On two consequences of CH established by Sierpiński

Abstract

We study the relations between two consequences of the Continuum Hypothesis discovered by Wacław Sierpiński, concerning uniform continuity of continuous functions and uniform convergence of sequences of real-valued functions, defined on subsets of the real line of cardinality continuum.

1 Introduction

In his classical treaty Hypothèse du continu [18] Wacław Sierpiński distinguished the following consequences of the Continuum Hypothesis (CH) (the notation is taken from [18]):

${C}_8$:

There exists a continuous function $f:E\rightarrow \mathbb {R}$, $E\subseteq \mathbb {R}$, $|E|=\mathfrak c$, not uniformly continuous on any uncountable subset of E.

${C}_9$:

There is a sequence of functions $f_n:E\rightarrow \mathbb {R}$, $E\subseteq \mathbb {R}$, $|E|=\mathfrak c$, converging pointwise but not converging uniformly on any uncountable subset of E.

Sierpiński established the equivalences of $C_9$ to several other statements, notably, to the existence of a matrix of sets of real numbers (called in [2] a BK-matrix), constructed under CH by Banach and Kuratowski [1] (statement $C_{11}$ in [18]).

Bartoszyński and Halbeisen [2] (see also [5]) proved that the existence of a BK-matrix is independent of CH. They also pointed out that the existence of a BK-matrix (hence statement $C_9$) is equivalent to the existence of a subset of $\mathbb {N}^{\mathbb {N}}$ (where $\mathbb {N}^{\mathbb {N}}$ denotes the set of functions $f:\mathbb {N}\rightarrow \mathbb {N}$ endowed with the topology of the countable product of the discrete space of natural numbers $\mathbb {N}=\{0,1,2,\ldots \}$) of cardinality $\mathfrak c$, intersecting each compact set in $\mathbb {N}^{\mathbb {N}}$ in an at most countable set (following [2] we shall call such sets K-Lusin), see [2, Proposition 1.1 and Lemma 2.3], cf. also [15].

Sierpiński [17] noticed that $C_8$ implies ${C}_9$ but he did not discuss the converse implication. However, in Topology I by Kuratowski [8], footnote (3) on page 533 suggests that the two statements are in fact equivalent. We are not aware of any publication addressing the implication $C_9 \Rightarrow C_8$ and this note is the result of our pondering on this matter.

We shall consider the following stratifications of statements $C_8$ and $C_9$ for uncountable cardinals $\kappa \le \lambda \le \mathfrak c$:

• $C_8(\lambda ,\kappa )$ There exist a set $E\subseteq \mathbb {R}$ of cardinality $\lambda$ and a continuous function $f\!\!:E \hbox { } \mathbb {R}$, which is not uniformly continuous on any subset of E of cardinality $\kappa$.

• $C_9(\lambda ,\kappa )$ There exists a set $E\subseteq \mathbb {R}$ of cardinality $\lambda$ (equivalently: for any set $E\subseteq \mathbb {R}$ of cardinality $\lambda$) and there is a sequence of functions $f_n:E\rightarrow \mathbb {R}$, converging on E pointwise but not converging uniformly on any subset of E of cardinality $\kappa$.

Clearly, statements $C_i$ (for $i=8,9$) are $C_i(\mathfrak c,\aleph _1)$ in our notation, and $C_i$ implies $C_i(\lambda ,\kappa )$ for all uncountable cardinals $\kappa \le \lambda \le \mathfrak c$.

In this note we prove (in ZFC) that, in particular:

• $C_8(\mathfrak c,\mathfrak c)\Leftrightarrow C_9(\mathfrak c,\mathfrak c)$ provided that the cardinal $\mathfrak c$ is regular, and under this assumption each of these statements is equivalent to the assertion $\mathfrak {d}=\mathfrak c$ (cf. Corollary 3.7),

• $C_8(\aleph _1,\aleph _1)\Leftrightarrow C_9(\aleph _1,\aleph _1)$, and each of these statements is equivalent to the assertion $\mathfrak {b}=\aleph _1$ (cf. Theorem 3.10).

Here $\mathfrak {d}$ and $\mathfrak {b}$ denote, as usual, the smallest cardinality of a dominating and, respectively, an unbounded family in $\mathbb {N}^{\mathbb {N}}$ corresponding to the ordering of eventual domination $\le ^*$ (cf. [5]).

An important role in our considerations is played by the notion of a K-Lusin set which we extend (cf. [2]) declaring that an uncountable subset E of a Polish space X is a $\kappa$-K-Lusin set in X, $\aleph _1\le \kappa \le \mathfrak c$, if $|E\cap K|<\kappa$ for every compact set $K\subseteq X$.

The existence of a $\kappa$-K-Lusin set of cardinality $\lambda$ in $\mathbb {N}^{\mathbb {N}}$ is equivalent to $C_9(\lambda ,\kappa )$ (cf. Theorem 2.3) and if $E\subseteq \mathbb {R}$, $|E|=\lambda$, is a witnessing set for $C_8(\lambda ,\kappa )$, then E is a $\kappa$-K-Lusin set in some $G_\delta$-extension of E (cf. Proposition 3.1).

However, it is not the case that every $\kappa$-K-Lusin set is a witnessing set for $C_8(\lambda ,\kappa )$. In particular, assuming CH, we show that there is a K-Lusin set in the irrationals of cardinality $\mathfrak c$ such that every continuous function $f:E \rightarrow \mathbb {R}$ is uniformly continuous on an uncountable subset of E (cf. Theorem 3.3). Our reasoning to that effect yields also that for every continuous function $f:X\rightarrow \mathbb {R}$ defined on a $G_\delta$-set X in the irrationals, there exists a closed copy of irrationals P in X such that f is uniformly continuous on P (cf. Theorem 3.2).

The paper is organized as follows.

In Sect. 2 we establish the equivalences of $C_9(\lambda ,\kappa )$ to several other statements, notably, to its topological counterparts (see Theorem 2.3).

Section 3 is devoted to $C_8(\lambda ,\kappa )$ and its relations to $C_9(\lambda ,\kappa )$ including proofs of the equivalence $C_8(\lambda ,\kappa )\Leftrightarrow C_9(\lambda ,\kappa )$ for $\kappa =\lambda =\mathfrak c$ (if the cardinal $\mathfrak c$ is regular) and $\kappa =\lambda =\aleph _1$. We will end the section by listing some additional set-theoretic assumptions under which the equivalence $C_8(\mathfrak c,\aleph _1)\Leftrightarrow C_9(\mathfrak c,\aleph _1)$ is true. Although the status of the implication $C_9(\mathfrak c,\aleph _1)\Rightarrow C_8(\mathfrak c,\aleph _1)$ remains unclear, these observations point out at difficulties in refuting it.

In Sect. 4 we gathered some comments and additional results related to the topic without proofs – we plan to present details elsewhere.

In this note $\mathbb {P}$ always denotes the set of irrationals of the unit interval [0, 1]. It is homeomorphic to the Baire space $\mathbb {N}^{\mathbb {N}}$ (cf. [7]).

