Understanding a proof about the density of continuous points of a point-wise limit function.

Question

This is an excerpt from a proof of a theorem.

Theorem 48.5 Suppose $X$ is a topological space and $(Y,d)$ is a metric space. Let $f_n:X\to Y$ be a series of continuous functions, and $f$ be its point-wise limit. If $X$ is a Baire space, there the points where $f$ is continuous are dense in $X$.

Proof. Given positive integer $N$ and $\epsilon > 0$, define $$ A_N(\epsilon) = \{x|d(f_n(x), f_m(x))\le\epsilon \textrm{ for all } n, m\ge N\}$$

Fix $\epsilon$, observe that $A_1(\epsilon)\subset A_2(\epsilon)\subset\cdots$. Now $X$ is the union of all of them. Observe that given $x_0\in X$, $x_0\in A_n(\epsilon)$ for a certain $N$.

Now, let $$U(\epsilon)=\bigcup_{N\in\mathbb Z_+} \mathrm{Int}A_N(\epsilon)$$ We will show that

1. $U(\epsilon)$ is a dense open set in $X$

2. [...]

To show that $U(\epsilon)$ is a dense open set in $X$, we must show that given a nonempty open subset (denote $V$) of $X$, there exists an $V\cap \mathrm{Int}A_N(\epsilon) $ that is nonempty. For this, observe that for all $N$, $V\cap A_N(\epsilon) $ is closed. Because $V$ is an open subspace of $X$, $V$ is a Baire space. So, there exists at least one set (denote $V\cap A_M(\epsilon) $) such that it contains a nonempty open subset of $V$. (Call that open subset $W$.) Since $V$ is open in $X$, $W$ is open in $X$ as well. Therefore, it is included in $V\cap \mathrm{Int}A_M(\epsilon) $.

[...](proof of 2.)

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I have difficulty understanding the highlighted line. It does not resemble to either the presumption or the conclusion of being a Baire space. How is it being deducted?

Answer 1

With the changes, I can figure it out, now.

• The $A_N(\epsilon)$ are closed by continuity of the metric $d$, functions $f_n$ and because the distance is allowed to take on $\epsilon$ as a value.
• Set $V_N := V \cap A_N(\epsilon)$. It is not correct to simply say $V_N$ is closed as you do in the proof. This is because "closed" by itself means as a subset of the ambient space $X$, and there is no reason this should be closed with respect to $X$. However, $V_N$ is closed as a subset of $V$.
• It was already noted that for any $x \in X, x \in A_N(\epsilon)$ for some $N$ (since $f_n(x)$ converges). So $\bigcup_N A_N(\epsilon) = X$, and also $\bigcup_N V_N = V$. In particular, the union has non-empty interior.
• If $\mathrm{Int}\ V_N = \emptyset$ for every $N$, then the $V_N$ would be closed nowhere-dense subsets of $V$, and therefore because $V$ is Baire, their union would have empty interior, which is false. Therefore, for at least one $M, \mathrm{Int}\ V_M$ is not empty.