \omega .$ Let $n\in \omega .$ Then there exists $m\ge 0$ such that ${\hat{\alpha }}_m <n \le {\hat{\alpha }}_{m+1}$ and $\Vert f|U_{\alpha , n}\Vert _{\infty } =\Vert f|X_m\Vert _{\infty }\le \alpha _m \le {\hat{\alpha }}_m<n.$ Thus

$$\begin{aligned} \forall f\in C_p(X) \exists \alpha \in \omega ^\omega \forall n\in \omega : \Vert f|U_{\alpha , n}\Vert _{\infty }\le n. \end{aligned}$$

Let $\alpha ,\beta \in \omega ^{\omega }$ with $\alpha \le \beta .$ Let $n\in \omega .$ Then there exist $m,k\ge 0$ such that ${\hat{\alpha }} _{m}<n\le {\hat{\alpha }}_{m+1}$ and ${\hat{\beta }}_{k}<n\le {\hat{\beta }}_{k+1}.$ Clearly ${\hat{\alpha }}\le {\hat{\beta }}$, so $m\ge k.$ Thus $U_{\alpha ,n}=X_{m}\supseteq X_{k}=U_{\beta ,n}.$

Let $\alpha ,\beta \in \omega ^{\omega }$ with $\alpha \preceq \beta .$ Then ${\hat{\alpha }}\preceq {\hat{\beta }},$ so there exists $v\in \omega$ such that ${\hat{\alpha }}_{n}\le {\hat{\beta }}_{n}$ for every $n\ge v$. Let $p={\hat{\beta }} _{v}+1.$ Let $n\ge p.$ Then there exist $m,k\ge 0$ such that ${\hat{\alpha }} _{m}<n\le {\hat{\alpha }}_{m+1}$ and ${\hat{\beta }}_{k}<n\le {\hat{\beta }}_{k+1}.$ Since $n>{\hat{\beta }}_{v}$ we infer that $k\ge v$. Hence ${\hat{\alpha }} _{k}\le {\hat{\beta }}_{k}<n,$ so $k\le m.$ Thus $U_{\beta ,n}=X_{k}\subseteq X_{m}=U_{\alpha ,n},$ so $U_{\beta ,n}\subseteq U_{\alpha ,n}$ for every $n\ge p.$ $\square$

Fact 26

A countable metrizable space X is scattered if and only if X is Polish. Indeed, if X is scattered, it is Polish by [26, Lemma 8.1, Theorem 1.3]. Conversely, if X is not scattered, it contains a closed copy of rationals $\mathbb {Q}$, so X is not Polish. This applies to illustrate the following example which will be used in the sequel.

Example 27

There exists a countable Polish subspace $\Gamma$ of $\mathbb {R}$ which is not open in its completion ${\hat{\Gamma }}$ and admits a nice framing such that for every $\alpha \in \omega ^{\omega }$ there exists $n\in \omega$ such that $U_{\alpha , n}$ is not functionally bounded.

Proof

Let $x_{n,k}=2^{-n}(1+2^{-k})$ for all $n,k\in \omega .$ Set $X_{n}=\{x_{n,k}:k\in \omega \}$ for $n\in \omega .$ The set $\Gamma =\bigcup _{n=1}^{\infty }X_{n}\cup \{0\}$ endowed with the topology induced from $\mathbb {R}$ is a metrizable and separable space. For any $n\in \omega$ the set $X_{n}$ is infinite, discrete and closed, so it is not functionally bounded in $\Gamma$. Note that $\Gamma$ is a Polish space by applying Fact 26.

Let $A_{0}=\emptyset$ and $A_{m}=\{x_{n,k}:1\le n,k\le m\}$ for $m\in \omega .$ Then $\bigcup _{m=0}^{\infty }A_{m}=\bigcup _{n=1}^{\infty }X_{n}.$ Let $\alpha =(\alpha _{n})\in \omega ^{\omega }.$ Let $\alpha _{0}=0$ and ${\hat{\alpha }}_{m}=\sum _{j=0}^{m}\alpha _{j}$ for $m\ge 0.$ Let $n\in \omega .$ Then there exists $m\ge 0$ such that ${\hat{\alpha }}_{m}<n\le \hat{\alpha }_{m+1}.$ Put $U_{\alpha ,n}=A_{0}$ if $n<\alpha _{1}$ and $U_{\alpha ,n}=A_{m}\cup Y_{\alpha _{1}}$ if $n\ge \alpha _{1}.$ Clearly $U_{\alpha ,n}$ is not functionally bounded in $\Gamma$, if $n\ge \alpha _{1},$ since $U_{\alpha ,n}\supseteq X_{n},$ $n\ge \alpha _{1}.$

