Bounded resolutions for spaces Cp(X)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C_{p}(X)$$\end{document} and a characterization in terms of X

An internal characterization of the Arkhangel’skiĭ-Calbrix main theorem from [4] is obtained by showing that the space Cp(X)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C_{p}(X)$$\end{document} of continuous real-valued functions on a Tychonoff space X is K-analytic framed in RX\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^{X}$$\end{document} if and only if X admits a nice framing. This applies to show that a metrizable (or cosmic) space X is σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document} -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 σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document}-boundedness of X is shown.


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 ω ω ) if and only if X is σ -compact (see for example [23,Theorem 9.6] and references therein).
Calbrix proved [11] that the analyticity of C p (X ) yields the σ -compactness of X for any Tychonoff space X . The converse fails in general. Nevertheless, as Okunev proved [27], if X is σ -bounded, C p (X ) is K -analytic-framed in R X , i. e., there exists a K -analytic space Z such that C p (X ) ⊆ Z ⊆ R X . This latter result motivated paper [4], where Arkhangel'skiȋ and Calbrix characterized cosmic σ -compact spaces by showing

Theorem 1 A cosmic space X is σ -compact if and only if C p (X ) is K -analytic-framed in
The proof of Arkhangel'skiȋ-Calbrix theorem depends on a result of Christensen about fundamental compact resolutions in metric spaces and Okunev's [28] about projectively σcompact 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 R X if and only if it has a bounded resolution.
We will show that C p (X ) is K -analytic-framed in 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 16, Corollary 14). The latter concept will be directly used to construct an (usc) map from ω ω into the compact sets of Z with C p (X ) ⊆ Z ⊆ 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 σ -compactness of X , and provide an alternative approach (independent of Arkhangel'skiȋ and Calbrix) to this fact. For example, we analyze situations when X admits a nice framing with a layer {U α,n : n ∈ ω} giving a sequence of compact sets covering X , and we introduce a class of spaces containing LindelöfČech-complete spaces for which a variant of the Arkhangel'skiȋ-Calbrix theorem is obtained (Theorem 32). Recall that if C p (X ) is K -analytic-framed in R X , the space X is projectively σ -compact, i. e. every continuous separable and metrizable image of X is σ -compact, [4,Theorem 2.3]. This motivated us to examine some versions of projectively σ -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 σ -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.

Bounded resolutions for spaces C p (X)
A covering F = {A α : α ∈ ω ω } of a set X is called a resolution for X if A α ⊆ A β whenever α ≤ β coordinatewise. If X is a topological space and the sets A α are compact, the family 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 λ > 0 such that λB ⊆ U . A resolution {A α : α ∈ ω ω } for E consisting of bounded sets is called a bounded resolution. If additionally every bounded set in E is contained in some A α , we say that the family {A α : α ∈ ω ω } 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 τ p and the compact-open topology τ 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 → C p (X ) be a continuous linear surjection; let {U n } ∞ n=1 be a decreasing base of neighborhoods of zero in F.
We refer also the reader to [19,Proposition 22] for a sufficient condition for C p (X ) to have a fundamental bounded resolution.
; X is pseudocompact if and only if C p (X ) does not contain a complemented copy of R ω (see [1,Section 4]); X is called σ -bounded if X = n X n and every X n is functionally bounded, i. e., every f ∈ C(X ) is bounded on X n . A special case of Theorem 2 is Proposition 4 (due to Uspenskiȋ, 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
Hence, the closure of nS in R X provides a sequence of compact sets in R X whose union K contains C (X ). 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 R ω , which is not covered by a sequence of functionally bounded sets.
One can ask whether a Tychonoff space X is σ -bounded if and only if C p (X ) does not contain a copy of R Y for some uncountable Y . The answer is negative: C p (R ω ) does not contain a copy of R Y for any set Y with |Y | > ℵ 0 . Indeed, since the weak* dual of C p (R ω ) is separable, C p (R ω ) admits a weaker metrizable locally convex topology, but R Y fails this property, whereas it is well known that C p (ω) = R ω is not σ -bounded. Nevertheless, the 'only if' part is true in general. In fact, if {B n } ∞ n=1 is a sequence of functionally bounded sets covering X , the sets A α = { f ∈ C(X ) : sup x∈B n | f (x)| ≤ α(n) ∀n ∈ ω} for α ∈ ω ω compose a bounded resolution for C p (X ). If C p (X ) contains a copy of 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 μ-space if every functionally bounded set in X is relatively compact.

