Undestanding an assumption in Baire's theorem proof.

Question

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.

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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.

Answer 1

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 $\operatorname{cl}V_n\subseteq\Omega_n$.

Answer 2

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.