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This might be an easy question, maybe the example I'm looking for is common knowledge. As always, recall that a topological space $X$ is scattered if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. It is known that any first countable, $T_3$, Lindelöf and scattered space is countable. Is there an example of an uncountable, first countable, Hausdorff, Lindelöf and scattered space?

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  • $\begingroup$ A Hausdorff space is scattered iff it is right separated, that is, it admits a is a well ordering whose initial segments are open. Hence a scattered Hausdorff space have equal Lindelöf number and cardinality. See here $\endgroup$ Commented Feb 17, 2022 at 15:18
  • $\begingroup$ @MathieuBaillif: I agree with your first sentence, but not your second. I think what you meant is that a scattered Hausdorff space has Lindelöf number at least as big as its cofinality. For example, $\omega_1+1$ is compact. $\endgroup$
    – Will Brian
    Commented Feb 17, 2022 at 16:17
  • $\begingroup$ Ah yes, of course, that's what I wanted to say (but not what I wrote). $\endgroup$ Commented Feb 17, 2022 at 21:29

2 Answers 2

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There exists an example for this question under Continuum Hypothesis, more precisely, under the assumption $\mathfrak b=\omega_1$. In this case by Theorem 10.2 in the van Douwen's survey paper ``The integers and Topology'', there exists an uncountable set $C\subseteq\mathbb R\setminus\mathbb Q$, which is concentrated at rationals in the sense that for every open set $U\subseteq\mathbb R$ with $\mathbb Q\subseteq U$ we the complement $C\setminus U$ is countable.

Consider the uncountable space $X=\mathbb Q\cup C$ endowed with the topology $\tau$ consisting of the sets $W\subseteq X$ such that for every $q\in W\cap\mathbb Q$ there exists $\varepsilon>0$ such that $\{x\in C:|x-q|<\varepsilon\}\subseteq W$.

It is easy to see that the space $(X,\tau)$ is (functionally) Hausdorff, first-countable, Lindelof and scattered. More precisely, the subspace $C=X\setminus\mathbb Q$ is open and discrete and $\mathbb Q$ is closed and discrete in $X$. So, $X$ is scattered of finite scattered height.

To see that $X$ is Lindelof, observe that for every open cover $\mathcal U$ there exists a countable subfamily $\mathcal V\subseteq\mathcal U$ such that $\mathbb Q\subseteq \bigcup\mathcal V$. Since $C$ is concentrated at $\mathbb Q$, the difference $C\setminus\bigcup\mathcal V$ is countable and hence we can choose a countable subfamily $\mathcal V'\subseteq \mathcal U$ such that $C\setminus \bigcup\mathcal V\subseteq\bigcup\mathcal V'$. Then $\mathcal V\cup\mathcal V'$ is a required countable subcover of $\mathcal U$, witnessing that $X$ is Lindelof.


The above example cannot be constructed in ZFC because of the following

Theorem. Under $\mathfrak b>\omega_1$, every scattered Lindelof $T_1$-space of countable pseudocharacter and finite scattered height is countable.

Prior writing down the proof of this theorem, let us recall the definition of the scattered height.

Given a subset $A$ of a topological space $X$, let $A^{(1)}$ be the set of nonisolated points of $A$. Observe that for a closed subset $A$ of $X$, the set $A^{(1)}$ is closed in $A$. Let $X^{(0)}=X$ and for every nonzero ordinal $\alpha$ let $$X^{(\alpha)}=\bigcap_{\beta<\alpha}\big(X^{(\beta)}\big)^{(1)}.$$ By transfinite induction it can be shown that $(X^{(\alpha)})_\alpha$ is a decreasing sequence of closed subsets of $X$, so it should stabilize at some ordinal $\alpha$. The smallest ordinal $\alpha$ such that $X^{(\alpha+1)}=X^{(\alpha)}$ is called the scattered height of $X$ and is denoted by $\hbar(X)$. A topological space $X$ is scattered if and only if $X^{(\hbar(X))}=\emptyset$.

Proof of Theorem. Assume that $\mathfrak b>\omega_1$. By Mathematical Induction we shall prove that for every natural number $n$, every Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)\le n$ is countable.