2 Uniform convergence of pointwise convergent sequences of functions and statement $C_9(\lambda ,\kappa )$

The following result is based on Sierpiński’s reasoning [16], cf. Remark 2.2(1) (an extension of this result is formulated in Sect. 4.3).

Theorem 2.1

For any Polish space X there is a sequence $f_1\ge f_2\ldots$ of continuous functions $f_n:X \hbox { } [0,1]$ which converges to zero pointwise but does not converge uniformly on any set with non-compact closure in X.

Proof

Since X embeds as a closed subspace in $[1,+\infty )^{\mathbb {N}_+}$ (cf. [7, Theorem 4.17]), where $\mathbb {N}_+=\{n\in \mathbb {N}:\ n>0\}$, it is enough to construct desired functions on $[1,+\infty )^{\mathbb {N}_+}$. So, with no loss of generality, we simply assume that $X=[1,+\infty )^{\mathbb {N}_+}$.

We begin with the Sierpiński functions $s_n:X \rightarrow \{0,1,\frac{1}{2},\frac{1}{3},\ldots \}$, $n = 1, 2, . . .,$ defined by (cf. Remark 2.2)

• (1) $s_n(x)= {\left\{ \begin{array}{ll} \frac{1}{\min \{i:\ x(i)\ge n\}}\quad &{}\hbox {if}\quad x(\mathbb {N}_+)\cap [n,+\infty )\ne \emptyset ,\\ 0 &{}\hbox {if} \quad x(\mathbb {N}_+)\subseteq [0,n). \end{array}\right. }$ We shall check that

• (2) $s_1\ge s_2\ge \ldots$ and $\lim \nolimits _{n\rightarrow \infty }s_n(x)=0$ for every $x\in X$,

• (3) for any $A\subseteq X$, if the sequence $(s_n)_{n=1}^{\infty }$ converges uniformly on A, then the closure $\bar{A}$ is compact.

The monotonicity in (2) is clear. If $x\in X$ and $p\in \mathbb {N}_+$ is given, then for any $n>\max \{x(i):\ i\le p\}$ we have $s_n(x)<\frac{1}{p}$, which gives the second part of (2).

To make sure that (3) is true, we shall follow closely Sierpiński [16]. Let $A\subseteq X$ and assume that the sequence $(s_n)_{n=1}^{\infty }$ converges uniformly on A. This means that for each $i\in \mathbb {N}_+$ there is a $\varphi (i)\in \mathbb {N}_+$ such that $s_m(x)<\frac{1}{i}$, whenever $m\ge \varphi (i)$ and $x\in A$.

By (1), for any $x\in X$ and $i\in \mathbb {N}_+$ we have $s_{\lfloor x(i)\rfloor }(x)\ge \frac{1}{i}$, and hence, $x(i)\le \varphi (i)$, for any $x\in A$. Therefore, A is contained in the compact set $\prod \nolimits _{i=1}^\infty [1,\varphi (i) ]\subseteq X$, and hence its closure $\bar{A}$ in X is compact.

Let us verify that for each $n\in \mathbb {N}_+$

• (4) the function $s_n$ is upper-semicontinuous,

i.e., for any $r>0$ the set $\{x\in X:\ s_n(x)<r\}$ is open in X. Since $s_n$ is bounded by 1, it is enough to consider $r\le 1$.

So let us fix $n\in \mathbb {N}_+$, $r\le 1$ and $a\in X$ with $s_n(a)<r$, and for any $p\in \mathbb {N}_+$, let us consider the open set $V_p$ defined by

• (5) $V_p=\{x\in X:\ x(i)<n\quad \hbox {for all}\quad i\le p \}$.

We shall show that we can always find p such that $V_p$ is a neighbourhood of a contained in the set $\{x\in X:\ s_n(x)<r\}$.

If $s_n(a)=0$, i.e., $a(i)<n$ for all $i\in \mathbb {N}_+$, cf. (1), then taking p such that $\frac{1}{p}<r$, we have $a\in V_p$ and $s_n(x)<\frac{1}{p}<r$ for every $x\in V_p$, cf. (1) and (5).

If $s_n(a)=\frac{1}{m}$, where $m=\min \{i:\ a(i)\ge n\}$, then since $s_n(a)<1$, we have $a(1)<n$. Hence $m\ge 2$ and let $p=m-1$. Then $p\ge 1$, $a\in V_p$ and for any $x\in V_p$, $s_n(x)\le \frac{1}{m}<r$.

Having checked (4), we apply a classical theorem of Hahn (cf. [4, 1.7.15(c)]) to get, for each n, continuous functions $f_{n,i}:X \rightarrow [0,1]$, $i=1,2,\ldots$, such that

• (6) $f_{n,1}\ge f_{n,2}\ge \ldots$ and $\lim \nolimits _{i\rightarrow \infty } f_{n,i}(x)=s_n(x)$ for every $x\in X$.

Finally, we define

• (7) $f_n(x)=\min \nolimits _{i,j\le n} f_{i,j}(x)$ for $x\in X$.

Clearly, the sequence $f_1\ge f_2\ge \ldots$ consists of continuous functions and converges pointwise to zero. Moreover, $f_n(x)\ge s_n(x)$ for any $n\in \mathbb {N}_+$ and $x\in X$. Consequently, for any $A\subseteq X$, if the sequence $(f_n)_{n=1}^{\infty }$ converges uniformly on A, then so does the sequence $(s_n)_{n=1}^{\infty }$ and hence by (3), $\bar{A}$ is compact.$\square$

Remark 2.2

• (1) The original Sierpiński functions were defined on $\mathbb {N}_+^{\mathbb {N}_+}$ by the formula:

  $$s_n(x)={\left\{ \begin{array}{ll} \frac{1}{\min \{i:\ x(i)= n\}}\quad &{}\hbox {if}\quad n\in x(\mathbb {N}_+),\\ 0 &{}\hbox {if} \quad n\notin x(\mathbb {N}_+). \end{array}\right. }$$

  Sierpiński was interested in neither regularity of the functions (in fact, $s_n$ are continuous on $\mathbb {N}_+^{\mathbb {N}_+}$) nor the monotonicity of the function sequence.

• (2) An approach similar to Sierpiński’s idea, in a different setting, was rediscovered by Pincirolli [11, Lemma 2 and Proposition 7].

Let us turn back to statement $C_9(\lambda ,\kappa )$. The following result provides some topological counterparts to $C_9(\lambda ,\kappa )$.

Theorem 2.3

For any uncountable cardinals $\kappa \le \lambda \le \mathfrak c$ the following are equivalent:

• (1) $C_9(\lambda ,\kappa )$,

• (2) there is a set $A\subseteq \mathbb {N}^{\mathbb {N}}$ of cardinality $\lambda$ and a sequence $g_1\ge g_2\ldots$ of continuous functions $g_n:A\rightarrow \mathbb {R}$, which converges to zero pointwise but does not converge uniformly on any set of cardinality $\kappa$,

• (3) there is a $\kappa$-K-Lusin set of cardinality $\lambda$ in $\mathbb {N}^{\mathbb {N}}$,

• (4) there is a Polish space X and a $\kappa$-K-Lusin set of cardinality $\lambda$ in X.

