There are known several constructions, using some additional set-theoretic assumptions, of nonmetrizable compact spaces without nonmetrizable zero-dimensional closed subspaces. Recently, Koszmider [9] constructed the first such example without such additional assumptions, and Plebanek [15] constructed a consistent example of such a space which is a Corson compact. So it is important to determine whether we can obtain such examples within some other known classes of compact spaces, for example Eberlein compact spaces.
Joel Alberto Aguilar has asked us the following question.
Problem 4.1
Let K be an Eberlein compact space of weight \(\kappa \). Does K contain a closed zero-dimensional subspace L of the same weight?
Probably, the most natural and interesting is the following simplified version of this question.
Problem 4.2
Let K be a nonmetrizable Eberlein compact space. Does K contain a closed nonmetrizable zero-dimensional subspace L?
We will show that the negative answer to this problem is consistent with ZFC, see Corollary 4.8. We do not know if the affirmative answer is also consistent with ZFC, see Problem 4.17.
We will also consider a more complex version of Problem 4.1:
Problem 4.3
Let \(\kappa \le \lambda \) be uncountable cardinal numbers, and let K be an Eberlein compact space of weight \(\lambda \). Does K contain a closed zero-dimensional subspace L of weight \(\kappa \)?
We will prove that, for every natural numbers \(1\le k\le n,\ n\ge 2\), the statement: each Eberlein compact space of weight \(\omega _n\) contains a closed zero-dimensional subspace L of weight \(\omega _k\) is independent of ZFC, see Corollaries 4.9 and 4.14.
We begin with a simple observation based on the following well known property of Corson compacta. Since we were not able to find a reference for it, we include a sketch of a proof of this fact.
Proposition 4.4
Let x be a nonisolated point of a Corson compact space K such that the character \(\chi (K,x) = \kappa \). Then K contains a copy of a one point compactification \(\alpha (\kappa )\) of a discrete space of cardinality \(\kappa \) with x as its point at infinity.
Proof
Without loss of generality we can assume that \(K\subseteq \Sigma (\Gamma )\), for some set \(\Gamma \), and, for each \(\gamma \in \Gamma \), there is \(x_\gamma \in K\) such that \(x_\gamma (\gamma ) \ne 0\). Translating K by the vector \(-x\), we can also assume that \(x= 0\) - the constant zero function in \(\Sigma (\Gamma )\). If \(\kappa = \omega \), then, by the Fréchet-Urysohn property of K, 0 is the limit of a sequence of distinct points of K, which gives the desired conclusion. If \(\kappa > \omega \), then, using transfinite induction of length \(\kappa \), and the fact that for compact spaces the pseudocharcter of a point is equal to the character, one can easily construct a set \(\{x_\alpha : \alpha \in \kappa \}\subset K\) of points with nonempty, pairwise disjoint supports. Then the subspace \(\{0\}\cup \{x_\alpha : \alpha \in \kappa \}\) of K is as desired. \(\square \)
Corollary 4.5
Let K be an Eberlein compact space with a point of character \(\kappa \). Then K contains a closed zero-dimensional subspace L of weight \(\kappa \). In particular, each Eberlein compact space of uncountable character contains a closed nonmetrizable zero-dimensional subspace L.
It is worth to recall here that, by Arhangel’skii’s theorem [6, Theorem 3.1.29], for an infinite compact space K, we have the estimate \(|K|\le 2^{\chi (K)}\). Hence we obtain the following.
Corollary 4.6
Let K be an Eberlein compact space of weight greater that \(2^\kappa \). Then K contains a closed zero-dimensional subspace L of weight \(\kappa ^+\). In particular, each Eberlein compact space of cardinality greater than continuum contains a closed nonmetrizable zero-dimensional subspace L.
Recall that the definition of a \((\lambda ,\kappa )\)-Luzin set appearing in the next lemma, can be found in Subsection 2.6.
Example 4.7
Let \(\kappa \le \lambda \) be uncountable cardinal numbers, and assume that there exists a \((\lambda ,\kappa )\)-Luzin set. Then, for each \(n\in \omega \) (\(n=\infty \)), there exists an n-dimensional nonmetrizable Eberlein compact space \(K_n\) of weight \(\lambda \) such that any closed subspace L of \(K_n\) of weight \(\ge \kappa \) has dimension n.
