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I was reviewing the proof of Baire's theorem I saw in class a few days ago, and there's an assumption that I didn't managed to see where it was used. I'll put here the proof given.

Theorem (Baire) : Let $(X,\tau)$ be a compact Hausdorff space. Then given a sequence $\{\Omega_n\}_{n \geq 0}$ of open dense sets, $\bigcap_{n \geq 1}\Omega_n$ is dense.

Proof: Without loss of generality, we can suppose that $\Omega_n \supseteq \Omega_{n+1}$ for all $n \geq 0$. $\color{blue}{(\ast)}$

Let $W \in \tau \setminus\{\varnothing\}$. We must show that $W \cap \left(\bigcap_{n \geq 1}\Omega_n\right)\neq \varnothing$. Recall that every compact Hausdorff space is $T_3$.

We have that $W \cap \Omega_0$ is a non-empty open set. So take $x_0 \in W \cap \Omega_0$. Then exists $V_0 \in \tau$ such that: $$x_0 \in V_0 \subseteq \overline{V_0}\subseteq W \cap \Omega_0.$$

Now $V_0 \cap \Omega_1$ is again a non-empty open set. So take $x_1 \in V_0 \cap \Omega_1$. Then exists $V_1 \in \tau$ such that: $$x_1 \in V_1 \subseteq \overline{V_1}\subseteq V_0 \cap \Omega_1.$$

Now $V_1 \cap \Omega_2$ is again a non-empty open set. Proceeding we get points $\{x_n\}_{n \geq 0}$ and open sets $\{V_n\}_{n \geq 0}$ such that: $$x_n \in V_n \subseteq\overline{V_n} \subseteq V_{n-1}\cap \Omega_n, \quad \forall\,n \geq 1.$$

By compactness of $(X,\tau)$, $\bigcap_{n \geq 0}\overline{V_n} \neq \varnothing$. Take $b$ in this intersection. So $b \in \overline{V_n}$ for all $n$. So: $$\overline{V_0}\subset W \cap \Omega_0 \implies b \in W \text{ and }b \in \Omega_0.$$Also, $$\overline{V_n}\subset V_{n-1}\cap \Omega_n \subset \Omega_n, \quad \forall\,n \geq 1 \implies b \in \Omega_n,\quad \forall\,n \geq 1.$$ So, $b \in W$ and $b \in \bigcap_{n \geq 0}\Omega_n$, hence $W \cap \left(\bigcap_{n \geq 1}\Omega_n\right)\neq \varnothing$ and we're done.


I can understand why we can assume $\color{blue}{(\ast)}$, but I don't see where $\color{blue}{(\ast)}$ was used in the proof. Can someone clarify this for me and give an input on this proof, please? Thanks.

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In fact $\color{blue}{(*)}$ isn’t really used in the proof. Since you choose $V_n$ so that $\operatorname{cl}V_n\subseteq V_{n-1}\cap\Omega_n$, an easy induction shows that for each $n\ge 0$ we have $\operatorname{cl}V_n\subseteq\bigcap_{k\le n}\Omega_k$ even if the sets $\Omega_k$ aren’t nested. Assuming that they are nested just means that we don’t have to do the induction: it makes the fact that $\operatorname{cl}V_n\subseteq \bigcap_{k\le n}\Omega_k$ an automatic consequence of the choice of $\operatorname{cl}V_n\subseteq\Omega_n$.

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    $\begingroup$ Congrats, you managed to beat me to the answer. I guess it's really time to hit the hay! :-) $\endgroup$ – Asaf Karagila Jun 9 '15 at 22:42
  • $\begingroup$ @Brian Thanks for the clarification! Just one more little thing... having $\overline{V_n}\subseteq \bigcap_{k \leq n}\Omega_k$ does not seems really relevant here, since I can conclude that $b \in \Omega_n$ by the $$\overline{V_n}\subset V_{n-1}\cap \Omega_n \subset \Omega_n, \quad \forall\,n \geq 1 \implies b \in \Omega_n,\quad \forall\,n \geq 1$$step, right? Is this necessary for the proof? $\endgroup$ – Ivo Terek Jun 9 '15 at 22:45
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    $\begingroup$ @Ivo: No, it’s not technically necessary; it just makes it more obvious that $b$ is where you want it. It’s the sort of thing that one tends (as Asaf notes) to do automatically, without really thinking about whether it’s needed or just convenient. $\endgroup$ – Brian M. Scott Jun 9 '15 at 22:49
  • $\begingroup$ Ok, everything is clear now. Thanks again. $\endgroup$ – Ivo Terek Jun 9 '15 at 22:52
  • $\begingroup$ @Ivo: You’re welcome. $\endgroup$ – Brian M. Scott Jun 9 '15 at 22:52
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Well, as for why you can make this assumption, that's easy.

  1. The intersection of two open dense sets is open and dense. (While the intersection of two dense sets need not be dense, in fact it can be empty, here we use the fact that the sets are open.)

  2. Replace each $\Omega_n$ by $\bigcap_{k\leq n}\Omega_k$.

And you're right that it's not being used. It's somewhat of a bad habit that you begin by making a lot of additional "WLOG assumptions", and sometimes you're not using all of them. This is not a big deal when it happens in class, because a lot of the time you don't read the proof from a piece of paper, but filling in the details from the idea that you remember. So starting with "WLOG ..." is usually helpful, even if you end up not needing that.

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  • $\begingroup$ Thanks for the answer! But will anything go wrong if I don't assume that the sets are nested and if I don't replace $\Omega_n$ with $\bigcap_{k \leq n}\Omega_k$? $\endgroup$ – Ivo Terek Jun 9 '15 at 22:46
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    $\begingroup$ No, nothing really would change in the proof. $\endgroup$ – Asaf Karagila Jun 9 '15 at 22:48
  • $\begingroup$ All right, thanks again! Welp, both answers were equally helpful, so my criterion for accepting will be time :P $\endgroup$ – Ivo Terek Jun 9 '15 at 22:52

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