# What can I say about $A^{c}$ if $A$ is not open

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 $$(i)$$?

$$(i) \Rightarrow (ii)$$. 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.
• @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. – Henno Brandsma May 11 at 11:14