# On the proof of Baire category theorem

I would like to ask about the proof of Baire Category theorem found on wolfram. The excerpt is as below:

Baire's category theorem, also known as Baire's theorem and the category theorem, is a result in analysis and set theory which roughly states that in certain spaces, the intersection of any countable collection of "large" sets remains "large." The appearance of "category" in the name refers to the interplay of the theorem with the notions of sets of first and second category.

Precisely stated, the theorem says that if a space S is either a complete metric space or a locally compact T2-space, then the intersection of every countable collection of dense open subsets of S is necessarily dense in S.

The above-mentioned interplay with first and second category sets can be summarized by a single corollary, namely that spaces S that are either complete metric spaces or locally compact Hausdorff spaces are of second category in themselves. To see that this follows from the above-stated theorem, let S be either a complete metric space or a locally compact Hausdorff space and note that if $\{E_i\}=\{E_i\}_{(i\in N)}$ is a countable collection of nowhere dense subsets of S and if $V_i$ denotes the complement in S of the closure $\bar{E_i}$ of $E_i$, then each set $V_i$ is necessarily dense in S. Because of the theorem, it follows that the intersection of all the sets $V_i$ must be nonempty (and indeed must be dense in S), thereby proving that S cannot be written as the union of the sets $E_i$. In particular, such spaces S cannot be written as the countable union of sets which are nowhere dense in themselves and are therefore second category sets relative to themselves.

1. My question would be the sentence in Boldface. Why is it that when the countable intersection of $V_i$ is not empty and dense, we can infer that S cannot be written as a countable union of nowhere dense set?

2. Also, why is the assumption of locally compact and Hausdorff needed as it is not used in the proof at all.

Thanks!

• The sentence Because of the theorem involves the hypotheses of the thm (complete metric or compact Haussdorff). For the bold sentence, suppose $S$ could be written that way and take complements. – Jesus RS Mar 26 '15 at 9:54

$$\bigcup \overline{E_i}=\bigcup (X\setminus V_i)=X\setminus\bigcap V_i\neq X$$
And since $\bigcup E_i\subseteq\bigcup\overline{E_i}$, if the above union is not everything, there's no way for the union of the $E_i$ to be everything.