Applying the above for the standard Luzin set we obtain that it is consistent that Problem 4.2 (Problem 4.1) has an negative answer.
Corollary 4.8
Assuming the existence of a Luzin set, there exists a nonmetrizable Eberlein compact space K without closed nonmetrizable zero-dimensional subspaces.
As we mentioned in Sect. 2.6, for each \(n\ge 1\), the existence of a \((\omega _n,\omega _1)\)-Luzin set in \(\mathbb {R}\) is consistent with ZFC, therefore we derive the following.
Corollary 4.9
For each \(n\ge 1\), it is consistent with ZFC that there exists an Eberlein compact space K of weight \(\omega _n\) without closed nonmetrizable zero-dimensional subspaces.
Construction of Example 4.7 Fix \(n\in \omega \) (\(n=\infty \)). Let X be a \((\lambda ,\kappa )\)-Luzin set in the cube \([0,1]^n\), see Sect. 2.6. We consider the following subspace of the Aleksandrov duplicate \(AD([0,1]^n)\) (see Sect. 2.8)
$$\begin{aligned} K = AD([0,1]^n){\setminus } [([0,1]^n{\setminus } X)\times \{1\}]\,. \end{aligned}$$
Since \(\dim (AD([0,1]^n)) = n\) and K contains a copy of the cube \([0,1]^n\), the compact space K is n-dimensional (cf. [7, Ch. 3]). By Proposition 2.8, K is uniform Eberlein compact.
Let L be a closed subspace of K of weight \(\ge \kappa \) and let
$$\begin{aligned} Z = L\cap ([0,1]^n\times \{1\}) = L\cap ([X\times \{1\}) = Y\times \{1\}\,. \end{aligned}$$
Since L has weight \(\ge \kappa \), the set Y must have the cardinality \(\ge \kappa \). Let \(T\subseteq Y\) be a subset of the same cardinality as Y, and without isolated points. By the definition of a \((\lambda ,\kappa )\)-Luzin set, T is not nowhere dense in \([0,1]^n\), hence its closure \(\mathrm{Cl}_{[0,1]^n}(T)\) has a nonempty interior in \([0,1]^n\). Therefore, \(\mathrm{Cl}_{[0,1]^n}(T)\) has dimension n. Since T is dense-in-itself, the closure of \(T\times \{1\}\) in \(AD([0,1]^n)\) (hence also in L) contains the set \(\mathrm{Cl}_{[0,1]^n}(T)\times \{0\}\), a topological copy of \(\mathrm{Cl}_{[0,1]^n}(T)\). It follows that \(\dim (L) = n\). \(\square \)
Now, we will switch to the consistent results giving an affirmative answer to some cases of Problem 4.3.
We start with the following technical lemma (the definitions of cardinal numbers used in this lemma can be found in Sect. 2.7).
Lemma 4.10
Let \(\kappa \) be a cardinal number of uncountable cofinality and assume that \(\mathfrak {b} > \kappa \). Let K be a compact subset of the product \(\mathbb {R}^\Gamma \) containing a subset X of cardinality \(\kappa \) such that, for some countable subset \(\Gamma _0\) of \(\Gamma \) and for all \(x\in X\), the sets \(\mathrm{supp}(x){\setminus } \Gamma _0\) are nonempty and pairwise disjoint. Then K contains a closed zero-dimensional subspace L of weight \(\kappa \).
Proof
From our assumption that \(\mathfrak {b} > \kappa \), it follows that also \(\mathrm {non}(\mathcal {M}) > \kappa \).
Let \(\Gamma _0\subseteq \Gamma \) be a countable set witnessing the property of the set X. Observe that the property of supports of points \(x\in X\), implies that X is a discrete subspace of K.
Without loss of generality we can assume that the set \(\Gamma _0\) is infinite, so we can enumerate it as \(\{\gamma _n: n\in \omega \}\). Let \(X_n = \{x(\gamma _n): x\in X\}\) for \(n\in \omega \). Since \(\mathrm {non}(\mathcal {M}) > \kappa \), each set \(X_n\) is meager. Therefore, for each \(n\in \omega \), we can find an increasing sequence \((C_n^k)_{k\in \omega }\) of closed nowhere dense subsets of \(\mathbb {R}\) such that \(X_n\subseteq \bigcup _{k\in \omega } C_n^k\). For each \(x\in X\), we define a function \(f_x: \omega \rightarrow \omega \) as follows
$$\begin{aligned} f_x(n) = \min \{k: x(\gamma _n)\in C_n^k\}\quad \text {for } n\in \omega . \end{aligned}$$
Since \(\mathfrak {b} > \kappa \), we can find a function \(g: \omega \rightarrow \omega \) such that \(f_x\le ^* g\) for all \(x\in X\). A routine refining argument, using uncountable cofinality of \(\kappa \), shows that there is a subset \(Y\subseteq X\) of size \(\kappa \) and a function \(h: \omega \rightarrow \omega \) such that \(f_x\le h\) for all \(x\in Y\).