For $n=0$, each scattered space $X$ with $\hbar(X)\le 0$ is empty and hence countable. Assume that for some positive integer number $n$, every Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)<n$ is countable. Take a Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)=n$. Then $X^{(n)}=\emptyset$ and $X^{(n-1)}$ is a nonempty discrete subspace of $X$. Being a closed subspace of the Lindelof space $X$, the discrete space $X^{(n-1)}$ is Lindelof and countable. Write $X^{(n-1)}$ as $\{x_k\}_{k\in\omega}$. Since $X$ has countable pseudocharacter, for every $k\in\omega$ there exists a decreasing sequence $(U_{k,m})_{m\in\omega}$ of open sets in $X$ such that $\bigcup_{m\in\omega}U_{k,m}=\{x_k\}$. Assuming that the space $X$ is uncountable, choose a transfinite sequence $\{y_\alpha\}_{\alpha\in\omega_1}$ consisting of pairwise disjoint elements of $X\setminus X^{(n-1)}$. For every $\alpha\in\omega_1$, let $f_\alpha:\omega\to\omega$ be the function assigning to each $k\in\omega$ the smallest number $f_\alpha(k)$ such that $y_\alpha\notin U_{k,f_\alpha(k)}$. The number $f_\alpha(k)$ exists since $y_\alpha\notin\{x_k\}=\bigcap_{m\in\omega}U_{k,m}$. Since $\mathfrak b>\omega_1$, there exists a countable subfamily $\mathcal F\subseteq\omega^\omega$ such that for every $\alpha\in\omega_1$ there exists a function $f\in\mathcal F$ such that $f_\alpha\le f$. For every $f\in\mathcal F$, consider the open neighborhood $U_f=\bigcup_{k\in\omega}U_{k,f(k)}$ of the set $\{x_k\}_{k\in\omega}=X^{(n-1)}$. The choice of $\mathcal F$ guarantees that $\{y_\alpha\}_{\alpha\in\omega_1}\cap\bigcap_{f\in\mathcal F}U_f=\emptyset$. By the Pigeonhole Principle, for some $f\in\mathcal F$, the set $\Omega=\{\alpha\in\omega_1:y_\alpha\notin U_f\}$ is uncountable. On the other hand, $Y=X\setminus U_f$ is a closed (and hence Lindelof) subspace of $X$. It follows from $Y\cap X^{(n-1)}=\emptyset$ that $\hbar(Y)<n$ and hence $Y$ is countable, by the inductive hypothesis. But this contradicts the uncountability of the set $\{y_\alpha\}_{\alpha\in\Omega}\subseteq Y$.


Can the theorem be generalized to scattered spaces of arbitrary scattered height?

Problem. Assume that $\mathfrak b>\omega_1$. Is every scattered Lindelof $T_1$-space of countable pseudocharacter countable?


Remark. Thanks to the intervention of Will Brian, I realized that the equality $\mathfrak d=\omega_1$ in the initial redaction of my answer can be replaced by the weaker equality $\mathfrak b=\omega_1$. So, now we have the following characterization.

Theorem. The following conditions are equivalent:

  1. $\mathfrak b>\omega_1$.

  2. Every Lindelof scattered $T_1$-space of countable pseudocharacter finite scattered height is countable.

  3. Every first-countable Lindelof scattered Hausdorff space of scattered height 2 is countable.

The answer of Will Brian shows that this theorem does not extend to scattered Lindelof spaces of infinite scattered height.

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  • $\begingroup$ This is a very interesting construction. For what it's worth, I think you can modify things slightly to make it work in the Cohen model. (I don't know how interesting this would be to you, and whether it's worth the trouble of writing down the details.) This shows that the existence of such a space is consistent with $\mathfrak{d} > \aleph_1$, but it does not answer your question about what happens when $\mathfrak{b} > \aleph_1$. $\endgroup$
    – Will Brian
    Commented Feb 19, 2022 at 13:29
  • $\begingroup$ @WillBrian Thank you for your comment. Concerning Cohen model, this is my Achilles heel: I do not know forcing, unfortunately. I thought a bit about what happens under $\mathfrak b>\omega_1$ and could only provide an answer for scattered spaces of finite scattered height. The inductive step breaks down at the first infinite ordinal: I do not know how to proceed if all derived sets $X^{(n)}$ are uncountable. $\endgroup$ Commented Feb 19, 2022 at 13:35
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The answer to the question is yes: $\mathsf{ZFC}$ alone proves there is an example of an uncountable, first countable, Hausdorff, Lindelöf, scattered space.