Proof

$(1)\Rightarrow (2)$. Subtracting from each function in $C_9(\lambda ,\kappa )$ the limit function, we get a sequence $f_n:E \rightarrow \mathbb {R}$, $|E|=\lambda$,

which converges to zero pointwise but does not converge uniformly on any set of cardinality $\kappa$. For every $n\in \mathbb {N}_+$ and $x\in E$ let $u_n(x)=\max \{|f_i(x)|:\ i\ge n\}$ (recall that $\lim \nolimits _{i\rightarrow \infty }f_i(x)=0$, hence the maximum is attained). Let us note that $u_1\ge u_2\ldots$ and $0\le |f_n|\le u_n$ for each n. The properties of the sequence $(f_n)_{n=1}^\infty$ yield readily that the sequence $(u_n)_{n=1}^\infty$ converges to zero pointwise on E but it does not converge uniformly on any subset of E of cardinality $\kappa$.

Let $h:E \rightarrow A$ be a bijection between E and a set $A\subseteq 2^{\mathbb {N}}$ such that all the functions $g_n=u_n\circ h^{-1}$ are continuous (for example, define h as the Marczewski characteristic function (cf. [10]) of a countable family $\{E_n:\ n\in \mathbb {N}\}$ of subsets of E, separating the points of E and containing all sets of the form $u_n^{-1}\bigl ((p,q)\bigr )$, where $n\in \mathbb {N}_+$ and $p<q$ are rationals). Then the sequence $g_1\ge g_2\ldots$ of continuous functions $g_n:A\rightarrow \mathbb {R}$ is as required.

$(2)\Rightarrow (3)$. Let us fix a set $A\subseteq \mathbb {N}^{\mathbb {N}}$ of cardinality $\lambda$ and a sequence $g_1\ge g_2\ldots$ of continuous functions $g_n:A \rightarrow \mathbb {R}$, which converges to zero pointwise but does not converge uniformly on any subset of A of cardinality $\kappa$. Let H be a $G_\delta$-set in $\mathbb {N}^{\mathbb {N}}$ with $A\subseteq H\subseteq \bar{A}$ and such that each $g_n$ extends to a continuous function $\tilde{g}_n: H \rightarrow \mathbb {R}$. Since $\tilde{g}_1\ge \tilde{g}_2\ge \ldots$, for any $x\in H$ we have

$$ \lim \limits _{n\rightarrow \infty }\tilde{g}_n(x)=0\ \Leftrightarrow \ \forall p\in \mathbb {N}_+\exists n\in \mathbb {N}_+\ \tilde{g}_n(x)<\frac{1}{p}, $$

so the set $G=\{x\in H:\ \lim \nolimits _{n\rightarrow \infty }\tilde{g}_n(x)=0\}$ is a $G_\delta$-set in $\mathbb {N}^{\mathbb {N}}$ containing A. Now, if $K\subseteq G$ is compact, then by the Dini theorem (see [4, Lemma 3.2.18]), the sequence $(\tilde{g}_n)_{n=1}^\infty$ converges uniformly on K, hence also $({g}_n)_{n=1}^\infty$ converges uniformly on $A\cap K$, and therefore $|A\cap K|<\kappa$.

Finally, let $w:G \rightarrow \mathbb {N}^{\mathbb {N}}$ embed G onto a closed subspace of $\mathbb {N}^{\mathbb {N}}$ (see [7, Theorem 7.8]) and let $E=w(A)$. Then $|E\cap K|<\kappa$ for every compact set $K\subseteq \mathbb {N}^{\mathbb {N}}$, as required.

$(3)\Rightarrow (4)$. This implication is trivial.

$(4)\Rightarrow (1)$. Let E be a subset of cardinality $\lambda$ of a Polish space X such that $|E\cap K|<\kappa$ for every compact set $K\subseteq X$. Theorem 2.1 provides us with a sequence $f_1\ge f_2\ldots$ of continuous functions $f_n:X \rightarrow [0,1]$ which converges to zero pointwise but does not converge uniformly on any set with non-compact closure in X. Consequently, the sequence $({f}_n)_{n=1}^\infty$ converges to zero pointwise on E but any set $M\subseteq E$ of cardinality $\kappa$ has a non-compact closure in X, so $({f}_n)_{n=1}^\infty$ does not converge uniformly on M. Clearly, this completes the proof of (1). $\square$

In two important cases, namely $\kappa = \lambda = \aleph _1$ and $\kappa = \lambda = \mathfrak c$, statement $C_9(\kappa ,\lambda )$ is characterized in terms of basic cardinal characteristics of the continuum, cf. [3].

Corollary 2.4

• (1) $C_9(\aleph _1,\aleph _1)\Leftrightarrow \mathfrak {b}=\aleph _1$,

• (2) $C_9(\mathfrak c,\mathfrak c)\Leftrightarrow \mathfrak {d}=\mathfrak c$, provided that the cardinal $\mathfrak c$ is regular (more precisely, $\mathfrak {d}=\mathfrak c$ implies $C_9(\mathfrak c,\mathfrak c)$ under no additional assumptions on $\mathfrak c$).

Proof

We shall repeatedly make use of Theorem 2.3.

(1). Assume $C_9(\aleph _1,\aleph _1)$ and let $E\subseteq \mathbb {N}^{\mathbb {N}}$ be a set of cardinality $\aleph _1$ whose intersection with every compact set $K\subseteq \mathbb {N}^{\mathbb {N}}$ is countable. Clearly, E is unbounded in $(\mathbb {N}^{\mathbb {N}},\le ^*)$, hence $\mathfrak {b}=\aleph _1$.

Conversely, if $\mathfrak {b}=\aleph _1$, then any subset of $\mathbb {N}^{\mathbb {N}}$ of the form $\{f_\alpha :\alpha < \mathfrak {b}\}$, where

• $\alpha<\beta <\mathfrak {b}$ implies $f_\alpha <^*f_\beta$,

• for every $f\in \mathbb {N}^{\mathbb {N}}$ there is $\alpha <\mathfrak {b}$ with $f_\alpha \nleq ^*f$,

has countable intersection with every compact $K\subseteq \mathbb {N}^{\mathbb {N}}$.

(2). Assume $C_9(\mathfrak c,\mathfrak c)$ and let $E\subseteq \mathbb {N}^{\mathbb {N}}$ be a set of cardinality $\mathfrak c$ such that $|E\cap K|<\mathfrak c$ for every compact set $K\subseteq \mathbb {N}^{\mathbb {N}}$. Let $\{g_\alpha :\alpha < \mathfrak {d}\}$ be a dominating set in $\mathbb {N}^{\mathbb {N}}$. In particular $E=\bigcup _{\alpha<\mathfrak {d}} \{f\in E:\ f<^* g_\alpha \}$ and the regularity of $\mathfrak c$ implies that $\mathfrak {d}=\mathfrak c$.