We define \(L = \mathrm{Cl}_K Y\).
The space L contains a discrete subspace Y of cardinality \(\kappa \), hence L has weight \(\ge \kappa \). Let \(\Gamma _1\) be a selector from the family \(\{\mathrm{supp}(x){\setminus } \Gamma _0: x\in Y\}\). Clearly, \(\Gamma _1\) has cardinality \(\kappa \). One can easily verify that the projection \(p: \mathbb {R}^\Gamma \rightarrow \mathbb {R}^{\Gamma _0\cup \Gamma _1}\) is one-to-one on L, hence \(w(L)\le \kappa \).
It remains to verify that L is zero-dimensional.
First, observe that each space \(C_n^k\), being closed nowhere dense in \(\mathbb {R}\) is zero-dimensional. Let \(\pi : L\rightarrow \mathbb {R}^\omega \) be defined by
$$\begin{aligned} \pi (x)(n) = x(\gamma _n)\quad \text {for } x\in L,\ n\in \omega . \end{aligned}$$
From our choice of Y and h it follows that
$$\pi (L)\subseteq \Pi _{n\in \omega } C_n^{h(n)}.$$
Since the product \(\Pi _{n\in \omega } C_n^{h(n)}\) is zero-dimensional, so is the space \(\pi (L)\). From the fact that the sets \(\mathrm{supp}(x){\setminus } \Gamma _0\), for \(x\in Y\), are pairwise disjoint, it follows that each fiber of \(\pi \) is either finite, or homeomorphic to a one point compactification of a discrete space, hence it is zero-dimensional. Therefore, by the theorem on dimension-lowering mappings [7, Theorem 3.3.10], L is zero-dimensional. \(\square \)
Lemma 4.11
Let \(k<n\) be natural numbers, Y be a set of cardinality \(\omega _k\), and \(\mathcal {C} = \{C_\alpha : \alpha < \omega _n\}\) be a family of countable subsets of Y. Then there exist a countable subset Z of Y and a subset S of \(\omega _n\) of cardinality \(\omega _n\) such that \(C_\alpha \subseteq Z\) for all \(\alpha \in S\).
Proof
Fix \(n\ge 1\). The case \(k=0\) is trivial. For \(k>0\), we proceed by induction on k. Without loss of generality we can assume that \(Y= \omega _k\). Since the cofinality of \(\omega _k\) is uncountable, we can find an \(\lambda <\omega _k\) and a subset R of \(\omega _n\) of cardinality \(\omega _n\) such that \(C_\alpha \subseteq \lambda \) for \(\alpha \in R\). Now, we can use the inductive hypothesis. \(\square \)
The following lemma is probably well known. We learned about it from Grzegorz Plebanek, who suggested to use it for the proof of Theorem 4.13. Its proof is based on an idea from the proof of Theorem 1.6 in [11].
Lemma 4.12
Let \(\Gamma \) be a set of cardinality \(\omega _n\), \(n\ge 2\), and \(\mathcal {A} = \{A_\alpha : \alpha < \omega _n\}\) be a family of countable subsets of \(\Gamma \), such that \(\bigcup \mathcal {A} = \Gamma \). Then there exist a countable subset B of \(\Gamma \) and a subset T of \(\omega _n\) of cardinality \(\omega _n\) such that the family \(\{A_\alpha {\setminus } B: \alpha \in T\}\) consists of nonempty, pairwise disjoint sets.
