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In many proofs (but not all) of the Baire Category theorem one requires on the n:th induction step that: $\overline{B(y_n,r_n)} \subset U_n \cap B(y_{n-1},r_{n-1})$ (where $\{ U_n \}_{n \geq}$ is a sequence of dense open sets). I wonder if this is a necessary step? Isn't it be enough to require: $B(y_n,r_n) \subset U_n \cap B(y_{n-1},r_{n-1})$ , and then show that the limit of $\{y_n\}$ belongs to $B(y_n,r_n) $?

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    $\begingroup$ If you don't require $\overline{B(y_n,r_n)} \subset U_n \cap B(y_{n-1},r_{n-1})$, how would you show that the limit doesn't lie in the boundary of the balls? $\endgroup$ – Daniel Fischer Apr 14 '18 at 13:50
  • $\begingroup$ Thanks for your reply! I'm not sure if I get this correctly. One wants to show that for an arbitrary $x$ there is a $y \in \cap U_n$ that also satisfies $y \in B(x,\epsilon)$. If one doesn't make the requirement above ($\overline{B(y_n,r_n} \subset B(y_{n-1},r_{n-1})\cap U_n$) then the limit of $\{y_n\}$ might end up on the boundary of $B(x,\epsilon)$ so one hasn't shown the density? $\endgroup$ – user202542 Apr 14 '18 at 14:34
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    $\begingroup$ It may end up on the boundary of $B(y_n,r_n)$, and that need not lie in $U_n$, it could lie in $\partial U_n$.. $\endgroup$ – Daniel Fischer Apr 14 '18 at 14:41
  • $\begingroup$ Ok, thanks! I was looking at Kreyszig's Functional Analysis where he proves the version that a complete metric space can't be written as a countable intersection of nowhere dense sets, and there he doesn't make the requirement above. Do you know why it works in that case? $\endgroup$ – user202542 Apr 14 '18 at 14:48
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    $\begingroup$ I don't know Kreyszig's proof. It may be that the $y_n$ and $r_n$ are chosen in such a way that the desired conclusion follows. It's also possible that it's an oversight and the proof as written isn't correct. $\endgroup$ – Daniel Fischer Apr 14 '18 at 15:02
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The reason (probably, I don't know what text you're using) is because to get the non-empty intersection, Cantor's theorem can be applied: if $C_n$ is a sequence of decreasing closed sets such that $\operatorname{diam}(C_n) \to 0$ then in a complete metric space $\cap_n C_n \neq \emptyset$. The $\overline{B(y_n, r_n)}$ can be used as $C_n$ provided the $r_n$ decrease to $0$, as $\operatorname{diam}(C_n) \le 2r_n$ in this case.

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