Theorem 5 ([22]) A Tychonoff space X is σ -compact if and only if X is a μ-space and there exists a metrizable locally convex topology
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 → Z and Z is metrizable but not separable. Choose a closed discrete set D in Z with |D| = ℵ 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 → Y onto a metrizable and separable space Y , which is not analytic. Indeed, such Y can be obtained as follows. Under (C H) we know that R contains 2 c subsets of the cardinality continuum, but only a continuum number of analytic subsets. So, one of those 2 c subsets Y is not analytic. Under (¬C H), take a subset Y ⊆ R of cardinality ℵ 1 . Then it is not analytic. Indeed, every uncountable analytic subset of R contains a copy of the Cantor set and hence has cardinality c.
The map f admits a (canonical) extension P f : P D → 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 (X ) 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 → δ y copy of Y in L(Y ) (the dual of C p (Y )) is closed in L (Y ) when the latter linear space is provided with the weak topology of the dual pair L (Y ) ,

Corollary 9 A paracompactČech-complete space X is σ -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 σ -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 From [4, p. 5200] the one-point Lindeöfication of an uncountable discrete space X is projectively σ -compact but is not σ -bounded. Even more can be shown.

Example 10 C p (X )
is not K -analytic framed in R X for the one-point Lindeöfication X of an uncountable discrete space.
Indeed, X is an ω-space, i. e. every continuous metrizable separable image of X is countable, see [3] or [2]. So, X is projectively σ -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.

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 α,n : (α, n) ∈ ω ω × ω} of closed subsets of X is called a framing if (i) for each α ∈ ω ω the layer {U α,n : n ∈ ω} is an increasing covering of X , and (ii) for every n ∈ ω one has that U β,n ⊆ U α,n , α ≤ β.

Theorem 13 The space C p (X ) has a bounded resolution if and only if there exists a framing
Proof If there is a framing {U α,n : (α, n) ∈ ω ω × ω} of the aforementioned characteristics, the sets compose a bounded resolution for C (X ). Indeed, each set A α is pointwise bounded by virtue of Lemma 12, since U α,n : n ∈ ω is an increasing covering of X by closed sets such that ω} is a framing satisfying the required property.
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

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.
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 σ -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 α : α ∈ ω ω } be a fundamental resolution of functions on X (we apply Corollary 14). Let X be a metric compactification (see [20] for details).
is the open ball at y and radius r . Clearly each K α is a compact subset of X \ X and 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 σ -compact. Now, Corollary 7 applies to get that X is σcompact. The converse follows from the fact, mentioned earlier, that if {B n } ∞ n=1 is a sequence of functionally bounded sets covering X , the sets 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 σ -compact. So, according to [27,Theorem 1.5], the space X is σ -compact. The converse is clear.
A regular space X is angelic if every relatively countably compact subset A of X is relatively compact and for every x ∈ 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 ω ω into compact subsets of some space Z where C p (X ) ⊆ Z ⊆ R X .

Theorem 16
If X has a nice framing, C p (X ) is K -analytic-framed in 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).
(2) For α ∈ ω ω let m(α) be the least integer such that ϕ is bounded on Thus for some W ⊆ ω ω the family {V α : α ∈ W } is a partition of ω ω on nonempty clopen subsets such that ϕ is bounded on V α for every α ∈ W . Let t α = sup ϕ(V α ) for α ∈ W . Let g : ω ω → [0, +∞) be the function such that g(β) = t α for any β ∈ V α , α ∈ W . Then g ≥ ϕ and g is locally constant, so it is continuous.

