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Consider the collection of those $\mathbb{R}$-valued functions on an interval $I\subseteq\mathbb{R}$, which have a dense set of points of continuity. I would expect this collection to be closed under taking uniform limits.

What are good ways to prove this? (Or in fact, what is a counterexample)?

Damian Reding
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2 Answers2

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Let $f_n:I\rightarrow\mathbb{R}$ converge uniformly to $f:I\rightarrow\mathbb{R}$. Let $C_n$ be the set of points of continuity of $f_n$. Then:

  • $C_n$ is $G_{\delta}$, i.e. it is the countable intersection of open sets. (See, for example, here.)

If we further assume that each $C_n$ is dense, then each $C_n$ is the countable intersection of dense open sets.

Now,

  • If $f_n$ is continuous at $x\in I$ for every $n$, then $f$ (its uniform limit) is continuous at $x$ as well.

Thus, the set of continuity points of $f$ contains $\bigcap\limits_{n\in\mathbb{N}}{C_n}$. Since each $C_n$ is the countable intersection of dense open sets, the intersection $\bigcap\limits_{n\in\mathbb{N}}{C_n}$ is also a countable intersection of dense open sets. By Baire's Theorem, this intersection is dense.

Community
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Joey Zou
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Say $f:I\to\Bbb R$. For $n=1,2\dots$ let $E_n$ be the set of $x\in I$ such that there exists $\delta>0$ with $|f(t)-f(s)|<1/n$ for every $s,t\in(x-\delta,x+\delta)$. If $x\in E_n$ and $\delta$ is a witness to this fact then $(x-\delta,x+\delta)\subset E_n$.

So $E_n$ is open. But the set of points where $f$ is continuous is precisely $\bigcap_n E_n$. So: For any $f$, the set of points where $f$ is continuous is a $G_\delta$ (a countable intersection of open sets).

The Baire Category Theorem says that a countable intersection of dense $G_\delta$'s is dense. So if each $f_n$ has a dense set of continuity then there exists a dense set where every $f_n$ is continuous.

David C. Ullrich
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