Proof
Without loss of generality we can assume that \(\Gamma = \omega _n\). For each \(\beta < \omega _n\) we pick an \(\alpha (\beta ) < \omega _n\) such that \(\beta \in A_{\alpha (\beta )}\). We consider the sets \(A_\alpha \) with the order inherited from \(\omega _n\). Since we have only \(\omega _1\) possible order types of these sets, we can find a countable ordinal \(\eta \), and a subset P of \(\omega _n\) of cardinality \(\omega _n\) such that, for all \(\beta \in P\), \(A_{\alpha (\beta )}\) has order type \(\eta \). For \(\gamma < \eta \), let \(\xi (\beta ,\gamma )\) be the \(\gamma \)-th element of \(A_{\alpha (\beta )}\). Since the union \(\bigcup \{A_{\alpha (\beta )}: \beta \in P\}\) contains an unbounded set P, and \(\omega _n\) has uncountable cofinality, there is \(\gamma < \eta \) such that the set \(\{\xi (\beta ,\gamma ): \beta \in P\}\) is unbounded in \(\omega _n\). Let \(\gamma _0\) be the smallest such \(\gamma \). Put
$$\begin{aligned} \delta = \sup \{\xi (\beta ,\gamma ): \beta \in P,\ \gamma <\gamma _0\} + 1\,. \end{aligned}$$
Using the definition of \(\gamma _0\), one can easily construct, by a transfinite induction of length \(\omega _n\), a subset Q of P of cardinality \(\omega _n\), such that the family \(\{A_{\alpha (\beta )}{\setminus } \delta : \beta \in Q\}\) consists of nonempty, pairwise disjoint sets. Finally, we can apply Lemma 4.11, for \(Y= \delta \) and the family \(\{A_{\alpha (\beta )}\cap \delta : \beta \in Q\}\), to find a countable subset B of \(\delta \) and a subset T of Q of cardinality \(\omega _n\), such that \((A_{\alpha (\beta )}\cap \delta )\subseteq B\) for \(\beta \in T\). \(\square \)
Theorem 4.13
Assume that \(\mathfrak {b} > \omega _n,\ n\ge 1\). Then each Corson compact space K of weight greater that \(\omega _1\) contains a closed zero-dimensional subspace L of weight equal to \(\min (w(K), \omega _n)\).
Corollary 4.14
For each \(n\ge 2\), it is consistent with ZFC that each Corson compact space K of weight \(\omega _n\) contains a closed zero-dimensional subspace L of the same weight.
Proof of Theorem 4.13
Let \(\lambda = w(K)\ge \omega _2\) and \(\kappa = \min (w(K), \omega _n)\). Let \(\eta = \omega _2\) if \(\kappa = \omega _1\), otherwise \(\eta = \kappa \). Without loss of generality we can assume that \(K\subseteq \Sigma (\lambda )\) and, for each \(\gamma \in \lambda \), there is \(x_\gamma \in K\) such that \(x_\gamma (\gamma ) \ne 0\). Pick a subset S of \(\lambda \) of cardinality \(\eta \) and put \(\Gamma = \bigcup \{\mathrm{supp}(x_\gamma ): \gamma \in S\}\). We apply Lemma 4.12 for \(\Gamma \) and the family \(\{\mathrm{supp}(x_\gamma ): \gamma \in S\}\), to obtain a countable subset B of \(\Gamma \) and subset \(T\subseteq S\) of cardinality \(\eta \) such that the family \(\{\mathrm{supp}(x_\gamma ){\setminus } B: \gamma \in T\}\) consists of nonempty, pairwise disjoint sets. If \(\kappa > \omega _1\), take \(X = \{x_\gamma : \gamma \in T\}\), otherwise pick a subset \(T_0\subseteq T\) of cardinality \(\omega _1\) and put \(X = \{x_\gamma : \gamma \in T_0\}\). Now, we can obtain the desired conclusion applying Lemma 4.10.
The construction from Example 4.7 and Lemma 4.10 motivated us to consider the following class of Eberlein compacta.
We say that a compact space K belongs to the class \(\mathcal {EC}_{\omega c}\) if, for some set \(\Gamma \) there is an embedding \(\varphi :K\rightarrow \mathbb {R}^\Gamma \) and a countable subset \(\Gamma _0\) of \(\Gamma \) such that, for each \(x\in K\), the set \(\mathrm{supp}(\varphi (x)){\setminus } \Gamma _0\) is finite. Since the product \(\mathbb {R}^{\Gamma _0}\) embeds into the Hilbert space \(\ell _2\) equipped with the pointwise topology, it easily follows that any compact space \(K\in \mathcal {EC}_{\omega c}\) is uniform Eberlein compact. It is clear that the class \(\mathcal {EC}_{\omega c}\) is preserved by the operations of taking finite products and closed subspaces. Example 4.16 below demonstrates that the countable power of a space from \(\mathcal {EC}_{\omega c}\) may not belong to this class.
