If $(X,\mathcal{\tau})$ is a topological space, a subset $A\subseteq (X,\mathcal{\tau})$ is said to have the property of Baire if it is expressible in the form $G\triangle M$ where $G\subseteq (X,\mathcal{\tau})$ is open and $M\subseteq (X,\mathcal{\tau})$ is meager. That is, $A\subseteq (X,\mathcal{\tau})$ has the property of Baire if there exists an open set $G\subseteq (X,\mathcal{\tau})$ such that $A\triangle G$ is meager.

Suppose now that $X=\{a,b\}$ is a discrete space so that $(X,\mathcal{\tau})=\{\emptyset,\{a\},\{b\},\{a,b\}\}$.

What does that concretely mean for the subset $A=\{a\}$ to have the property of Baire? I mean what could be the sets $G$ and $M$ is that particular case?

I know that, in that topology, the only one set which is meager is the empty set $\emptyset$. Does that mean that $G$ can only be $A$ itself (because $A\triangle A=\emptyset$), or could $G$ be something else?


The definition of meager is the countable union of nowhere dense sets. In a finite space, it just means a nowhere dense set. But the only nowhere dense set in a discrete finite space is $\varnothing$.

So yes, the only witness that $A$ has the Baire property is $A$ itself.

  • $\begingroup$ Thanks for the answer. Is there any possibility though that $G$ be something else, e.g. its boundary (since $\{a\}$ is open-closed)? $\endgroup$ – user60264 Jan 1 '15 at 17:15
  • $\begingroup$ Sorry, my comment was erroneous. I would rather think of $A=\text{cl}(A)$ and $G=\text{int}(A)$. Then $A\triangle G$ is the boundary of $A$ which is nowhere dense. $\endgroup$ – user60264 Jan 1 '15 at 17:25
  • $\begingroup$ If you want to understand it by constructing a nontrivial example, you either have to switch to an infinite set, or use non-discrete topologies on finite sets. $\endgroup$ – Asaf Karagila Jan 1 '15 at 17:48

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