# proof for one equivalent statement of Baire category theorem

First, let’s assume Baire’s category theorem and prove the alternative statement.
We have $B=\bigcup_{n=1}^{\infty}B_{n}$, with $\textrm{int}(\overline{B_{k}})=\emptyset\;\forall k\in\mathbf{N}$.
Then

 $X=X-\textrm{int}(\overline{B_{k}})=\overline{X-\overline{B_{k}}}\;\forall k\in% \mathbf{N}$

Then $X-\overline{B_{k}}$ is dense in $X$ for every $k$. Besides, $X-\overline{B_{k}}$ is open because $X$ is open and $\overline{B_{k}}$ closed. So, by Baire’s Category Theorem, we have that

 $\bigcap_{n=1}^{\infty}(X-\overline{B_{n}})=X-\bigcup_{n=1}^{\infty}\overline{B% _{n}}$

is dense in $X$. But $B\subset\bigcup_{n=1}^{\infty}\overline{B_{n}}\Longrightarrow X-\bigcup_{n=1}^% {\infty}\overline{B_{n}}\subset X-B$, and then $X=\overline{X-\bigcup_{n=1}^{\infty}\overline{B_{n}}}\subset\overline{X-B}=X-% \textrm{int}(B)\Longrightarrow\textrm{int}(B)=\emptyset$.

Now, let’s assume our alternative statement as the hypothesis, and let $(B_{k})_{k\in N}$ be a collection of open dense sets in a complete metric space $X$. Then $\textrm{int}(\overline{X-B_{k}})=\textrm{int}(X-\textrm{int}(B_{k}))=\textrm{% int}(X-B_{k})=X-\overline{B_{k}}=\emptyset$ and so $X-B_{k}$ is nowhere dense for every $k$.

Then $X-\overline{\bigcap_{n=1}^{\infty}B_{n}}=\textrm{int}(X-\bigcap_{n=1}^{\infty}% B_{n})=\textrm{int}(\bigcup_{n=1}^{\infty}X-B_{n})=\emptyset$ due to our hypothesis. Hence Baire’s category theorem holds.
QED

Title proof for one equivalent statement of Baire category theorem ProofForOneEquivalentStatementOfBaireCategoryTheorem 2013-03-22 14:04:52 2013-03-22 14:04:52 gumau (3545) gumau (3545) 8 gumau (3545) Proof msc 54E52