Conversely, if $\mathfrak {d}=\mathfrak c$, then any subset of $\mathbb {N}^{\mathbb {N}}$ of the form $\{g_\alpha :\alpha < \mathfrak c\}$, where

• $\alpha<\beta <\mathfrak c$ implies $g_\beta \not \le ^*g_\alpha$,

• for every $f\in \mathbb {N}^{\mathbb {N}}$ there is $\alpha <\mathfrak c$ with $f <^*g_\alpha$,

has the property that $|E\cap K|<\mathfrak c$ for every compact $K\subseteq \mathbb {N}^{\mathbb {N}}$.$\square$

The proof of $(4)\Rightarrow (1)$ in Theorem 2.3 yields also the following result.

Corollary 2.5

For any uncountable cardinals $\kappa \le \lambda \le \mathfrak c$, if E is a $\kappa$-K-Lusin set of cardinality $\lambda$ in a Polish space X, then there exists a sequence $f_1\ge f_2\ldots$ of continuous functions $f_n:E\rightarrow \mathbb {R}$, which converges to zero pointwise but does not converge uniformly on any subset of E of cardinality $\kappa$.

Remark 2.6

As was mentioned in the introduction, the notion of a K-Lusin set was introduced by Bartoszyński and Halbeisen [2], where it was pointed out that a reasoning of Banach and Kuratowski [1], establishing under CH the existence of a BK-Matrix, actually shows that the existence of a BK-matrix is equivalent to the existence of a K-Lusin set of cardinality $\mathfrak c$. Earlier, Sierpiński [16] proved that a BK-Matrix exists if and only if $C_9$ holds. Combining these two results, we get the equivalence "$C_9 \Leftrightarrow \ \hbox {there exists a K-Lusin set of cardinality }\mathfrak c$“ which was obtained in Theorem 2.3 by a different reasoning.

3 Uniform continuity of continuous functions and statement $C_8(\lambda ,\kappa )$

Let us now turn our attention to statement $C_8(\lambda ,\kappa )$. Sierpiński [17] proved that $C_8$ implies $C_9$ and his argument can be easily adapted to establish a more general implication concerning $C_i(\lambda ,\kappa )$ (for $i=8,9$). Instead of repeating the argument of Sierpiński we present a proof based on Theorem 2.3 which gives some additional information about the involved sets.

Proposition 3.1

For any uncountable cardinals $\kappa \le \lambda \le \mathfrak c$:

$$ C_8(\lambda ,\kappa )\Rightarrow C_9(\lambda ,\kappa ). $$

Moreover, if a set $E\subseteq \mathbb {R}$, $|E|=\lambda$, together with a continuous function $f:E \rightarrow \mathbb {R}$ witness $C_8(\lambda ,\kappa )$, then there is a $G_\delta$-set G in $\mathbb {R}$ such that $E\subseteq G$ and E is a $\kappa$-K-Lusin set in G.

Proof

Let us extend f to a continuous function $\tilde{f}:G\rightarrow \mathbb {R}$ over a $G_\delta$-set $G\subseteq \mathbb {R}$. Now, if $K\subseteq G$ is compact, then the extension $\tilde{f}$ is uniformly continuous on K, hence so is f on $E\cap K$. Therefore, $|E\cap K|<\kappa$, as f is not uniformly continuous on any set of cardinality $\kappa$. This shows that the equivalent to $C_9(\lambda ,\kappa )$ statement, formulated in Theorem 2.3(4), is true, completing the proof.

In the rest of this note we investigate the possibility of reversing the above implication, at least for some pairs of uncountable cardinals $\kappa \le \lambda \le \mathfrak c$.

In view of Proposition 3.1, a related question is whether a $\kappa$-K-Lusin set E in $\mathbb {P}$ always carries a continuous function $f:E \rightarrow \mathbb {R}$, which is not uniformly continuous on any set of cardinality $\kappa$. The negative answer (cf. Theorem 3.3) is a consequence of the following general result, closely related to the “limit systems” of Hurewicz [6].

Theorem 3.2

Let X be a Polish non $\sigma$-compact space and let d be a compatible completely bounded metric on X. Then for every continuous function $f:X\rightarrow \mathbb {R}$ there exists a closed copy of irrationals P in X such that f is uniformly continuous on P in the metric d.

Proof

Let $(\hat{X},\hat{d})$ be the completion of (X, d); then $\hat{X}$ is compact, d being totally bounded. Since X is not $\sigma$-compact, by a theorem of Hurewicz (see [7, Theorem 7.10]), X contains a closed in X copy of the irrationals G. Let $\rho$ be a complete metric on G.

We shall use generalized Hurewicz systems in the setting considered in [12, Section 2.4] and [13, Section 2]. Namely, we shall define a pair of families: $(U_s)_{s\in \mathbb {N}^{<\mathbb {N}}}$ of subsets of G, and $(x_s)_{s\in \mathbb {N}^{<\mathbb {N}}}$ of points in $\hat{X}$ with the following properties (the closures are taken in $\hat{X}$, $B_{\hat{d}}(x_s,\varepsilon )=\{x\in \hat{X}:\hat{d}(x_s,x)<\varepsilon \}$ and for $A\subseteq G$, $\textrm{diam}_{\rho } (A)$ or $\textrm{diam}_{d} (A)$ stand for the diameter with respect to $\rho$ or d):

• (1) $U_s$ is relatively open in G, $U_s\ne \emptyset$,

• (2) $\textrm{diam}_{\rho } (U_s) \le 2^{-length(s)}$,

• (3) $\overline{U_s}\cap \overline{U_t}=\emptyset$ for distinct $s,\ t$ of the same length,

• (4) $\overline{U_{s \hat{\ }i}}\cap G\subseteq U_s$,

• (5) $x_s \in \overline{U_s}\setminus G$,

• (6) $x_s\not \in \overline{U_{s \hat{\ }i}}$ for any $i\in \mathbb {N}$,

• (7) each neighbourhood of $x_s$ in $\hat{X}$ contains all but finitely many $U_{s\hat{\ }i}$; in particular, $\lim _i\textrm{diam}_d (U_{s\hat{\ }i})=0$,

• (8) $\textrm{diam}(f(U_{s\hat{\ }i}))\le 2^{i}$ for any $i\in \mathbb {N}$,

• (9) if $c_i \in U_{s\hat{\ }i}$ for each $i\in \mathbb {N}$, then the sequence $(f(c_i))_{i\in \mathbb {N}}$ is convergent in $\mathbb {R}$.

To define sets $U_s$ and points $x_s$ we proceed as follows.

Let $U_{\emptyset }$ be a non-empty relatively open set in G such that f is bounded on $U_{\emptyset }$ and $\textrm{diam}_{\rho } (U_{\emptyset }) \le 1$.

At the inductive step let $n\ge 0$ and assume that we have already defined $U_s$ and $x_t$ for $s\in \mathbb {N}^{\le n}$ and $t\in \mathbb {N}^{< n}$ satisfying the required conditions. Fix s with $length(s)=n$ and pick $x_s\in \overline{U_s}\setminus G$ arbitrarily (this is possible since G does not contain compact sets with non-empty interior). Let us choose points $a_n\in U_s$ such that $\lim _n a_n = x_s$ and the sequence $(f(a_n))_n$ is convergent (first, we choose $b_n\in U_s$ so that $\lim _n b_n = x_s$ and then, using the fact that the sequence $(f(b_n))_n$ is bounded, we choose its convergent subsequence). Next, using the continuity of f on G, let us enlarge each $a_n$ to its open neighbourhood $U_{s\hat{\ }n}$ in G so that relevant instances of conditions (1)–(8) are satisfied. Then (8) and the fact that the sequence $(f(a_n))_n$ is convergent readily yield (9).

