In my studies of functional analysis I have come across this two statements of the Baire Category Theorem.
- Let $(X,d)$ be a complete metric space. Let $U_n$ be a dense, open set for each $n \in \Bbb N$. Then $\bigcap_{n=1}^\infty U_n$ is dense.
And
- Let $(X.d)$ be a complete metric space. $A_0, A_1,..$ are closed subsets. If $\bigcup_{n=1}^\infty A_n $ contains an open ball, then at least one of the $A_n$ contains an open ball.
Now I am struggling with how to show $1.\Rightarrow 2.$ ,
unfortunately our professor didn't proved this one and we can't use the assertion:
Let $(X,d)$ be a complete metric space. Let $U_n$ be a closed subset, without interior point for each $n \in \Bbb N$. Then $\bigcup_{n=1}^\infty U_n$ has no interior point.
Which is equivalent to $1.$ and $2.$
Maybe it's a silly question or it's obvious to see, but thanks in advance!!
Best Answer
If $\bigcup_{n=1}^\infty A_n$ contains an open ball, then its interior is not empty. On the other hand,$$\left(\bigcup_{n=1}^\infty A_n\right)^\complement=\bigcap_{n=1}^\infty A_n^\complement,$$which is in intersection of a sequence of open sets. If no $A_n$ contains an open ball, then every $A_n$ has empty interior and therefore every $A_n^\complement$ is dense. But then $\bigcap_{n=1}^\infty A_n^\complement$ is dense, which is impossible, since $\bigcup_{n=1}^\infty A_n$ contains an open ball.