For closed $(S_n)_{n\in\mathbb{N}}$ in $\mathbb{R}^n$ and $S:=\bigcup_{n\in\mathbb{N}}S_n$, if $S°\neq0 $ then $S_n°\neq0$ for some $n \in \mathbb{N}$

Question

For a sequence $(S_n)_{n\in \mathbb{N}}$ of closed sets in $\mathbb{R}^n$ with the euclidian metric and $S:=\bigcup\limits_{n\in \mathbb{N}} S_n$, if $S° \neq 0 $ then $S_n° \neq 0$ for some $n \in \mathbb{N}$. (This is supposedly a version of the so-called "Baire category theorem" on $\mathbb{R}^n$.)

I am supposed to show this using a proof by contradiction, by assuming that all $S_n°$ are empty. The hint is given that then you can find a sequence of nested closed balls $B_{n+1} \subset B_n \subset S$ with $B_n \cap S_n = \emptyset$. I can't see how you arrive at this. All I can see is that when you assume $S°$ to be nonempty, then $S\backslash S°$ is closed (because its complement to $S$ is $S°$ and open). But how can I construct this sequence of closed balls?

Answer 1

If I recall correctly, this is a formulation of Baire's theorem. Here's a sketch of the proof: suppose the contrary, that is $(S_n)° = \emptyset$ for all $n$. Since $S^° \neq \emptyset$, we can take $U \subseteq S$ open. Now, $S_1$ has no interior, so it can't be that $U \subseteq S_1$, and in particular, $U \setminus S_1$ is open and not empty: it therefore contains an open set $U_1$, and we can take a closed ball $B_1 \subseteq U_1$ inside it. Again, since the interior of $S_2$ is empty, it can't be that $(B_1)° \subseteq S_2$, and in particular, $(B_1)° \setminus S_2$ is open and non empty, hence it contains an closed ball $B_3$, and so on.

This process gives a sequences of decreasing closed balls $(B_n)_{n\in\mathbb{N}}$, with

$$ B_i \subseteq (S_i)^c, \ B_i \subseteq S \ (\forall i \in \mathbb{N}) $$

Moreover, in this process we could have replaced each ball for one contained in it, of a smaller radius, such that $diam(B_n) \to 0$. Therefore, for this choice of closed balls, $\cap_{n\geq1} B_n$ is not empty. But this is absurd: if $x \in \cap_{n\geq1} B_n$, $x$ is in $S = \cup_{n \geq 1}S_n$ but at the same time it is not in any $S_i$, precisely because $B_i \subseteq (S_i)^c$.

Answer 2

This is a version of the Baire category theorem, whose standard proof goes as follows. Since $S$ has nonempty interior, find an open ball included in $S$. Shrink that ball to a closed ball $B_0$, with the same center.

If the interior of $B_0$ is entirely included in $S_0$, then that means $S_0$ has nonempty interior, and so you're done. If, on the other hand, there is a point $p_0$ in the interior of $B_0$ but outside $S_0$, then a whole open ball around that point is $\subseteq (B_0)^\circ-S_0$, since $S_0$ is closed. Choose such an open ball and shrink it to a closed ball $B_1$.

Now repeat the preceding paragraph with all subscripts increased by $1$ (i.e., work with $B_1$ and $S_1$ to either get what you want, nonempty interior of $S_1$, or get $p_1$, etc.) ending with a closed ball $B_2\subseteq (B_1)^\circ-S_1$.

Continue inductively. If, at any stage, there is no $p_n\in(B_n)^\circ-S_n$, then you're done. Otherwise, the process continues forever and you get a decreasing sequence of closed balls. There's a point in the intersection of all these closed balls. It's not in any $S_n$ (because $B_{n+1}$ was chosen to be disjoint from $S_n$), but it's in $S$ (because $B_0\subseteq S$). That contradicts the fact that $S=\bigcup_nS_n$.