Let

• (10) $P = \bigcap _n \bigcup \{U_s: \hbox {length(s)} = n\}.$

be the copy of the irrationals determined by the generalized Hurewicz system $(U_s)_{s\in \mathbb {N}^{<\mathbb {N}}}$, $(L_s)_{s\in \mathbb {N}^{<\mathbb {N}}}$ ([13, Section 2]), where $L_s=\{x_s\}$ for each $s\in \mathbb {N}^{<\mathbb {N}}$. In particular, $\overline{P} = P \cup \{x_s: s\in \mathbb {N}^{<\mathbb {N}}\}$, so $P=\overline{P}\cap G$ is closed in G, and hence also in X.

We claim that for each $s\in \mathbb {N}^{<\mathbb {N}}$

• (11) $\inf \limits _{\varepsilon > 0} \textrm{diam} \Bigl (f\bigl (B_{\hat{d}}(x_s,\varepsilon )\cap P\bigr )\Bigr ) =0.$

To justify the claim, let us fix $s\in \mathbb {N}^{<\mathbb {N}}$ and for each $i\in \mathbb {N}$ let us pick $c_i\in U_{s\hat{\ }i}$. By (9), $\lim \nolimits _{i\rightarrow \infty } f(c_i)=r$, and let J be an arbitrary open interval containing r. From (7) and (8) we get $i_0$ such that $f(U_{s\hat{\ }i})\subseteq J$, whenever $i>i_0$.

Now, we can find an $\varepsilon > 0$ such that $B_{\hat{d}}(x_s,\varepsilon )\cap P$ is disjoint from any $U_t$ with $t\ne s$ and $length(t)=length(s)$, i.e., cf. (10),

• (12) $W= B_{\hat{d}}(x_s,\varepsilon )\cap P\subseteq B_{\hat{d}}(x_s,\varepsilon )\cap U_s$.

For suppose that for every $\varepsilon >0$, (12) is false. This allows us to define a sequence $(z)_{n\in \mathbb {N}}$, $z_n\in P$, converging to $x_s$ such that the set $Z=\{z_n:n\in \mathbb {N}\}$ is disjoint from $U_s$.

By (3) and (5), $Z\cap U_t$ is finite for any $t\in \mathbb {N}^{<\mathbb {N}}$ with $t\ne s$ and $length(t)=length(s)$. Then, since $Z\subseteq U_\emptyset$, it follows that we can find $t\in \mathbb {N}^{<\mathbb {N}}$ such that $length(t)<length(s)$, $Z\cap U_t$ is infinite but $Z\cap U_{t\hat{\ }i}$ is finite for each $i\in \mathbb {N}$. By (7), $\lim \nolimits _{n\rightarrow \infty } z_n=x_t$, however, by (3) and (6), $x_t\ne x_s$, and this contradiction completes the justification of (12).

Next, by appealing to (6), we can make $\varepsilon$ still smaller to ensure that $B_{\hat{d}}(x_s,\varepsilon )\cap P$ omits also all $U_{s\hat{\ }i}$ with $i\le i_0$. Consequently, cf. (10) and (12), $W\subseteq \bigcup \nolimits _{i>i_0}U_{s\hat{\ }i}$, hence $f(W)\subseteq J$.

This completes the justification of the claim, and let us note that (11) means exactly that the oscillation of f|P at any point of $\overline{P}\setminus P=\{x_s: s\in \mathbb {N}^{<\mathbb {N}}\}$ is zero. This guarantees that f|P can be extended continuously over $\overline{P}$. By compactness of $\overline{P}$ this extension is uniformly continuous, and in effect we get uniform continuity of f on P.$\square$

As a corollary we obtain that not every $\kappa$-K-Lusin set of cardinality $\lambda$ in the irrationals is a witnessing set for $C_8(\lambda ,\kappa )$ (cf. [17, a remark at the end of the paper]).

Theorem 3.3

If $\mathfrak {d}=\mathfrak c$, then there is a $\mathfrak c$-K-Lusin set E in $\mathbb {P}$ of cardinality $\mathfrak c$ such that every continuous function $f:E \rightarrow \mathbb {R}$ is uniformly continuous on a subset of E of cardinality $\mathfrak c$.

In particular, assuming CH, there is a K-Lusin set $E\subseteq \mathbb {P}$ of cardinality $\mathfrak c$ such that every continuous function $f:E \rightarrow \mathbb {R}$ is uniformly continuous on an uncountable subset of E.

Proof

We list all compact sets in $\mathbb {P}$ as $( K_\alpha :\ \alpha <\mathfrak c)$, and all closed copies of irrationals in $\mathbb {P}$ as $( P_\alpha : \ \alpha <\mathfrak c)$, where each closed copy of irrationals P in $\mathbb {P}$ appears in this transfinite sequence $\mathfrak c$-many times.

Then we inductively pick

$$ x_\alpha \in P_\alpha {\setminus }\Bigl (\bigcup _{\beta<\alpha }K_\beta \cup \{x_\beta : \beta <\alpha \} \Bigr ), $$

the choice being made possible by the assumption $\mathfrak {d}=\mathfrak c$ which means that $\mathbb {P}$ is not covered by any collection of less that $\mathfrak c$-many its compact subsets (cf. [3, Theorem 2.8]).

We let $E=\{x_\alpha :\alpha <\mathfrak c\}$. Let us notice that E is a $\mathfrak c$-K-Lusin set E in $\mathbb {P}$ and

• (1) E intersects each closed copy of irrationals in $\mathbb {P}$ in a set of cardinality $\mathfrak c$.

To see that E is a set we are looking for, let $f:E\rightarrow \mathbb {R}$ be a continuous function, and let X be a $G_\delta$-set in $\mathbb {P}$ containing E such that f extends to the (uniquely defined) continuous function $\hat{f}:X\rightarrow \mathbb {R}$.

Now, E being a $\mathfrak c$-K-Lusin set in $\mathbb {P}$, it cannot be covered by countably many compact sets in $\mathbb {P}$. Consequently, X is a non $\sigma$-compact Polish space contained in [0, 1], so by Theorem 3.2 there exists a closed copy of irrationals P in X such that $\hat{f}$ is uniformly continuous on P (in the metric inherited from [0, 1]). Shrinking P, if necessary, we may assume that P is closed also in $\mathbb {P}$. By (1), $|E\cap P|=\mathfrak c$ and we conclude that f is uniformly continuous on $E\cap P$.$\square$

On the other hand, we have the following result.

