Can we prove the existence of such set $F$ of separable space $E$ by contradiction?

Question

Let $E$ be complete and there are no isolated points in $E$. I work on a question to show that there exists a subset $F$ satisfying two properties (1) the $F^{\circ}$ is empty and (2) $F$ is not meagre.

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I would like to check my proof. My proof is as follows.

Update: I follow the comments and change my proof.

There is a countable open base $\{U_n\}_{n=1}^{\infty}$. We take $x_1\in U_1$ and construct the set of $\{x_1,x_2,\dots\}$ inductively as follows. We take $x_2\in U_2\setminus span\{x_1\}$, $x_3\in U_3\setminus span\{x_1,x_2\}$, $\dots$, $x_n\in U_n\setminus span\{x_1,\dots, x_{n-1}\}$, $\dots$.

Set $A=\{x_1,x_2,\dots \}=\cup_{i=1}^{\infty}\{x_i\}$. Then I need to show that $(A^c)^{\circ}=\emptyset$ and is not meagre.

But I am not sure if we can say

(1) Since every open sets intersects $A$, then $A$ is dense. Thus, $A^c$ is nowhere dense. Then $int(\bar{A^c})=\emptyset$.

(2) Can we say $A^c$ is not meagre?

Answer 1

So a Banach space is completely metrizable. Since we are assuming it also separable, it is Polish (which I mention only for reference as there is a lot of literature on the topic of Polish spaces; e.g. descriptive set theory). As mentioned in the comments, in (1), if $F$ is assumed to have empty interior, then the only open set $U$ for which $U \subseteq F$ is $U = \emptyset$, so it can't reach the intended contradiction with the union of the $B$s.

Another edit: As pointed out in the comments, the approach below doesn't work, in general. I do believe it to work if you assume, additionally, that $X^\ast$ is infinite. (whoops, again) MW's answer contains the right condition.

Edit: In fact, the contradiction approach doesn't seem to be much more helpful since the contradictory hypothesis is of the form "for every subset $F$, $F$ is either $P_1$ or $P_2$ where $P_1$ and $P_2$ are the stated properties. However sets having both of those properties exist all the time. For example, any closed nowhere dense set has non-empty interior is of first category. As long as we have a space that is not meager in itself (true of any complete metrizable space), then the entire space is an example of a subset which has non-empty interior and is of second category. So the contradictory approach somehow needs to go against the "for all" which boils down to just needing to produce a set that contradicts the assumption, which was the original task.

At any rate, I think a more direct approach works for completely metrizable spaces that are separable and aren't just a countable set of isolated points, generally. Consider a completely metrizable and separable space $X$ that is not a countable union of isolated points. Then $X^\ast$ defined to be $X$ without isolated points is still a Baire space since a closed subspace of a completely metrizable space is completely metrizable. Also by metrizability, separability is hereditary, so we can consider a countable dense set $D = \{ x_n : n \in \mathbb N \} \subseteq X^\ast$. Note that, for each $n \in \mathbb N$, $U_n := X^\ast \setminus \{ x_n \}$ is open and dense in $X^\ast$ since $X^\ast$ has no isolated points. Then consider $F := \bigcap_{n\in\mathbb N} U_n$. As mentioned above, $X^\ast$ satisfies the Baire Category Theorem. Hence, not only is $F$ dense, but it cannot be of first category. Moreover, $F$ has empty interior since it misses a dense subset of $X^\ast$.

Answer 2

[Edited to remove some incorrect statements.]

This fact does not require any facts about the linear structure of the space. The fact that $E$ is complete and there are no isolated points in $E$ is more than enough, since this means every countable subset is meagre (i.e., first category), since the singletons which comprise it are nowhere dense.

That is, if $D$ is countable and dense in $E$, then $F:=E\backslash D$ has no interior, but must be nonmeagre, since otherwise $E=F\cup D$ would be a union of two meagre sets, and would hence be meagre, a contradiction to the Baire Category theorem.

Update

Regarding your proof, the logic of the method isn't exactly wrong (to complete the proof you would invoke Baire Category as in the previous argument to argue that $A^c$ cannot be meager, as then $E=A\cup A^c$ would be meager), however, there was no need to go the roundabout route of constructing your own countable dense subset when you already have one by definition of separability.

Remark

The condition that there are no isolated points in $E$ is stronger than we need - a necessary and sufficient condition for the statement to hold in a complete separable metric space $E$ is that there is some nonempty open $U\subseteq E$ containing no isolated points.

To see sufficiency, note that in such a space, we can pick such a $U$ and let $D$ be our countable dense set, then $D\cap U$ is meagre and and $F:=U\backslash D$ must be nonmeagre, since by Baire category a nonempty open subset of a complete space must be nonmeagre.

To see necessity, note that if every open set contains an isolated point, then the set $U$ of isolated points is a dense open set in $E$, and any $F\subseteq E$ that has empty interior must not intersect $U$, hence its closure must be contained in $E\backslash U$, which is nowhere dense, hence $F$ is meagre.