Equivalent formulations of Baire's Lemma / Theorem

Question

I use the following notation $int_Z(A)$ for the interior of $A \subset Z$ in a metric space $Z$.

Now I have the following problem understanding a proof:

Let $(M,d)$ be complete. Then we have that statement i) implies ii).

i) If $M= \bigcup_{j=0}^{\infty}A_j$ with $A_j$ all closed, then there exists a $j_0$ with $int_M(A_{j_0}) \neq \emptyset$.

ii) If $int_M(\bigcup_{j=0}^{\infty}A_j) \neq \emptyset$ with closed $A_j$'s, then there exist a $j_0$ with $int_M(A_{j_0}) \neq \emptyset$

I came as close to understanding the proof as this:

Define the metric space $(X,d)=(\overline{B_{r}(x)},d)$ where $B_{2r}(x) \subset int(\bigcup_{j=0}^{\infty}A_j) \subset \bigcup_{j=0}^{\infty}A_j$ (since it has non-empty interior). Then by using (i) on the sets $X\cap A_j$ we get a $j_0$ with $int_{X}(X\cap A_{j_0})\neq \emptyset$.

Now how do I show that I have $int_M(A_{j_0}) \neq \emptyset$, if I only know that $int_{X}(X\cap A_{j_0})\neq \emptyset$?

Thanks a lot in advance!

Answer 1

Let $p\in int_X(X\cap A_{j_0})=S\cap X$ where $S$ is open in $M$. Let $b>0$ such that $B_b(p)\subset S.$

Since $p\in X=\overline {B_r(x)}$ we may take some $q\in B_b(p)\cap B_r(x).$

Let $c>0$ be small enough that $B_c(q)\subset B_b(p)$ and $B_c(q)\subset B_r(x).$

Then $B_c(q)\subset B_b(p)\subset S$ and $B_c(q)\subset B_r(x)\subset X.$ Therefore $B_c(q)\subset S\cap X=int_X(X\cap A_{j_0}).$

Addendum : More simply: In any space $M,$ if $S$ is open in $M$ and $X=\overline Y$, then $S\cap X\ne \phi \implies S\cap Y\ne \phi.$..... So in the Q, with $Y=B_r(X)$ we have $\phi\ne S\cap Y\subset S\cap X=int_X(X\cap A_{j_0}),$ and $S\cap Y$ is open in $M.$

Answer 2

I suppose that $X$ is the closed ball $\overline{B(x,r)}$, let $Y=\overline{B(x,r/2)}$, if $Y\cap A_{j_0}$ has a non empty interior considered as a subspace of the metric space $\overline{B(x,r/2)}$, it implies there exists an open ball $B(x',r')$ of $Y$ (that you can contained also in the interior of $X$ since if $y\in B(x',r')\subset \overline{B(x,r/2)}, d(x,y)\leq r/2<r$ ) contained in the interior $ Y\cap A_{j_0}\subset B(x,r)\cap A_{j_0}$, but an open subset of the interior of $X$ is an open subset of $M$ since it is the intersection $U\cap int(X)$ where $U$ is open in $M$.