Theorem 3.4

If $\mathfrak {d}=\mathfrak c$, then there is a $\mathfrak c$-K-Lusin set E in $\mathbb {P}$ of cardinality $\mathfrak c$ and a continuous function $f:E \rightarrow \mathbb {R}$, which is not uniformly continuous on any set of cardinality $\mathfrak c$.

In particular, we have $\mathfrak {d}=\mathfrak c\Rightarrow C_8(\mathfrak c,\mathfrak c)$.

Our proof will be based on the following proposition.

Proposition 3.5

Let $f:\mathbb {P}\rightarrow [0,1]$ be a continuous function such that the closure $\overline{G(f)}$ in $[0,1]^2$ of the graph G(f) of f intersects each $\{q\}\times [0,1]$, $q\in \mathbb {Q}\cap [0,1]$, in an uncountable set. Then $\mathbb {P}$ cannot be covered by less than $\mathfrak {d}$ sets on which f is uniformly continuous.

Proof of Proposition 3.5

Let ${\mathcal {A}}$ be a collection of subsets of $\mathbb {P}$ such that $|{\mathcal {A}}|<\mathfrak {d}$ and f is uniformly continuous on each $A\in {\mathcal {A}}$. We will show that $\mathbb {P}\setminus \bigcup {\mathcal {A}}\ne \emptyset$.

Let $A\in {\mathcal {A}}$. Then f|A being uniformly continuous extends continuously over the closure of A in [0, 1], and let $K_A$ be the graph of this extension.

For each $q\in \mathbb {Q}\cap [0,1]$, $K_A$ intersects $\{q\}\times [0,1]$ in at most a singleton, and hence $|\bigcup \nolimits _{A\in {\mathcal {A}}}\bigl (K_A\cap (\{q\}\times [0,1])\bigr )|<\mathfrak {d}$. If V is an open neighbourhood of (t, f(t)), $t\in \mathbb {P}$, in $[0,1]^2$, then there are non-empty open intervals $I_1, \ I_2$ in [0, 1] such that $t\in I_1$, $I_1\times I_2\subseteq V$ and $f(I_1\cap \mathbb {P})\subseteq I_2$. It follows that for every $q\in I_1\cap \mathbb {Q}$, $\overline{G(f)}\cap (\{q\}\times \bar{I}_2)=\overline{G(f)}\cap (\{q\}\times [0,1])$, so by the properties of f, $|\overline{G(f)}\cap (\{q\}\times I_2)|=\mathfrak c$. Consequently, the set

$$ H=\overline{G(f)}\cap (\mathbb {Q}\times [0,1]){\setminus } \bigcup \limits _{A\in {\mathcal {A}}}K_A $$

is dense in $\overline{G(f)}$.

Since G(f) is a $G_\delta$-set dense in $\overline{G(f)}$, by the Baire theorem, each $F_\sigma$-set covering G(f) must hit H, and by the Kechris-Louveau-Woodin theorem (see [7, Theorem 21.22]), we get a Cantor set $C\subseteq G(f)\cup H$ such that $P=C\cap G(f)$ is a copy of the irrationals, closed in G(f).

For each $A\in {\mathcal {A}}$, $K_A$ being compact, the set $K_A\cap P=K_A\cap C$ is compact and since $|{\mathcal {A}}|<\mathfrak {d}$ it follows that $P\setminus \bigcup \nolimits _{A\in {\mathcal {A}}}K_A\ne \emptyset$ (cf. [3, Theorem 2.8]). In effect, $G(f) \setminus \bigcup \nolimits _{A\in {\mathcal {A}}}K_A\ne \emptyset$ which proves that $\mathbb {P}\not \subseteq \bigcup {\mathcal {A}}$.$\square$

With Proposition 3.5 in hand, we can easily get Theorem 3.4.

Proof of Theorem 3.4

Let $f:\mathbb {P}\rightarrow [0,1]$ be as in Proposition 3.5 (see Example 3.6).

Let us list as $(F_\alpha :\ \alpha <\mathfrak c)$ all closed sets in $\mathbb {P}$ on which f is uniformly continuous. Since $\mathfrak {d}=\mathfrak c$, by the assertion of Proposition 3.5, we can inductively pick points

$$ x_\alpha \in \mathbb {P}{\setminus }\Bigl (\bigcup \limits _{\beta<\alpha } F_\beta \cup \{x_\beta :\beta<\alpha \}\Bigr ), \alpha <\mathfrak c, $$

and finally let $E=\{x_\alpha :\alpha <\mathfrak c\}$.

Then if $A\subseteq E$ and f is uniformly continuous on A, the closure of A in $\mathbb {P}$ is listed as some $F_\alpha$, and hence $|A|<\mathfrak c$.

Likewise, every compact set K in $\mathbb {P}$ is on the list, so $|E\cap K|<\mathfrak c$ which shows that E is a $\mathfrak c$-K-Lusin set in $\mathbb {P}$.$\square$

For the sake of completeness we recall an example given by Kuratowski and Sierpiński in [9], of a function satisfying the assertion of Proposition 3.5.

Example 3.6

Let $\psi :[0,1]\rightarrow [0,1]$ be given by the formula

$$ \psi (t)=\sum _{n=1}^{\infty } \frac{\phi (t-q_n)}{2^n}, $$

where $\phi (t)=|\sin (\frac{1}{t})|$ for $t\ne 0$ and $\phi (0)=0$, and $(q_1, q_2,\ldots )$ is an injective enumeration of $\mathbb {Q}\cap [0,1]$.

Then $f=\psi |([0,1]\setminus \mathbb {Q})$ satisfies the assertion of Proposition 3.5. To see this, let us fix $q_n$, and let

$$ \sigma (t)=\sum _{m\ne n} \frac{\phi (t-q_m)}{2^m} \quad \hbox {for}\ t\in [0,1]. $$

Then $\sigma$ is continuous at $q_n$,

$$ \psi (t)=\sigma (t) + \frac{\phi (t-q_n)}{2^n}, $$

and the definition of $\phi$ yields that

$$ \overline{G(f)}\cap (\{q_n\}\times [0,1])= \{q_n\}\times [\sigma (t), \sigma (t)+2^{-n}]. $$

By combining Theorem 3.4 with Proposition 3.1 and Corollary 2.4, we immediately get the following corollary.

Corollary 3.7

If the cardinal $\mathfrak c$ is regular, then the following statements are equivalent:

• (1) $C_8(\mathfrak c,\mathfrak c)$,

• (2) $C_9(\mathfrak c,\mathfrak c)$,

• (3) $\mathfrak {d}=\mathfrak c$.

Without any additional set-theoretic assumptions we have (in ZFC) the following weaker result.

Theorem 3.8

For any uncountable cardinal $\kappa <\mathfrak c$,

$$ C_9(\mathfrak c,\kappa )\Rightarrow C_8(\mathfrak c,\mathfrak c). $$

Before giving the proof of Theorem 3.8 let us recall that a subset S of a Polish space X is $\kappa$-concentrated in X on a set $D\subseteq X$, $\aleph _1\le \kappa \le \mathfrak c$, if $|S\setminus U|<\kappa$ for every open in X set $U\subseteq X$ that contains D.

