A topological space $X$ is concentrated on a set $D$ iff for any open set $G$ if $D\subseteq G$, then $X\setminus G$ is countable.


What is an example of a separable metrizable (uncountable) meager (meaning a countable union of nowhere dense subsets, also known as a set of first category) space $X$ such that $X$ is concentrated on countable dense set?



What is an example of a meager space X such that X is concentrated on countable dense set?
**ADDED LATER**


The answer to your question is that there is such a space $X$ if and only if $\mathfrak{b} = \aleph\_1$.


*If $\mathfrak{b} = \aleph\_1$, then there is such a space.*


To see this, first note that $\mathfrak{b} = \aleph\_1$ if and only if there is an uncountable subset of the irrationals concentrated on $\mathbb Q$. This is proved by van Douwen in section 10 of his article in the *Handbook of Set Theoretic Topology*. By modifying his argument slightly, we can show that if $\mathfrak{b} = \aleph\_1$ then there is an uncountable subset of the Cantor space concentrated on a countable subset of the Cantor space.


[The proof of this goes as follows. If $\mathfrak{b} = \aleph\_1$, then one can construct via transfinite recursion a length-$\omega\_1$ sequence $\langle f\_\alpha :\, \alpha < \omega\_1 \rangle$ of functions that is unbounded with respect to $\leq^\*$, but also with the property that $\alpha < \beta$ implies $f\_\alpha \leq^\* f\_\beta$. This means that a subset of this sequence is $\leq^\*$-bounded if and only if it is countable. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. It is also fairly well known that the irrationals are homeomorphic to $C \setminus D$, where $C$ is the Cantor space and $D$ is some countable relatively dense subset of $C$. Let $Y$ be the image of your sequence of functions under some homeomorphism $\omega^\omega \rightarrow C \setminus D$. If $U$ is an open set containing $D$, then $C \setminus U$ is a compact subset of $C$. It's not too difficult to show that every compact subset of $\omega^\omega$ is $\leq^\*$-bounded above by some function (or even more -- it is $\leq$-bounded), and this means $C \setminus U$ contains only countably many points of $Y$.]


Let $Y$ be, as in the previous paragraph, an $\aleph\_1$-sized subset of the Cantor space $C$ that concentrates on a countable $D \subseteq C$. Let $X = Y \cup \mathbb Q$. As a subspace of the reals, this set $X$ meets all your requirements.


*A proof that if there is a space $X$ as described in your post then $\mathfrak{b} = \aleph\_1$.*


Suppose there is an uncountable separable metrizable space $X$ that is concentrated on a countable set $D \subseteq X$. Let $\{ d\_0, d\_1, d\_2, \dots \}$ be an enumeration of $D$. For each $x \in X \setminus D$, define a function $f\_x : \omega \rightarrow \omega$ by setting $f\_x(n) = \min\{ k :\, \mathrm{dist}(x,d\_n) > \frac{1}{k} \}$.


Let $Y$ be a subset of $X \setminus D$ with $|Y| = \aleph\_1$. The set of functions $\{ f\_x :\, x \in Y \}$ is an $\aleph\_1$-sized subset of $\omega^\omega$, and I claim it is unbounded with respect to $\leq^\*$. In other words, I claim this set of functions witnesses $\mathfrak{b} = \aleph\_1$.


To see this, suppose instead there is some $g \in \omega^\omega$ such that $f\_x \leq^\* g$ for all $x \in Y$. By a pigeonhole argument, there is some function $h \in \omega^\omega$, differing from $g$ in only finitely many places, such that $f\_x \leq h$ for uncountably many $x \in Y$. Let $U = \bigcup\_{n \in \omega} B\_{1/h(n)}(d\_n)$. This is an open set containing $D$, and our definition of the $f\_x$'s ensures that $x \notin U$ if and only if $f\_x \leq h$. Thus there are uncountable many $x$'s with $x \notin U$, contradicting the fact that $Y$ concentrates on $D$.


Let me point out that this argument is essentially contained in an earlier MO post by Taras Banakh found [here](https://mathoverflow.net/questions/416331/example-of-an-uncountable-scattered-space-with-some-properties). (A version of the argument seems to be in van Douwen's article too, and I don't know whether it really originated there either, but anyway I first learned it from Taras' post.)


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**ORIGINAL POST**


Here is a consistent example of a subspace of $\mathbb R$ with these properties.


First, there is an uncountable subset $X$ of $\mathbb R$ concentrating on $\mathbb Q$ if and only if $\mathfrak{b} = \aleph\_1$. This is proved in chapter 10 of van Douwen's article in the *Handbook of Set-Theoretic Topology*. Note that $X$ concentrates on $\mathbb Q$ if and only if every subset of $X$ does, so if $\mathfrak{b} = \aleph\_1$ then there is an $\aleph\_1$-sized set of reals concentrating on $\mathbb Q$.


Now suppose $\aleph\_1 = \mathfrak{b} < \mathrm{non}(\mathcal M)$ (the least size of a non-meager subset of $\mathbb R$). This situation is consistent -- it happens in the random real model, for example. In such a model, let $X$ be an $\aleph\_1$-sized set concentrating on $\mathbb Q$. Adding countably many points to $X$ if necessary, we may (and do) assume $X$ is dense in $\mathbb R$. Then (viewed as a subspace of $\mathbb R$) it meets all your requirements.
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*A sketch of an argument as to why $\mathfrak{b} = \aleph\_1$ implies there is an $\aleph\_1$-sized set concentrating on $\mathbb Q$*:


Since you may not have access to van Douwen's article, here is a sketch of the idea I quoted above. If $\mathfrak{b} = \aleph\_1$, then one can construct via transfinite recursion a length-$\omega\_1$ sequence $\langle f\_\alpha :\, \alpha < \omega\_1 \rangle$ of functions that is unbounded with respect to $\leq^\*$, but also has the property that $\alpha < \beta$ implies $f\_\alpha \leq^\* f\_\beta$. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. Let $X$ be the image of your sequence of functions under any such homeomorphism. If $U$ is an open set containing $\mathbb Q$, then $\mathbb R \setminus U$ is a $\sigma$-compact subset of the irrationals. It's not too difficult to show that every $\sigma$-compact subset of $\omega^\omega$ is $\leq^\*$-bounded above by some function, and this means $\mathbb R \setminus U$ can only contain countably many points of $X$.


