Artificial intelligent assistant

The only meagre and open set is the empty set? Suppose that A is a meagre set in a topological space X. I think that the answer of my question is yes, because $\mathring {A}$ is empty. Am I right?

This is not true in general. For example, in $\mathbb{Q}$ with its usual topology, for any set $U\subseteq \mathbb{Q}$, we have $U = \bigcup_{q\in U} \\{q\\}$. This is a countable union, and each singleton $\\{q\\}$ is nowhere dense, so $U$ is meager. Thus _every_ open set is meager.

Of course, in a Baire space (a space where the conclusion of the Baire category theorem holds, e.g. any locally compact Hausdorff space or any complete metric space), every meager set has empty interior, so the statement is true.

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