Let $f_n$ be an equicontinuous sequence in C(X) with X compact. Show that the set of all $x\in X$ for which $f_n(x)$ converges is closed.

Question

This question is from Chapter 11 "The Space of Continuous Functions" in the section on equicontinuity of NL Carother's "Real Analysis." Here is the precise question:

Let $X$ be a compact metric space, and let $f_{n}$ be an equicontinuous sequence in $C(X)$. Show that $C=\text{\{}x\in X:(f_{n}(x))$ converges} is a closed set in $X$.

Note: $f_{n}$ is an equicontinuous sequence in C(X). For every $\epsilon>0$ there is a $\delta>0$ such that whenever $x,y\in X$ satisfy $d(x,y)<\delta$ then we have that $|f(x)-f(y)|<\epsilon$ for every $f\in\{f_{n}\}$.

Here is my attempt: $X$ is a compact metric space and $f_{n}$ is equicontinuous and so for every $\epsilon>0$ there is a $\delta>0$ such that $d(x,y)<\delta$ then $|f(x)-f(y)|<\epsilon.$ Take $x_{1},x_{2},...,x_{i}$ to be a $\delta-net$. $f_{n}$ is uniformly continuous since it is equicontinuous on a compact metric space. And so if $x_{1},x_{2},...,x_{n}$ are a $\delta-net$ for $X$ then use a diagonalization argument to find a subset (call it $F$) of $x_{i}$ that converges for every $f_{n}$. This subset is closed since the compliment is the subset of all $x_{i}$ for which $f_{n}$ does not converge for some $n$ is open.

I am stuck because I didn't actually answer the question. I might be on the right track, but I don't understand how to show that the compliment is open. I just stated it because I'm not sure what the steps are to get to that conclusion.

Answer 1

Let $(x_k)_{k>0}$ be a sequence in $C$ that converges against a limit $y \in X$. Since every compact metric space is complete, it suffices to show that $(f_n (y))_{n>0}$ is Cauchy in $X$. To this end, consider

$$ |f_{n}(y) - f_{m}(y)| \leq |f_{n}(y) - f_{n}(x_k)| + |f_{n}(x_k) - f_{m}(x_k)| + |f_{m}(x_k) - f_{m}(y)|$$

Given $\epsilon > 0$, we want to find an $N$ such that $|f_{n}(y) - f_{m}(y)| \leq \epsilon$ for $n,m \geq N$. Because $x_k \to y$ and the $f_n$ are equicontinous, we can find a $k_0$ such that $|f_{n}(y)-f_{n}(x_k)| \leq \epsilon/3$ for all $n \in \mathbb{N}$ if only $k \geq k_0$. As $x_{k_0} \in C$, the sequence $(f_{n}(x_{k_0}))_{n > 0}$ is Cauchy in $X$ and therefore there is a $N$ such that $|f_{n}(x_{k_0}) - f_{m}(x_{k_0})| \leq \epsilon/3$ for $n,m \geq N$. We conclue that

$$|f_{n}(y) - f_{m}(y)| \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} +\frac{\epsilon}{3} \leq \epsilon, \quad m,n \geq N$$ which shows that $(f_{n}(y))_{n>0}$ is Cauchy in $X$ and as $X$ is complete, it converges and hence $y \in C$.

Answer 2

Since $X$ is a metric space, a subset of $X$ is closed iff it is sequentially closed.

Let $M = \{x \in X: \{ f_n(x) \} \text{ converges} \}$. If $M$ is empty then it is closed. Otherwise, take a sequence $\{x_k\}$ in $M$ which converges to $x \in X$. To prove that $M$ is closed, it suffices to prove that $x \in M$; that is we need to show that $\{ f_n(x) \}$ converges. Since every compact metric space is complete, to do so, it suffices to shows that $\{ f_n(x) \}$ is a Cauchy sequence.

Take $m, n, k \in \mathbb N$. Then we see $$\lvert f_m(x) - f_n(x) \rvert \le \lvert f_m(x) - f_m(x_k) \rvert + \lvert f_m(x_k) - f_n(x_k) \rvert + \lvert f_n(x_k) - f_n(x) \rvert.$$ Take $\epsilon > 0$. By equicontinuity, there is a $\delta > 0$ so that $\lvert f(x) - f(y) \rvert < \epsilon / 3$ whenever $d(x,y) < \delta$ and $f \in \{f_n\}$. For this $\delta$, since $x_k \to x$, we can find $K \in \mathbb N$ so that $d(x,x_k) < \delta$ when $k > K$. Then $$\lvert f_m(x) - f_m(x_k) \rvert , \lvert f_n(x) - f_n(x_k) \rvert < \epsilon / 3$$ when $k > K$. Choose one specific $k > K$. Then for this $k$, since $\{f_n(x_k)\}$ is a convergent sequence (and thus a Cauchy sequence), we can find $N \in \mathbb N$ so that $m,n > N$ gives $$\lvert f_m(x_k) - f_n(x_k) \rvert < \epsilon / 3.$$ Then for $m,n > N$, we see $$\lvert f_m(x) - f_n(x) \rvert < \epsilon$$ which shows that $\{f_n(x)\}$ is Cauchy (hence $x \in M$) and completes the proof.