First proof of Theorem 16 By Corollary 14 fix a fundamental resolution of functions
Clearly, for any x ∈ X the function ω ω → [0, +∞), α → g(α, x) is locally constant. Moreover for any function f ∈ C p (X ) there is an g(α, x) for every x ∈ X . For any α ∈ ω ω the set F α = x∈X [−g(α, x), g(α, x)] in R X is compact. Put Z = {F α : α ∈ ω ω }. Then C p (X ) ⊆ Z ⊆ R X . Using the continuity of g with respect to the first variable it is easy to see that F : α → F α is an upper semi-continuous (usc) set-valued map from ω ω with compact values in Z . Thus C p (X ) is K -analytic-framed in R X .
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 α,n : (α, n) ∈ ω ω × ω} for X .
Proof Let {U α,n : (α, n) ∈ ω ω × ω} be a nice framing for X . For each (α, n) ∈ ω ω × ω, define the closed set Observe that W α,n ⊆ W α,n+1 for each α ∈ ω ω and W β,n ⊆ W α,n for each n ∈ ω whenever α ≤ β. We claim that n∈ω W α,n = X for each α ∈ ω ω . Indeed, suppose otherwise that there exists x / ∈ n∈ω W α,n for some α ∈ ω ω . For every n ∈ ω choose β n ∈ ω ω with β n (i) = α (i) for 1 ≤ i ≤ n such that x / ∈ U β n ,n . Put γ := sup{β n : n ∈ ω}. Then, for every n ∈ ω, β n ≤ γ and hence x / ∈ U γ,n since U γ,n ⊆ U β n ,n by the definition of framing. Hence x / ∈ n∈ω U γ,n = X , a contradiction. All this means that {W α,n : (α, n) ∈ ω ω × ω} is a framing for X . Note that the family {W α,n : (α, n) ∈ ω ω × ω} is countable since W α,n depends only on α(1), . . . , α(n). Finally, if f ∈ C (X ), by Theorem 13 there is γ ∈ ω ω such that | f (x)| ≤ n for every x ∈ U γ,n and all n ∈ ω. So | f (x)| ≤ n, x ∈ W γ,n , n ∈ ω Second proof of Theorem 16 By Lemma 19 let F = U α,n : (α, n) ∈ ω ω × ω be a countable nice framing for X . First we prove that C p (X ) is Lindelöf -framed in R X . Let us say that a function f ∈ R X is F -bounded if for each x ∈ X there exists (α, n) ∈ ω ω × ω such that x ∈ U α,n and f (U α,n ) ⊆ [−n, n] . Let us denote by Z the subset of R X consisting of all F -bounded functions on X . We claim that C (X ) ⊆ Z . Indeed, if f ∈ C (X ) there exists δ ∈ ω ω such that f (U δ,n ) ⊆ [−n, n] for every n ∈ ω. Since {U δ,n : n ∈ ω ω } covers X , given x ∈ X there exists m ∈ N N with x ∈ U δ,m and f (U δ,m ) ⊆ [−m, m], which shows that f ∈ Z . Thus C (X ) ⊆ Z , as stated. Now we prove that Z is a Lindelöf -space. If R designates the usual two points compactification of R , then R X is a compactification of Z .
The sets L α,n are compact since they are closed in R X , and compose a countably family because the framing F is countable. Choose f ∈ Z and g ∈ R X \ Z . As g ∈ R X \ Z , there exists y ∈ X such that g U α,n [−n, n] for each (α, n) ∈ ω ω ×ω for which y ∈ U α,n . Due n] , so that f ∈ Z , and B α ⊆ Z .

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 σ -bounded and Y metric, and assume that there exists a continuous linear surjection from C p (X ) onto C p (Y ). Then Y is σ -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 σ -compact but X is not σ -compact (as X contains a closed copy of ω ω ).

