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The Baire Category Theorem - Let $X$ be a complete metric space

a.) If $\{U_n\}_1^\infty$ is a sequence of open dense subsets of $X$, then $\bigcap_1^\infty U_n$ is dense in $X$.

b.) $X$ is not a countable union of nowhere dense sets.

The name for this theorem comes from Baire's terminology for sets: If $X$ is a topological space, a set $E\subset X$ if of the first category, or meager, according to Baire, if $E$ is a countable union of nowhere dense sets; otherwise $E$ is of the second category.

Problem 5.3.27 from Folland's Real Analysis: There exist meager subsets of $\mathbb{R}$ whose complements have Lebesgue measure zero

Attempted proof: Let $X$ be a topological space in $\mathbb{R}$ and $E\subset X$ be of the first category (meager). Set $\{x_k\}$ to be an enumeration of the rational numbers, let $$E_n = \bigcup_{k=1}^{\infty} \left(x_k - \frac{1}{2^{k-1} n},x_k + \frac{1}{2^{k-1}n}\right)$$ Consider the set $E = \cap_{1}^{\infty} E_n$, then $$m(E_n) \leq \sum_{k=1}^{\infty}\frac{1}{2^k n} = \frac{1}{n}$$ hence $m(E) = 0$

I am not sure if I am right, any suggestions is greatly appreciated.

Eugene Zhang
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Wolfy
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    I'm not sure about your first sentence. You don't seem to need $X$ for anything (and what does "let $X$ be a topological space in $\mathbb R$" mean exactly?), and you later construct $E$. I think you could get rid of the first sentence entirely and the proof will be correct. You might add a sentence to explain why the complement of $E$ is meager. –  Mar 20 '16 at 00:31
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    Is your computation of $m(E_n)$ correct? As I understad we are bounding as follows: $$ m(E_n) \leq \sum_k m\left(x_k - \frac{1}{2^{k - 1}n}, x_k + \frac{1}{2^{k - 1}n} \right) = \sum_k \frac{2}{2^{k - 1}n} = \frac 2n (1 + \frac 12 + \frac 14 + \ldots) = 4/n. $$ Am I missing something? Of course, that doesn't change the conclusion. – Danilo Gregorin Afonso Jul 26 '20 at 15:00

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Yes, your proof is right. You need to prove that $E^c$ is meager.

Since $E_n$ is open, $E_n^c$ is closed. $E_n$ is dense in $\Bbb{R}$ for $\Bbb{Q}\subset E_n$. Thus $E_n^c$ is nowhere dense. So $E^c=\bigcup_{n=1}^{\infty}E_n^c$ is meager.

Eugene Zhang
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