One can also easily verify that, for a metrizable compact space M the Aleksandrov duplicate AD(M) belongs to the class \(\mathcal {EC}_{\omega c}\) (cf. the proof of Proposition 2.8), hence all spaces constructed in such a way as in Example 4.7, are in this class. One can even show that, for metrizable compacta \(M_n\), the countable product of spaces \(AD(M_n)\) is in \(\mathcal {EC}_{\omega c}\). In particular, the product \([AD(2^\omega )]^\omega \) belongs to \(\mathcal {EC}_{\omega c}\). Note, that by the remarkable result of Dow and Pearl [5] this product is an example of a homogeneous nonmetrizable Eberlein compact space. The first such example was given by Jan van Mill in [18]. The structure of these two examples seems to be closely related, but we do not know if they are homeomorphic.
Theorem 4.15
Assuming that \(\mathfrak {b} > \omega _1\), each nonmetrizable compact space \(K\in \mathcal {EC}_{\omega c}\) contains a closed nonmetrizable zero-dimensional subspace L.
Proof
Let \(K\in \mathcal {EC}_{\omega c}\). Without loss of generality we can assume that, for some set \(\Gamma \) and its countable subset \(\Gamma _0\), K is a subset of \(\mathbb {R}^\Gamma \) such that, for each \(x\in K\), the set \(\mathrm{supp}(x){\setminus } \Gamma _0\) is finite. Since K is nonmetrizable, obviously the set \(\Gamma \) must be uncountable. We can also assume that, for each \(\gamma \in \Gamma \), there is \(x_\gamma \in K\) such that \(x_\gamma (\gamma ) \ne 0\). For each \(\gamma \in \Gamma {\setminus } \Gamma _0\), the set \(F_\gamma = \mathrm{supp}(x_\gamma ){\setminus } \Gamma _0\) is finite and nonempty. Using the \(\Delta \)-system lemma we can find a finite set \(A\subseteq \Gamma \) and a set \(S\subseteq (\Gamma {\setminus } \Gamma _0)\) of size \(\omega _1\) such that, for any distinct \(\alpha ,\beta \in S\), \(F_\alpha \cap F_\beta = A\). By enlarging \(\Gamma _0\) to \(\Gamma _0\cup A\), we can assume that A is empty. Now, we can apply Lemma 4.10 for the set \(X=\{x_\gamma : \gamma \in S\}\). \(\square \)
In the light of Corollary 4.5 and Theorem 4.15 it seems natural to ask whether every first-countable Eberlein compact space belongs to the class \(\mathcal {EC}_{\omega c}\). Unfortunately, this is not the case.
Example 4.16
There exists a first-countable uniform Eberlein compact space which does not belong to the class \(\mathcal {EC}_{\omega c}\).
Proof
Our construction uses the following modification L of the Aleksandrov duplicate AD([0, 1]) of the unit interval (roughly speaking, we replace isolated points of AD([0, 1]) by copies of [0, 1]). The space L is similar to the space X(C) used by van Mill in [18] to construct an example of a homogeneous nonmetrizable Eberlein compact space.
Let \(S= [0,1]\cup \{2\}\). For any \(t\in [0,1],\ u\in [1,2]\), we define functions \(f_t: S\rightarrow [0,2],\ g_{t,u}: S\rightarrow [0,2]\) by
$$\begin{aligned} f_t(s)= & {} {\left\{ \begin{array}{ll} 0\quad \text {if }\; s\in [0,1]\,,\\ t\quad \text {if }\; s = 2\,; \end{array}\right. }\\ g_{t,u}(s)= & {} {\left\{ \begin{array}{ll} u\quad \text {if }\; s=t\,,\\ 0\quad \text {if }\; s\in [0,1],\ s\ne t\,,\\ t\quad \text {if }\; s = 2\,. \end{array}\right. }\\ \end{aligned}$$
We consider
$$\begin{aligned} L = \{f_t: t\in [0,1]\}\cup \{g_{t,u}: t\in [0,1],\ u\in [1,2]\} \end{aligned}$$
as a subspace of the cube \([0,2]^S\). One can easily verify that L is closed in \([0,2]^S\). Since the cardinality of supports of functions \(f_t\) and \(g_{t,u}\) is bounded by 2, L is an uniform Eberlein compact space. It is also easy to observe that the space L is first-countable.