Let us assume that $C\subseteq \mathbb {R}$ is the Cantor set, Q is a countable dense set in C and $P=C\setminus Q$.

Let us note that if $S\subseteq P$, then S is $\kappa$-concentrated on Q in C if and only if S is a $\kappa$-K-Lusin in P. (cf. [2, Proposition 3.4]).

We shall need the following lemma.

Lemma 3.9

Suppose that there is a set $S\subseteq P$ of cardinality $\mathfrak c$ $\kappa$-concentrated on Q in C. Then there is a set $H\subseteq S\times C$ such that for every compact set K in $C\times C$ with $K\cap (Q\times C)$ countable, we have $|H\cap K|<\mathfrak c$.

Proof of Lemma 3.9

We list all compact sets in $C\times C$ with $K\cap (Q\times C)$ countable as $( K_\alpha : \alpha <\mathfrak c)$. Then we inductively pick distinct points $s_\alpha \in S$ and $t_\alpha \in C$ such that $(s_\alpha ,t_\alpha )\not \in \bigcup \nolimits _{\beta <\alpha }K_\beta$ for $\alpha <\mathfrak c$.

At stage $\alpha <\mathfrak c$, let us notice that $\bigcup \nolimits _{\beta <\alpha }K_\beta \cap (Q\times C)$ has cardinality less than $\mathfrak c$, and therefore, there is $t\in C$ such that $\bigcup \nolimits _{\beta <\alpha }K_\beta \cap (Q\times \{t\})=\emptyset$. Since $S\times \{t\}$ is $\kappa$-concentrated on $Q\times \{t\}$ in $C\times \{t\}$, it follows that, for each $\beta <\alpha$, $|K_\beta \cap (S\times \{t\})|<\kappa$. Therefore,

$$ \left| \bigcup \limits _{\beta<\alpha }\bigl (K_\beta \cap (S\times \{t\})\bigr )\right| <\mathfrak c, $$

so we can pick $s\in S$, distinct from $s_\beta$ for all $\beta <\alpha$, such that $(s,t)\notin \bigcup \nolimits _{\beta <\alpha }K_\beta$, and we let $(s_\alpha ,t_\alpha )=(s,t)$.

Finally, we let $H=\{(s_\alpha ,t_\alpha ):\alpha <\mathfrak c\}$.$\square$

With Lemma 3.9 in hand, we can now prove Theorem 3.8 as follows.

Proof of Theorem 3.8

Let us assume that statement $C_9(\mathfrak c,\kappa )$ is true. With the help of Theorem 2.3 and the fact that P is a homeomorphic copy of $\mathbb {N}^{\mathbb {N}}$, we get a $\kappa$-K-Lusin S in P of cardinality $\mathfrak c$.

There is a continuous map $\phi :C\times C \rightarrow C$ such that $\phi |(P\times C)$ is a homeomorphism onto $G=\phi (P\times C)$, and the set $D=\phi (Q\times C)$ is countable and disjoint from G (a simple argument to this effect is given in [14, Lemma 4.2]).

Since S is $\kappa$-concentrated on Q in C there is, by Lemma 3.9, a set $H\subseteq S\times C$ such that for every compact set K in $C\times C$ with $K\cap (Q\times C)$ countable, we have $|H\cap K|<\mathfrak c$. Let $E=\phi (H)$ and $f=\phi ^{-1}|E: E\rightarrow H$. Upon an embedding of $C\times C$ in $\mathbb {R}$, we can consider f as a function from a subset of $\mathbb {R}$ of cardinality $\mathfrak c$ into $\mathbb {R}$ and we are going to prove that it is a witness that statement $C_8(\mathfrak c,\mathfrak c)$ is true.

Aiming at a contradiction, assume that f|A is uniformly continuous (with respect to any metric compatible with the topology of $C\times C$) on a set $A\subseteq E$ of cardinality $\mathfrak c$ and let $B=f(A)=\phi ^{-1}(A)$. Then, since $\phi |B:B\rightarrow A$ is also uniformly continuous, the function f|A extends to a homeomorphism $\tilde{f}: \bar{A}\rightarrow \bar{B}$, where $\bar{A}$ and $\bar{B}$ are the closures of A and B in C and $C\times C$, respectively (cf. [4, Theorem 4.3.17]). Since D is countable and $\tilde{f}^{-1}=\phi |\bar{B}$ injectively maps $\bar{B}\cap (Q\times C)$ into $\bar{A}\cap D$, $\bar{B}\cap (Q\times C)$ is also countable, and we have $\bar{B}=K_\alpha$ for some $\alpha <\mathfrak c$. This, however, is impossible, as by Lemma 3.9, we have $|H\cap K_\alpha |<\mathfrak c$, but on the other hand $B\subseteq H\cap K_\alpha$ has cardinality $\mathfrak c$. $\square$

A similar reasoning shows, in particular, that also the statements $C_8(\aleph _1,\aleph _1)$ and $C_9(\aleph _1,\aleph _1)$ are equivalent.

Let us recall that E is a $\lambda '$-set in X if for every countable set L in X, L is a relative $G_\delta$-set in $E\cup L$; by a theorem of Sierpiński, there is (in ZFC) an uncountable $\lambda '$-set in $\mathbb {R}$, cf. [8].

Theorem 3.10

For any uncountable cardinal $\nu \le \mathfrak c$, the existence of a $\lambda '$-set T of cardinality $\nu$ in the Cantor set $C\subseteq \mathbb {R}$ and a K-Lusin set S in $P=C\setminus Q$ of cardinality $\nu$, where Q is a countable dense set in C, implies $C_8(\nu ,\aleph _1)$. Consequently, the existence of a $\lambda '$-set of cardinality $\nu$ in the Cantor set C implies that $C_8(\nu ,\aleph _1)\Leftrightarrow C_9(\nu ,\aleph _1)$ and hence,

we have (in ZFC) $C_8(\aleph _1,\aleph _1)\Leftrightarrow C_9(\aleph _1,\aleph _1),$ and each of the statements is equivalent to the assertion $\mathfrak {b}=\aleph _1$.

Proof

Let H be the graph of a bijection from S onto T. Since S is concentrated in C on Q, it follows that H is concentrated in $C\times C$ on $Q\times C$. Indeed, if U is an arbitrary open set in $C\times C$ containing $Q\times C$, and $D=(C\times C) \setminus U$, then $V=C\setminus \hbox {proj}_1(D)$ is an open set in C containing Q (where $\hbox {proj}_1$ is the projection of $C\times C$ onto the first axis). Thus V contains all but countably many points of S, and consequently, the set $H\setminus U$ is countable.

As in the proof of Theorem 3.8, let $\phi :C\times C \rightarrow C$ be a continuous map such that $\phi |(P\times C)$ is a homeomorphism onto $G=\phi (P\times C)$, and the set $D=\phi (Q\times C)$ is countable and disjoint from G.

Let $E=\phi (H)$ and $f=\phi ^{-1}|E: E\rightarrow H$. Upon an embedding of $C\times C$ in $\mathbb {R}$, we consider f as a function from a subset of $\mathbb {R}$ of cardinality $\nu$ into $\mathbb {R}$ and we shall prove that it is a witness that statement $C_8(\nu ,\aleph _1)$ is true.