The example I will describe has scattered height $\omega$. By the second part of Taras' answer, having infinite scattered height is a necessary feature of the example, because we are not assuming $\mathfrak{b} = \aleph_1$.

Recall that a Bernstein set is a subset $X$ of $\mathbb R$ such that both $X$ and its complement have nonempty intersection with every uncountable closed subset of $\mathbb R$. Such subsets of $\mathbb R$ can be proved to exist using the Axiom of Choice. In fact, the standard proof of the existence of Bernstein sets can be modified slightly to get a partition of $\mathbb R$ into infinitely many Bernstein sets:

Lemma: There is a partition of $\mathbb R$ into a countably infinite number of sets, such that each of these sets has nonempty intersection with every uncountable closed subset of $\mathbb R$.

(Actually, it is possible to partition $\mathbb R$ into $\mathfrak{c}$ Bernstein sets. But to construct our example, we want a partition into just $\aleph_0$ Bernstein sets.)

Let $X_0, X_1, X_2, X_3, \dots$ be a countable collection of pairwise disjoint Bernstein sets, as described in the lemma.

The underlying set of our example is $\mathbb R$, endowed with the topology generated by the following basic open sets. Given $x \in \mathbb R$, there is a unique $n$ such that $x \in X_n$; we take the basic open neighborhoods of $x$ to be sets of the form $\{x\} \cup \left( U \cap \bigcup_{i < n}X_i \right)$, where $U$ is a rational open interval in $\mathbb R$ containing $x$ (i.e., $U$ is a basic open neighborhood of $x$ in the normal Euclidean topology on $\mathbb R$).

This space is clearly uncountable, because its underlying set is $\mathbb R$. It is first countable, because the neighborhood bases described above are all countable. This space is Hausdorff because rational Euclidean-open intervals are still open in this topology (i.e., it refines the usual topology on $\mathbb R$). It is scattered because for any $Y \subseteq \mathbb R$, there is some minimal $n$ such that $Y \cap X_n \neq \emptyset$, and any point of $Y \cap X_n$ is isolated in $Y$.

Finally, it remains to show this topology is Lindelöf. Roughly, the idea is that for each $n$, we can use countably many of the neighborhoods of the points on level $n$ of our space (i.e., points in $X_n$) to reach down and cover all but countably many of the points on lower levels.

Suppose $\mathcal U$ is an open cover for this space. By shrinking the elements of $\mathcal U$ if necessary, we may (and do) assume that each member of $\mathcal U$ is a basic open set as described above. For each $n$, let $\mathcal U_n$ denote the set of all those $U \in \mathcal U$ such that $|U \cap X_n| = 1$; and for each $U \in \mathcal U_n$, let $I_U$ denote the (unique) rational Euclidean-open interval such that $U \cap \bigcup_{i < n}X_i = I_U \cap \bigcup_{i < n}X_i$.

Fix $n \in \omega$ with $n > 0$. For each Euclidean-open interval $I \in \{ I_U :\, U \in \mathcal U_n \}$, choose some particular $U(I) \in \mathcal U_n$ such that $I_{U(I)} = I$. The set of all rational Euclidean-open $I$'s is countable, so the set of all such $U(I)$'s is a countable subset of $\mathcal U_n$. Let us denote this countable subset of $\mathcal U_n$ by $\mathcal V_n$. Now observe that $W_n = \bigcup \{ I(U) :\, U \in \mathcal V_n \}$ is a Euclidean-open subset of $\mathbb R$ containing $X_n$. Because $X_n$ is a Bernstein set, this means $\mathbb R \setminus W_n$ is countable. In particular, this implies that $\bigcup \mathcal V_n$ contains all but countably many points of $\bigcup_{i < n}X_i$.

Now unfix $n$. By the previous paragraph, $\bigcup_{n \in \omega \setminus \{0\}} \mathcal V_n$ is a countable subset of $\mathcal U$ that covers all but countably many points of our space. Adding in one more open set for each of the points not covered already by these sets, we obtain a countable subcover of $\mathcal U$.

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  • $\begingroup$ Very nice example. My congratulations! $\endgroup$ Commented Feb 22, 2022 at 18:52
  • $\begingroup$ I join in the congratulations, what a beautiful example. I had already accepted Taras' answer and I wouldn't like taking it from him (unless he deems it appropriate), but your answer definitely deserves an honorable mention. If I could accept both answers, I would. $\endgroup$
    – Peluso
    Commented Feb 23, 2022 at 16:24

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