Strong framings, -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 σ -compact cover of X . We prove however the following Theorem 24 Let M be the class of metrizable and separable spaces with a nice framing.
(1) If X ∈ M, then X admits a nice framing such that for each α ∈ ω ω the layer {U α,n : n ∈ ω} consists of compact sets. (2) If X ∈ M is non-Polish, then X admits also a nice framing such that for each α ∈ ω ω there exists n ∈ ω such that U α,n is not compact. (3) There exists a countable Polish space ∈ M with the conclusion like in item (2).
First we show some auxiliary results. The first one, when dealing with X ∈ M, asserts that X admits a nice framing each layer {U α,n : n ∈ ω} consists of compact sets. We need also the following concept. For α, β ∈ ω ω we write α β, if there exists m ∈ ω such that α n ≤ β n for every n ≥ m. A nice framing is said to be a strong framing, if for all α, β ∈ ω ω with α β there exists p ∈ ω such that U β,n ⊆ U α,n for every n ≥ p.
Let α, β ∈ ω ω with α β. Thenα β , so there exists v ∈ ω such thatα n ≤β n for every Then there exist m, k ≥ 0 such thatα m < n ≤α m+1 and β k < n ≤β k+1 . Since n >β v we infer that k ≥ v. Henceα k ≤β k < n, so k ≤ m. Thus U β,n = X k ⊆ X m = U α,n , so U β,n ⊆ U α,n for every n ≥ p. Conversely, if X is not scattered, it contains a closed copy of rationals 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 of R which is not open in its com-pletionˆ and admits a nice framing such that for every α ∈ ω ω there exists n ∈ ω such that U α,n is not functionally bounded.

Lemma 28 If a metrizable space Z is not open in its completionẐ , then Z has a closed copy of . Hence a separable metrizable non-Polish space contains a closed copy of .
Proof Assume Z is not open inẐ . Then there exist z 0 ∈ Z and a sequence (z n ) n ⊆Ẑ \Z that is convergent to z 0 inẐ . We can assume that z n = z m , if n = m. Let s n = inf m =n d(z n , z m ), n ∈ ω, where d is the metric inẐ . Clearly, s n > 0 for any n ∈ ω. Let (r n ) n be a sequence of positive numbers that is convergent to 0 such that r n < 2 −1 s n , n ∈ ω. Clearly, the balls KẐ (z n , r n ), n ∈ ω, are pairwise disjoint. For every n ∈ ω there exists a sequence (z n,m ) m ⊆ Z ∩ KẐ (z n , r n ) which is convergent to z n and such that z n,m = z n,k , if m = k. Set Z n = {z n,m : m ∈ ω} for n ∈ ω and Z 0 = ∞ n=1 Z n ∪ {z 0 }. Clearly Z 0 is a closed subspace of Z and the map h : → Z 0 such that h(0) = z 0 and h(x n,m ) = z n,m for all n, m ∈ ω is a homeomorphism.
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 completionX . Then X admits a nice framing no layer of it forms a σ -compact cover. In particular, every separable metrizable space which is non-Polish enjoys this property.
Proof of Theorem 24 X ∈ M is σ -compact by Theorem 15. (1) follows from Proposition 25.

More about strong framings
We introduce a class of Tychonoff spaces containing the LindelöfČech-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 ∈ 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
Clearly, R ω has countable functional boundedness, since a subset B ⊆ R ω is functionally bounded if and only if the canonical projections π n : R ω → R, (x 1 , x 2 , x 3 , . . .) → 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 R ω . Recall (see [13, 5.5.9(a)]) that X is LindelöfČech-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 ⊆ Y . Hence, if X is a μ-space, the following assertions are equivalent: (1) X has countable functional boundedness.
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 Let A ⊆ 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 α ) 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 μ -space, then the preimage of any compact set of Y is compact in X , so T is perfect. Hence X is a LindelöfČech-complete space. Each LindelöfČech-complete space has countable functional boundedness. Finally, recall that C p (X ) isČech-complete if and only if X is countable and discrete, see [30, S.265].
Next theorem characterizes those σ -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 σ -bounded if and only if it has a strong framing. If {U α,n : (α, n) ∈ ω ω × ω} is a strong framing, there is α ∈ ω ω with all U α,n functionally bounded.
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 paracompactČech-complete space X is σ -compact if and only if it has a strong framing.

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 μ-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 μ-space space X such that C p (X ) is K -analytic framed in R X but X is not σ -compact.
The following problems have been posed in [4].

Problem 34 ([4]) Let X be a Tychonoff space.
(1) Is X σ -compact if X is Lindelöf and C p (X ) is K -analytic-framed in R X ?
(2) Let C p (X ) be K -analytic-framed in R X . Is X a σ -bounded space?
(3) Let X be a Lindelöf space such that C p (X ) is K -analytic. Is X a σ -compact space?
( ∞ ) p is covered by the sequence [−n, n] ω of compact sets. By construction of X it is clear (by applying [3,Proposition 3.4]) that C p (X ) is not projectively σ -compact.
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