Let $f\in C_{p}(\Gamma ).$ Since $\sup Y_{n}=\sup X_{n}=3\cdot 2^{1-n}\rightarrow _{n}0,$ there exists $s\in \omega$ with $\Vert f|Y_{s}\Vert _{\infty }<|f(0)|+1<s.$ Let $\alpha =(\alpha _{n})\in \omega ^{\omega }$ with $\alpha _{1}\ge s$ and $\alpha _{k}\ge \Vert f|A_{k}\Vert _{\infty }$ for $k\in \omega .$ If $n<\alpha _{1}$, then $U_{\alpha ,n}=\emptyset$, so $\Vert f|U_{\alpha ,n}\Vert _{\infty }=0<n.$ Let $n\ge \alpha _{1}.$ Then $\Vert f|A_{m}\Vert _{\infty }\le \alpha _{m}\le \hat{ \alpha }_{m}<n$ and $\Vert f|Y_{\alpha _{1}}\Vert _{\infty }\le \Vert f|Y_{s}\Vert _{\infty }<s\le \alpha _{1}\le n.$ Hence $\Vert f|U_{\alpha ,n}\Vert _{\infty }\le n.$ Clearly, $U_{\alpha ,n}\subseteq U_{\alpha ,n+1}$ for all $\alpha \in \omega ^{\omega },\,n\in \omega$ and $\bigcup _{n=1}^{\infty }U_{\alpha ,n}=\bigcup _{n=1}^{\infty }X_{n}\cup \{0\}=\Gamma$ for all $\alpha \in \omega ^{\omega }.$ Let $\alpha ,\beta \in \omega ^{\omega }$ with $\alpha \le \beta .$ If $n<\beta _{1}$ then $U_{\alpha ,n}\supseteq \emptyset =U_{\beta ,n}.$ Let $n\ge \beta _{1}.$ Then there exist $m,k\ge 0$ such that ${\hat{\alpha }}_{m}<n\le {\hat{\alpha }} _{m+1}$ and ${\hat{\beta }}_{k}<n\le {\hat{\beta }}_{k+1}.$ Clearly ${\hat{\alpha }} \le {\hat{\beta }}$, so $m\ge k.$ Thus $U_{\alpha ,n}=A_{m}\cup Y_{\alpha _{1}}\supseteq A_{k}\cup Y_{\beta _{1}}=U_{\beta ,n}.$ Thus $\{U_{\alpha ,n}:(\alpha ,n)\in \omega ^{\omega }\times \omega \}$ is a nice framing in $\Gamma$ such that for every $\alpha \in \omega ^{\omega }$ there exists $n\in \omega$ such that $U_{\alpha ,n}$ is not functionally bounded. Note that ${\hat{\Gamma }}\setminus \Gamma =\{2^{-n}:n\in \omega \}$. $\square$

Lemma 28

If a metrizable space Z is not open in its completion ${\hat{Z}}$, then Z has a closed copy of $\Gamma$. Hence a separable metrizable non-Polish space contains a closed copy of $\Gamma$.

Proof

Assume Z is not open in ${\hat{Z}}$. Then there exist $z_{0}\in Z$ and a sequence $(z_{n})_{n}\subseteq {\hat{Z}}\setminus Z$ that is convergent to $z_{0}$ in ${\hat{Z}}.$ We can assume that $z_{n}\ne z_{m}$, if $n\ne m.$ Let $s_{n}=\inf _{m\ne n}d(z_{n},z_{m}),n\in \omega ,$ where d is the metric in ${\hat{Z}}$. Clearly, $s_{n}>0$ for any $n\in \omega .$ Let $(r_{n})_{n}$ be a sequence of positive numbers that is convergent to 0 such that $r_{n}<2^{-1}s_{n},\,n\in \omega .$ Clearly, the balls $K_{{\hat{Z}} }(z_{n},r_{n}),n\in \omega ,$ are pairwise disjoint. For every $n\in \omega$ there exists a sequence $(z_{n,m})_{m}\subseteq Z\cap K_{{\hat{Z}} }(z_{n},r_{n})$ which is convergent to $z_{n}$ and such that $z_{n,m}\ne z_{n,k}$, if $m\ne k.$ Set $Z_{n}=\{z_{n,m}:m\in \omega \}$ for $n\in \omega$ and $Z_{0}=\bigcup _{n=1}^{\infty }Z_{n}\cup \{z_{0}\}.$ Clearly $Z_{0}$ is a closed subspace of Z and the map $h:\Gamma \rightarrow Z_{0}$ such that $h(0)=z_{0}$ and $h(x_{n,m})=z_{n,m}$ for all $n,m\in \omega$ is a homeomorphism. $\square$

The next result follows from Lemma 28 and Example 27.

