While reading the section about Baire Spaces in the book "Espaços Métricos" by Elon Lages Limas I found the following statement.

Elon Lages Lima "Espaços Métricos"

I shall translate from portuguese after giving some context.

The author motivates the definition of a meagre set as something analogous to the notion of a set of measure zero. Therefore he refers to it as "insignificant" (in a topological sense of course). Firstly he points out that they should be preserved when one takes subsets and countable unions.

Then he gives an example to show it wouldn't suffice to require such a set to have empty interior since $\text{int}( \mathbb{Q})=\emptyset$ and $\text{int}( \mathbb{R} \setminus \mathbb{Q})=\emptyset$ but $\mathbb{Q} \cup \mathbb{R} \setminus \mathbb{Q}=\mathbb{R}$.

Translation: A better idea would be to consider a set $X\subset M$ whose closure $\overline X$ has empty interior in $M$. In fact, if $\text {int}( \overline X)=\emptyset $ and $\text {int}( \overline Y)=\emptyset $ then $$\text {int}( \overline {X\cup Y})=\text {int}( \overline X \cup \overline Y) (\color {red}\subset) \supset \text {int}( \overline X) \cup \text {int}( \overline Y)=\emptyset$$

I highlighted in red color where I think the author made a mistake.

The author then says that it is not true that if $\text {int}( \overline {X_n})=\emptyset$ for all $n \in \mathbb {N}$ then $X=\cup X_n$ still has the property that $\text {int}( \overline X)=\emptyset$. (Simply take the rationals as an example). Actually, if what I quoted above was right it would violate this statement and that's why this caught my eye.

My question if indeed I stumbled upon a mistake in the book. Also, I was enjoying the way the author was motivation the definition and I don't see why, intuitively (using the idea of "insignificant" topological set), we require that the property holds when we take countable unions. Why not uncountable?

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    $\begingroup$ You can find more information about the subject in chapter 6 of Bachman & Narici's Functional Analysis. $\endgroup$ Dec 3, 2017 at 18:14

2 Answers 2


Let $X$ and $Y$ be subsets of a space $S$ with int$(\bar X)=\emptyset=\text{int}(\bar Y).$ If $T$ is any non-empty open set of $S$ then $T\not \subset \bar X$ so $T^*=T\cap (S$ \ $\bar X)$ is a not empty . But since $T^*$ is a non-empty open set and $\text{int}(\bar Y)=\emptyset,$ we have $T^* \not \subset \bar Y,$ so $$\emptyset \ne T^*\cap (S \setminus \bar Y)= T \setminus (\bar X \cup \bar Y) =T \setminus \overline {X \cup Y}.$$ So $T\not \subset \overline {X\cup Y}.$ Therefore $\text{int}(\bar X \cup \bar Y)= \text{int}(\overline {X\cup Y})=\emptyset.$

I think the mistake in the book is a "typo". But after correcting it, there are no details (in the book) about why $\text{int}(\bar A \cup \bar B)\subset \text{int}(\bar A) \cup \text{int}(\bar B).$


Yes, you are right: it should be $\supset$ instead of $\subset$. As a result, his proof of the fact that the union of two meagre sets is again meagre is wrong (the statement is true, however).

Note that what you are calling “meagre set” here is not the usual definition of such sets. The sets that you're dealing with here are usually called “nowhere dense sets”.

The reason why we don't consider arbitrary unions is because any set is the union of its points and, usually, singletons are meagre sets.

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    $\begingroup$ I mentioned meagre sets since nowhere dense sets are preliminary for its definition. Now I understood the intention of the author... He wanted to show that the finite union of nowhere dense sets are nowhere dense, whereas this fails for countable unions. $\endgroup$
    – aadcg
    Dec 3, 2017 at 17:53

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