We will show that the space \(K= L^\omega \) has the required property. Clearly, it is enough to show that K does not belong to the class \(\mathcal {EC}_{\omega c}\).
Suppose the contrary, i.e., there exist a set \(\Gamma \), its countable subset \(\Gamma _0\), and an embedding \(\varphi :K\rightarrow \mathbb {R}^\Gamma \) such that, for each \(x\in K\), the set \(\mathrm{supp}(\varphi (x)){\setminus } \Gamma _0\) is finite. We will treat K as a subset of the product \([0,2]^{\omega \times S}\), namely we identify the sequence \((x_n)_{n\in \omega }\in L^\omega \) with the function \(x: \omega \times S\) defined by \(x(n,s) = x_n(s)\) for \(n\in \omega \), \(s\in S\). Let \(\pi : \mathbb {R}^\Gamma \rightarrow \mathbb {R}^{\Gamma _0}\) be the projection, and let \(\psi = \pi \circ \varphi \). By the Tietze extension theorem we can extend \(\psi \) to a continuous map \(\Psi : [0,2]^{\omega \times S}\rightarrow \mathbb {R}^{\Gamma _0}\). It is well known that such a map depends on countably many coordinates, i.e., there is a countable subset \(T\subseteq \omega \times S\) and a continuous map \(\theta : [0,2]^T\rightarrow \mathbb {R}^{\Gamma _0}\) such that \(\Psi = \theta \circ \rho \), where \(\rho \) denotes the projection of \([0,2]^{\omega \times S}\) onto \([0,2]^T\) (cf. [6, 2.7.12]). Denote the restriction \(\rho \upharpoonright K\) by \(\upsilon \). Clearly, \(\theta \circ \upsilon = \psi = \pi \circ \varphi \). This implies that, for any \(y\in \psi (K)\), we have \(\upsilon ^{-1}(\theta ^{-1}(y)) = \varphi ^{-1}(\pi ^{-1}(y))\). In particular, this means that the sets \(\upsilon ^{-1}(\theta ^{-1}(y))\) and \(\pi ^{-1}(y)\cap \varphi (K)\) are homeomorphic. Observe that the set \(\pi ^{-1}(y)\cap \varphi (K)\) can be treated as a subspace of the space \(\sigma (\mathbb {R}^{\Gamma {\setminus } \Gamma _0})\), consisting of functions from \(\mathbb {R}^{\Gamma {\setminus } \Gamma _0}\) with finite supports. The space \(\sigma (\mathbb {R}^{\Gamma {\setminus } \Gamma _0})\) is strongly countable-dimensional, i.e, is a countable union of closed finite-dimensional subspaces, cf. [8, proof of Proposition 1]. Therefore, the space \(\pi ^{-1}(y)\cap \varphi (K)\) is also strongly countable-dimensional.
Pick a point \(t\in [0,1]\) such that \((\omega \times \{t\})\cap T = \emptyset \). Let \(x = (x_n)_{n\in \omega }\in K\) be the constant sequence, where \(x_n= f_t\) for all n, and let \(y = \psi (x)\). One can easily verify that the set \(\upsilon ^{-1}(\upsilon (x))\subseteq \upsilon ^{-1}(\theta ^{-1}(y))\), contains the product \(\{g_{t,u}: u\in [1,2]\}^\omega \) homeomorphic to the Hilbert cube \([0,1]^\omega \). Since the Hilbert cube is not strongly countable-dimensional (which follows easily from the Baire Category Theorem), the subspace \(\upsilon ^{-1}(\theta ^{-1}(y))\) is not strongly countable-dimensional, a contradiction. \(\square \)
Let us note that using a very similar argument as above one can show that the unit ball B of the Hilbert space \(\ell _2(\omega _1)\) equipped with the weak topology is an example of an uniform Eberlein compact space which does not belong to the class \(\mathcal {EC}_{\omega c}\). This is much simpler example than above one, but it is not first-countable.
Corollaries 4.9 and 4.14 indicate that probably the most interesting and left open case of Problem 4.3 is the following
Problem 4.17
Is it consistent that every Eberlein compact space K of weight \(\omega _1\) contains a closed zero-dimensional subspace L of the same weight?