Proposition 29

Let X be a metrizable space with a nice framing and which is not open in its completion ${\hat{X}}$. Then X admits a nice framing no layer of it forms a $\sigma$-compact cover. In particular, every separable metrizable space which is non-Polish enjoys this property.

Proof of Theorem 24

$X\in \mathfrak {M}$ is $\sigma$-compact by Theorem 15. (1) follows from Proposition 25. (2) follows from Proposition 29 and (3) follows from Example 27. $\square$

5 More about strong framings

We introduce a class of Tychonoff spaces containing the Lindelöf Čech-complete spaces which are naturally related to the subject of the previous section. One may define a cardinal function b on X as the least cardinality of a set A in C(X) such that a set B in X is functionally bounded if f(B) is bounded for every $f\in A$. We call this cardinal b(X) the functional boundedness of X.

Definition 30

We say that a Tychonoff space X has countable functional boundedness if $b\left( X\right) =\aleph _{0}$, that is, if there exists a sequence $\left\{ f_{n}\right\} _{n=1}^{\infty }\subseteq C(X)$ such that a set $B\subseteq X$ is functionally bounded if all $f_{n}$ are bounded on B.

Clearly, $\mathbb {R}^{\omega }$ has countable functional boundedness, since a subset $B\subseteq \mathbb {R}^{\omega }$ is functionally bounded if and only if the canonical projections $\pi _{n}:\mathbb {R}^{\omega }\rightarrow \mathbb {R},(x_{1},x_{2},x_{3},\ldots )\rightarrow x_{n},$ are bounded on B . By Tietze-Urysohn’s Theorem any closed subspace of a space that has countable functional boundedness, has countable functional boundedness. Hence each Polish space has countable functional boundedness, as it is homeomorphic to a closed subspace of $\mathbb {R}^{\omega }$. Recall (see [13, 5.5.9(a)]) that X is Lindelöf Čech-complete if and only if X can be mapped onto a Polish space under a perfect map. We prove the main result of this section.

Theorem 31

X has countable functional boundedness if and only if there exists a continuous map T from X onto a Polish space Y such that $T^{-1}(A)$ is functionally bounded for each functionally bounded $A\subseteq Y$. Hence, if X is a $\mu$-space, the following assertions are equivalent:

• (1) X has countable functional boundedness.

• (2) X is a Lindelöf Čech-complete space.

If $C_{p}(X)$ is a $\mu$-space, $C_{p}(X)$ has countable functional boundedness if and only if $C_{p}(X)$ is isomorphic to $\mathbb {R}^{\omega }$.

Claim (for $C_{p}(X)$) holds for example if X is metrizable [2, 3.4.12 Theorem], so a metric separable X has countable functional boundedness if and only if X is Polish.

Proof of Theorem 31

If X has countable functional boundedness, it admits a fundamental resolution consisting of functionally bounded sets. Indeed, set $K_{\alpha }=\{x\in X:|f_{n}(x)|\le \alpha (n)\}$, $\alpha \in \omega ^{\omega }$, where $\left\{ f_{n}\right\} _{n=1}^{\infty }\subseteq C(X)$ is as in the definition. Define a map $T:X\rightarrow \mathbb {R}^{\omega }$, $T(x)=(f_{n}(x))_{n=1}^{\infty }\in \mathbb {R}^{\omega }$, $x\in X$. Let $A\subseteq T(X)$ be functionally bounded. By properties of T and X the set $T^{-1}(A)$ is functionally bounded. Hence, since T(X) is metrizable and separable, the closure (in T(X)) of the sets $T(K_{\alpha })$ compose a fundamental compact resolution. By Christensen’s theorem [23, Theorem 6.1] the image $Y=T(X)$ is Polish. The converse is clear since any Polish space has countable functional boundedness. If additionally X is a $\mu$ -space, then the preimage of any compact set of Y is compact in X, so T is perfect. Hence X is a Lindelöf Čech-complete space. Each Lindelöf Čech-complete space has countable functional boundedness. Finally, recall that $C_{p}(X)$ is Čech-complete if and only if X is countable and discrete, see [30, S.265]. $\square$

Next theorem characterizes those $\sigma$-bounded spaces that have countable functional boundedness. In contrast to nice framings, each strong framing in a space X with countable functional boundedness has a layer consisting of bounded sets.

