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In this page: https://en.wikipedia.org/wiki/Meagre_set

The complement of a meagre set is a comeagre set or residual set.

I am asking if a comeagre set is still closed in the (usually larger) topological space. If no, is there is any examples of that property.

Safwane
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  • Hardly ever (if ever). –  Jan 15 '20 at 15:53
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    It might help to realize that $\mathbb R$ can be expressed as the disjoint union of a meager set and a Lebesgue measure zero set (see here), and thus a co-meager set can have Lebesgue measure zero (equivalently, a meager set can be such that its intersection with every open interval has Lebesgue measure equal to the length of that interval). In fact, in $\mathbb R$ a co-meager set can be MUCH smaller, having Hausdorff dimension zero (see here for stronger results in more general spaces). – Dave L. Renfro Jan 15 '20 at 16:39

2 Answers2

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The topological spaces $X$ where the only closed comeagre subset is $X$ itself are called Baire spaces. It is a very large class of spaces which includes all locally compact Hausdorff spaces and all complete metric spaces.

On the other hand, in $\Bbb Q$ all subsets are (co)meagre.

Added: Also, notice that a topological space has some non-closed comeagre subset if and only if it isn't discrete. In fact, let $x\in X$ be a non-open point. Then $X\setminus\{x\}$ is comeagre and non-closed. On the other hand, if $X$ is discrete then $X$ is the only comeagre subset of $X$.

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No. For instance, $\mathbb R\setminus\mathbb Q$ is comeagre in $\mathbb R$, but it is not a closed subset.

José Carlos Santos
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