Aiming at a contradiction, assume that f|A is uniformly continuous on an uncountable set $A\subseteq E$, let $B=f(A)=\phi ^{-1}(A)$ and

extend the function f|A to a homeomorphism $\tilde{f}: \bar{A}\rightarrow \bar{B}$, where $\bar{A}$ and $\bar{B}$ are the closures of A and B in C and $C\times C$, respectively (cf. the proof of Theorem 3.8).

Let us notice that E is concentrated on D in C and $\bar{A}\cap D\ne \emptyset$. It is easy to see that this implies that A is concentrated on $\bar{A}\cap D$ in $\bar{A}$, hence also B is concentrated on $L=\tilde{f}(\bar{A}\cap D)$ in $\bar{B}$. Clearly, L is a countable subset of $\bar{B}\subseteq C\times C$ and B is concentrated on L also in $C\times C$. Therefore, $B'= \hbox {proj}_2(B)$ is concentrated on $L'=\hbox {proj}_2(L)$ in C (where $\hbox {proj}_2$ is the projection of $C\times C$ onto the second axis). Since H is the graph of an injection, $B'$ is uncountable, and it follows that $L'$ is a countable set in C which is not a $G_\delta$-set in $B'\cup L'$. This, however, contradicts the fact that T is a $\lambda '$-set in C and $B'\subseteq T$.

The assertion, stating the equivalence of statements $C_8(\nu ,\aleph _1)$ and $C_9(\nu ,\aleph _1)$ assuming the existence of a $\lambda '$-set of cardinality $\nu$ in the Cantor set follows now from Theorem 2.3 and Proposition 3.1. Indeed, $C_9(\nu ,\aleph _1)$ implies that there is also a K-Lusin set in $P=C\setminus Q$ of cardinality $\nu$, where Q is a countable dense set in C (cf. Theorem 2.3), which by what we have already proved, yields $C_8(\nu ,\aleph _1)$. The converse implication is always true (see Proposition 3.1).

The final assertions follow now from Corollary 2.4(1) and the existence (in ZFC) of a $\lambda '$-set of cardinality $\aleph _1$ in the Cantor set.$\square$

While the status of the implication $C_9 \Rightarrow C_8$, the central topic of this note, remains unclear, the following conditions, sufficient for the validity of $C_9 \Rightarrow C_8$, hint at difficulties in finding a model of ZFC where, possibly, $C_9$ is true but $C_8$ is false.

Proposition 3.11

Any of the following three conditions implies that $C_9$ implies $C_8$ (and then $C_8$ and $C_9$ are equivalent):

• (1) no K-Lusin set in $\mathbb {N}^{\mathbb {N}}$ has cardinality $\mathfrak c$ (in particular, this is so if $\mathfrak {b}>\aleph _1$ or $\mathfrak {d}<\mathfrak c$),

• (2) there exists a Lusin set in $\mathbb {R}$ of cardinality $\mathfrak c$,

• (3) there exists a $\lambda '$-set in the Cantor set of cardinality $\mathfrak c$.

Proof

The non-existence of K-Lusin sets in $\mathbb {N}^{\mathbb {N}}$ of cardinality $\mathfrak c$ makes $C_9$ false by Theorem 2.3.

Similarly, the existence of a Lusin set in $\mathbb {R}$ of cardinality $\mathfrak c$ makes $C_8$ true (see [18, proof of Théorème 6 on page 45]).

The existence of a $\lambda '$-set of cardinality $\mathfrak c$ in the Cantor set C implies that $C_8(\mathfrak c,\aleph _1)\Leftrightarrow C_9(\mathfrak c,\aleph _1)$, by Theorem 3.10. $\square$

4 Comments

In this section we present some additional results related to the subject of this note; their proofs will be given in a forthcoming paper.

4.1 Mappings into compact spaces

One can show that statement $C_8(\lambda ,\kappa )$ is equivalent to its analogue concerning functions with ranges in compact metric spaces.

Proposition 4.1

For any uncountable cardinals $\kappa \le \lambda \le \mathfrak c$ if the cofinality of $\lambda$ is uncountable, then the following are equivalent:

• (1) $C_8(\lambda ,\kappa )$,

• (2) there is a set $E\subseteq [0,1]$ of cardinality $\lambda$, such that for every uncountable compact metric space Y there is a continuous function $f:E\rightarrow Y$, which is not uniformly continuous on any subset of E of cardinality $\kappa$.

Moreover, if a set E satisfies property (2) above for a continuous function $f:E\rightarrow Y$ into an uncountable compact metric space Y, then so does it for a certain real-valued continuous function on E.

However, the latter is no longer true when we replace $E\subseteq [0,1]$ by an arbitrary separable metrizable space, as shown by the following result.

Proposition 4.2

Assuming that no family of less than $\mathfrak c$ meager sets covers $\mathbb {R}$, there exists a set of positive dimension $E\subseteq [0,1]^\mathbb {N}$ such that

• (1) there is a continuous function $f:E \rightarrow [0,1]^\mathbb {N}$, which is not uniformly continuous on any subset of E of cardinality $\mathfrak c$,

• (2) each continuous function $g:E\rightarrow \mathbb {R}$ is constant on a subset of E of cardinality $\mathfrak c$.

4.2 Mappings into non-compact spaces

One can show that also statement $C_9(\lambda ,\kappa )$ is equivalent to an analogue of statement $C_8(\lambda ,\kappa )$ but concerning functions with ranges in non-$\sigma$-compact Polish spaces.

Proposition 4.3

For any uncountable cardinals $\kappa \le \lambda \le \mathfrak c$ the following are equivalent:

• (1) $C_9(\lambda ,\kappa )$,

• (2) there is a set $E\subseteq [0,1]$ of cardinality $\lambda$ such that for every non-$\sigma$-compact Polish space Y there is a continuous function on E which is not uniformly continuous (with respect to any complete metric on Y) on any subset of E of cardinality $\kappa$.

Moreover, any $\kappa$-K-Lusin set E of cardinality $\lambda$ in $\mathbb {P}$ has property (2) above. In particular, with the help of Theorem 3.3 and Proposition 4.1, it follows that, under CH, there exists a K-Lusin set E in $\mathbb {P}$ of cardinality $\mathfrak c$ such that E admits a continuous function $f:E\rightarrow \mathbb {R}^\mathbb {N}$ which is not uniformly continuous on any uncountable subset of E, but each continuous map $g:E\rightarrow [0,1]^\mathbb {N}$ is uniformly continuous on an uncountable subset of E.

4.3 A characterization of completeness

One can show that the existence of a function sequence described in Theorem 2.1 characterizes completeness of a separable metrizable space X.

In fact, the following more general result can be obtained (for terminology see [4]).

Proposition 4.4

Let X be a Hausdorff space. Then X is a Čech-complete Lindelöf space if and only if there is a sequence $f_1\ge f_2\ge \ldots$ of continuous functions $f_n:X \rightarrow [0,1]$ converging pointwise to zero but not converging uniformly on any closed non-compact set in X.