Theorem 32

A space X with countable functional boundedness is $\sigma$-bounded if and only if it has a strong framing. If $\{U_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ is a strong framing, there is $\alpha \in \omega ^{\omega }$ with all $U_{\alpha ,n}$ functionally bounded.

Proof

Since X has countable functional boundedness, there exists a sequence $\left\{ f_{n}\right\} _{n=1}^{\infty }\subseteq C(X)$ as mentioned in the definition. If $\{U_{\alpha ,n}:\left( \alpha ,n\right) \in \omega ^{\omega }\times \omega \}$ is a strong framing, for any $k\in \omega$ there exists $\alpha ^{k}\in \omega ^{\omega }$ such that $\Vert f_{k}|U_{\alpha ^{k},n}\Vert _{\infty }\le n$ for every $n\in \omega .$ Let $\alpha _{n}=\max \{\alpha _{n}^{k}:1\le k\le n\}$ for $n\in \omega .$ Then $\alpha =(\alpha _{1},\alpha _{2},\alpha _{3},\ldots )\in \omega ^{\omega }$ and $\alpha _{n}^{k}\le \alpha _{n}$ for all $k,n\in \omega$ with $n\ge k$ . Hence $\alpha ^{k}\preceq \alpha$ for every $k\in \omega .$ Thus for every $k\in \omega$ there exists $n_{k}\in \omega$ such that $U_{\alpha ^{k},n}\supseteq U_{\alpha ,n}$ for every $n\ge n_{k}.$ Hence for any $k\in \omega$ we have $\Vert f_{k}|U_{\alpha ,n}\Vert _{\infty }\le \Vert f_{k}|U_{\alpha ^{k},n}\Vert _{\infty }\le n$ for every $n\ge n_{k}.$ The sequence $(U_{\alpha ,n})_{n=1}^{\infty }$ is increasing, so $\Vert f_{k}|U_{\alpha ,n}\Vert _{\infty }<\infty ,$ $k,n\in \omega .$ Thus $U_{\alpha ,n}$, with $n\in \omega$, are functionally bounded. The converse follows from Proposition 25. $\square$

A direct consequence of above Theorem 32 is Corollary 33. Note only that, by applying [22, Remark 3.1 (i)], paracompact X is Lindelöf if X has a nice framing.

Corollary 33

A paracompact Čech-complete space X is $\sigma$-compact if and only if it has a strong framing.

6 Around two problems

Being motivated by Proposition 3 one can formulate a natural question (*):

Is it true that $C_{p}(X)$ has a bounded resolution if and only if $C_{p}(X)$ admits a stronger metrizable locally convex topology?

This problem has been also posed in [17, Problem 9.3]. We show that this question has a negative solution by applying Example 35 below. Observe first that the following claims are equivalent.

• (i) There exists a $\mu$-space such that the space $C_{p}(X)$ admits a bounded resolution but does not admit a stronger metrizable locally convex topology.

• (ii) There exists a $\mu$-space space X such that $C_{p}(X)$ is K-analytic framed in $\mathbb {R}^{X}$ but X is not $\sigma$-compact.

Indeed, $(i)\Rightarrow (ii)$: We apply [16] (see Corollary ) and Theorem 5 to get that X is not $\sigma$-compact. $(ii)\Rightarrow (i)$: Apply again Theorem 5.

The following problems have been posed in [4].

Problem 34

([4]) Let X be a Tychonoff space.

• (1) Is X $\sigma$-compact if X is Lindelöf and $C_{p}(X)$ is K-analytic-framed in $\mathbb {R}^{X}$?

• (2) Let $C_{p}(X)$ be K-analytic-framed in $\mathbb {R}^{X}$. Is X a $\sigma$-bounded space?

• (3) Let X be a Lindelöf space such that $C_{p}(X)$ is K-analytic. Is X a $\sigma$-compact space?

Example 35, due to Leiderman [24], shows that the above problems (including question (*)) have negative solutions. Later on, Banakh and Leiderman recalled this again in [10, Proposition 3.8, Remark 3.9]. The present version of Example 35 provides a slightly stronger claim than the original one from [24].

