In the proof of the Baire category theorem(for non-empty Banach Spaces), I cannot understand the following

Baire Category Theorem: A non-empty Banach Space cannot be a countable union of nowhere dense sets

The proof is trying to show by contradiction that a Banach space $X$ cannot be written as a countable union of nowhere dense sets(rare sets). The proof begins by assuming that suppose $\exists$ a sequence $ (A_n)_{n\in \mathbb{N}}$ of nowhere dense sets such that $$X=\bigcup_{n \in \mathbb{N}} A_n$$ The proof says let $x_1 \notin A_1$ and $B_{r_1}(x_1)$ be a ball such that $B_{r_1}(x_1)\cap \overline{A_1}={\phi}$ and $r_1<1$

I am stuck on the first step. Why can we choose such a $r_1$ and why does such an open ball exist. Does a non-empty Banach space always contain an open set ? Could someone explain the fist part of the proof?

  • $\begingroup$ $A_1$ is nowhere dense. Hence $\overline{A_1}$ has emtpy interior. In particular, $X\setminus \overline{A_1} \neq \varnothing$, since $\overset{\Large\circ}{X} = X$. $\endgroup$ – Daniel Fischer Jun 20 '15 at 14:30
  • $\begingroup$ @DanielFischer What do you mean by the dot over $X$? I havent seen this notation $\endgroup$ – user3503589 Jun 20 '15 at 14:33
  • $\begingroup$ It means the interior, dual to how $\overline A$ means closure. $\endgroup$ – Arthur Jun 20 '15 at 14:34
  • $\begingroup$ The interior of a set. Is $\operatorname{int} X$ the notation you know? $\endgroup$ – Daniel Fischer Jun 20 '15 at 14:34
  • 1
    $\begingroup$ In every topological space $T$, the set $T$ is both, open and closed. So in a non-empty topological space, a nowhere dense set is never dense in the whole space, for the whole space has non-empty interior (the whole space). $\endgroup$ – Daniel Fischer Jun 20 '15 at 15:04

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