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I'm stuck in understanding the proof of the following theorem given during a course:

Let $X$ be a Baire space, and $(Y,d)$ a metric space. Let $f_n:X\to Y$ be a sequence of continuous function, pointwise converging to $f:X\to Y.$ Then $f$ is continuous on a dense set.

Proof: Let $N,k\in \mathbb{N}^*.$ Let $A_{N,k}:=\bigcap_{n,m\ge N}\{x\in X\ |\ d(f_n(x),f_m(x))\leq \frac1k\}.$

Then by continuity every $A_{N,k}$ is a closed set and by pointwise convergence $X=\bigcup_{N\in\mathbb{N}^*} A_{N,k}.$

So for every $k\geq 1$ we have that $\Omega_k=\bigcup_{N\in\mathbb{N}^*} A_{N,k}^\mathrm{o}$ is an open dense subset of $X.$

By the Baire property then, we have that $\bigcap_{k\ge 1}\Omega_k$ is a dense set, and that $f$ is continuous right on that set. QED.

What I really don't understand is the reason why the set $\Omega_k$ should be dense in the space $X:$ the fact that $X$ is the union of $A_{N,k}$ just implies that one of these sets has non-empty interior (by the Baire property), which is totally different from what I want.

Riccardo Ceccon
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1 Answers1

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Every open subset of a Baire space is again a Baire space. If you apply that then you'll find that $\Omega_k$ is dense.

Given a nonempty open set $O$ look at the family $\{A_{N,k}\cap O:N\in\mathbb{N}\}$; because $O$ is Baire at least one member must have interior in $O$, but because $O$ is open that means for such an $N$ we have $O\cap A_{N,k}^\circ\neq\emptyset$.

hartkp
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