Example 35

There is a Lindelöf $\Sigma$-space X with a unique non-isolated point and:

• (1) $C_p(X)$ is K-analytic.

• (2) X lacks a compact resolution, so X is not $\sigma$-compact and does not have countable functional boundedness.

• (3) Every continuous metrizable image of X is countable.

• (4) X has a nice framing; no nice framing has a layer with functionally bounded sets.

Remark 36

Leiderman’s example [24] is based on Talagrand’s paper [29] who constructed a space X with a unique non-isolated point which is a Lindelöf $\Sigma$ but not K-analytic. Leiderman proved that $C_{p}(X)$ is K-analytic. (3) follows from: Every disjoint covering of X by $G_{\delta }$-sets is countable. Item (2) follows from [22, Lemma 2.3]: X is K-analytic if and only if X is a $\mu$-space and X has a compact resolution. Clearly X does not have countable functional boundedness by Theorem 31.

Note that if X is both separable and is a continuous image of a metrizable space, the conclusion in (1) of Problem 34 still may fails.

Example 37

There exists a separable Tychonoff space X not being a $\mu$-space and

• (1) X is a continuous compact-covering image of a metric space.

• (2) X does not admit a compact resolution, in particular X is not $\sigma$-compact.

• (3) There exists a $\sigma$-compact space L such that $C_{p}(X)\subseteq L\subseteq \mathbb {R}^{X}$ but $C_{p}(X)$ is not K -analytic. Hence X admits a nice framing.

• (4) $C_{p}(X)$ admits a quotient map onto the $\sigma$-compact subspace $(\ell _{\infty })_{p}=\{(x_{n})\in \mathbb {R}^{\omega }:\sup _{n}|x_{n}|<\infty \}$ of $\mathbb {R}^{\omega }$, but $C_{p}(X)$ is not projectively $\sigma$-compact.

Proof

Denote the family of all infinite subsets of a countable set X by $[X]^{\omega }$. Set $\omega ^{*}=\beta \omega \setminus \omega$. For each $A\in [\omega ]^{\omega }$, choose an ultrafilter $u_{A}\in \omega ^{*}$ in the closure of A in $\beta \omega$. Let $X=\omega \cup \{u_{A}:A\in [\omega ]^{\omega }\}$ be topologized as a subspace of $\beta \omega$.

Proof of (1): It is known (Haydon [21]) that X is pseudocompact (separable) with cardinality of continuum and all compact subspaces of X are finite. Clearly, X is a continuous compact-covering image of a metrizable space by [25, Theorem 1.1].

Proof of (2): Assume X admits a compact resolution $\{K_{\alpha }:\alpha \in \omega ^{\omega }\}$. Since X is uncountable, some $K_{\alpha }$ is infinite, [23, Proposition 3.7], a contradiction.

Proof of (3): By Proposition 4 the space $C_{p}(X)$ has the first property. For the next one, assume $C_{p}(X)$ is K-analytic. Then by [18, Corollary 3.4] the Banach space $C^{b}(X)$ of continuous bounded real-valued functions on X equipped with the Banach topology $\xi$ generated by the norm $\Vert f\Vert =\text {sup}_{x\in X}|f(x)|$ is weakly K-analytic, i. e., the weak topology $\sigma$ of $C^{b}(X)$ is K-analytic. Hence the weak topology of $C^{b}(X)$ admits a compact resolution [23, Proposition 3.10]. Since X is separable, $C_{p}(X)$ admits a weaker metrizable topology. But then $\sigma$ is analytic by [12, Theorem 15]. Hence $C^{b}(X)=C(\beta X)$ is separable, impossible as $\beta X$ is non-metrizable. X is not a $\mu$ -space: Otherwise $C_{p}(X)=C_{k}(X)$ is barrelled by [23, Proposition 2.15], so by the closed graph theorem the identity map $I:C_{k}(X) \rightarrow (C(X),\xi )$ is continuous; hence X is compact, a contradiction.

Proof of (4): Since X is pseudocompact containing $\omega$, $C^{*}$-embedded into X, we apply [9, Theorem 1] to get a quotient map from $C_{p}(X)$ onto the subspace $(\ell _{\infty })_{p}$ of $\mathbb {R}^{\omega }$. Clearly $(\ell _{\infty })_{p}$ is covered by the sequence $[-n,n]^{\omega }$ of compact sets. By construction of X it is clear (by applying [3, Proposition 3.4]) that $C_{p}(X)$ is not projectively $\sigma$-compact. $\square$