Let $X$ be some topological space:

I want to prove that

$i)\Rightarrow ii)$ where:

$i)$ Let $A \neq \varnothing$ and $ A\subset X$ open, then $A$ is non-meagre.

$ii)$ Let $A\subset X$ be meagre, then $X \setminus A$ is dense in $X$.

My idea:

Let $A$ be meagre, by definition there is a $(A_{n})_{n} \subset X$ so that $A_{n}$ is nowhere dense $\forall n \in \mathbb N$, i.e. $(\overline{A_{n}})^{c}$ is dense $\forall n \in \mathbb N$ and of course $\bigcup_{n \in \mathbb N}A_{n}=A$

Note that $(\overline{A_{n}})^{c}=X\setminus \overline{A_{n}}$ and I also am aware that $X\setminus A=\bigcap_{n\in \mathbb N}X\setminus \overline{A_{n}}$. Note that $X\setminus \overline{A_{n}}$ by definition open as $\overline{A_{n}}$ is closed. But, somewhere I will need to use assumption $(i)$ at some stage. Since $A$ is meagre, by $(i)$ it follows that $A$ is not open.

What follows from the fact that $A$ is not open. I am sure $A$ is not necessarily closed (e.g. clopen sets). Any ideas of how I can use $(i)$?


This direction is quite simple, you jst ahve to be systematic in what you're showing and what you know:

$(i) \Rightarrow (ii)$. So we assume $(i)$ holds. To show $(ii)$, let $A$ be meagre. Let $O$ be an arbitrary non-empty and open set. We need to show it intersects $X\setminus A$ (to see that the latter set is dense).

If $O \cap (X\setminus A) = \emptyset$, then $O \subseteq A$ and as $A$ is meagre, $O$ is meagre as a subset of a meagre set. This contradicts $(i)$ so the assumption that $O \cap (X\setminus A) = \emptyset$ was false and so $O \cap (X\setminus A) \neq \emptyset$ and so $X \setminus A$ is dense (as $O$ was arbitrary) and $(ii)$ has been shown.

The reverse implication is quite similar and also easy: assume $(ii)$ and suppose $A$ is open non-empty and also meagre. Then by $(ii)$ $X\setminus A$ is dense but it's a proper closed set and so not dense at all. So $A$ cannot be meagre and all non-empty open sets are non-meagre and $(i)$ holds.

  • $\begingroup$ Thank you, but I still do not know what can be said about a set that is NOT open. It is not necessarily closed, is it? $\endgroup$ – SABOY May 11 at 10:34
  • 1
    $\begingroup$ @SABOY NO, you cannot say that: sets are not "doors". We only know that when $A$ is not open, $X\setminus A$ is not closed. $\endgroup$ – Henno Brandsma May 11 at 11:14

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