# Arzela-Ascoli net question

Let $$X$$ be a compact metric space. Let $$C(X)$$ denote the space of real-valued continuous functions on $$X$$. A commonly given corollary to the Arzela-Ascoli theorem is:

Proposition: If $$f_n$$ is an equicontinuous sequence in $$C(X)$$ converging pointwise to $$f \in C(X)$$, then actually $$f_n \to f$$ uniformly.

In order to prove this, I first proved a simple lemma

Lemma: Suppose $$x$$ is a point and $$S$$ is a sequence in a compact metric space $$M$$. If every convergent subsequence of $$S$$ converges to $$x$$, then $$S$$ converges to $$x$$.

Proof: Suppose for contradiction that $$S$$ does not converge to $$x$$. Then, there is a subsequence $$S'$$ of $$S$$ whose terms are bounded away from $$x$$. Being a sequence in a compact metric space, $$S'$$ has a convergent subsequence $$S''$$. Since $$S''$$ is a convergent subsequence of $$S$$, it converges to $$x$$ by hypothesis. But, this is impossible since the terms of $$S'$$ should be bounded away from $$x$$.

which is applied as follows.

Proof of proposition: Let $$M$$ be the uniform closure of $$\{f_n : n \in \mathbb{N} \} \cup \{f\}$$. By the Arzela-Ascoli, $$M$$ is compact for the uniform norm. Consider now $$f_n$$ as a sequence in $$M$$. Any subsequence of $$f_n$$ which converges uniformly must converge to $$f$$ (since $$f_n \to f$$ pointwise). So, by the lemma, $$f_n \to f$$ uniformly.

What I am slightly unsure of is whether these arguments carry over, mutatis mutandis, for nets?

Question: If $$X$$ is a compact metric space, $$C(X)$$ is the space of real-valued continuous functions on $$X$$, and $$f_i$$ is an equicontinuous net in $$C(X)$$ converging pointwise to $$f \in C(X)$$, does $$f_i \to f$$ uniformly?

I think the answer is yes, but I am not totally comfortable with the concept of a subnet, so it is difficult to be certain.

Edit: Thinking about it more, it seems the point requiring clarification is the following one.

Claim: If a net $$(x_i)_{i \in I}$$ in a metric space $$M$$ does not converge to a point $$x \in M$$, then there is a subnet that is bounded away from $$x$$. That is, there is an $$\epsilon > 0$$ and a cofinal, increasing function $$\varphi : J \to I$$ out of a directed set $$J$$ such that $$d( x_{\varphi(j)}, x) \geq \epsilon$$ for all $$j \in J$$.

Since $$(x_i)$$ does not converge to $$x$$ above, we know there exists an $$\epsilon >0$$ such that, for all $$i_0 \in I$$, there exists $$i \geq i_0$$ with $$d(x_i,x) \geq \epsilon$$. To find a subnet, it seems the obvious thing to do is try $$J = \{ i \in I : d(x_i,x) \geq \epsilon\}$$ and $$\varphi$$ the inclusion. Now, the above condition says exactly that $$J$$ is cofinal in $$I$$. And, obviously the inclusion map is increasing... so I guess that settles things?

Edit 2:

I guess in my first edit I forgot to verify that $$J$$ was, itself, a directed set. But it seems a cofinal set $$J$$ in a directed set $$I$$ is automatically a directed set. Any finite subset of $$J$$ has an upper bound in $$I$$, which has then an upper bound in $$J$$.

The answer is yes, and we don't need to consider any subnets to see that. We have

Theorem: Let $X$ a topological space, and $H \subset C(X)$ an equicontinuous subset. Then on $H$

• the topology of pointwise convergence on a (given) dense subset,
• the topology of pointwise convergence, and
• the topology of uniform convergence on (quasi)compact subsets of $X$

coincide.

The theorem holds more generally with a uniform space in place of $\mathbb{R}$. Since a union of finitely many equicontinuous subsets is equicontinuous, and a singleton is evidently equicontinuous, $H = \{f_i : i \in I \} \cup \{f\}$ is an equicontinuous subset of $C(X)$, whence on $H$, pointwise convergence coincides with (quasi)compact convergence. Since $X$ is assumed compact, that is uniform convergence.

Proof: (of the theorem)

It is clear that (quasi)compact convergence implies pointwise convergence, and that in turn implies pointwise convergence on any dense subset. So it remains to see that pointwise convergence on a dense subset implies compact convergence.

Let $D$ be a dense subset. Let $\varepsilon > 0$ and $K$ be any quasicompact subset of $X$. For every $x \in K$, there exists an open neighbourhood $U_x$ of $x$ such that for all $y \in U_x$ and every $h\in H$ we have $\lvert h(y) - h(x)\rvert < \varepsilon/5$, by the equicontinuity. By the quasicompactness of $K$, the cover $\{ U_x : x \in K\}$ has a finite subcover, say $U_1,\, \dotsc ,\, U_n$, with corresponding points $x_1,\, \dotsc ,\, x_n \in K$. For every $1 \leqslant k \leqslant n$, there is a $d_k \in D \cap U_k$.

For every $p \in H$, the neighbourhood $W = \{h \in H : \lvert h(d_k) - p(d_k)\rvert < \varepsilon/5,\, 1\leqslant k \leqslant n\}$ in the topology of pointwise convergence on $D$ is contained in the neighbourhood $V(K,\varepsilon; p) = \{ h \in H : \bigl(\forall x\in K\bigr)( \lvert h(x) - p(x)\rvert <\varepsilon) \}$ of $p$ in the topology of (quasi)compact convergence:

Let $h \in W$ and $x\in K$. Pick a $k$ such that $x \in U_k$. Then

\begin{align} \lvert h(x) - p(x)\rvert &\leqslant \lvert h(x) - h(x_k)\rvert + \lvert h(x_k) - h(d_k)\rvert\\ &\qquad + \lvert h(d_k) + p(d_k)\rvert + \lvert p(d_k) - p(x_k)\rvert + \lvert p(x_k) - p(x)\rvert\\ &< 5\varepsilon/5 = \varepsilon. \end{align}

Added by OP: To help fix ideas, I'll attempt to paraphrase this nice answer for a more restricted setting below.

Claim: Let $X$ be a compact Hausdorff space and let $\mathscr{F} \subset C(X)$ be an equicontinuous subset. Fix $\epsilon > 0$. Then, there exist points $x_1,\ldots, x_n \in X$ such that, if $f,g \in \mathscr{F}$ have $|f(x_i) - g(x_i)| < \epsilon/3$ for $i=1,\ldots, n$, then $\|f-g\|_\text{uniform} < \epsilon$.

Proof: By equicontinuity, for every $x \in X$, there is a neighbourhood $U_x$ of $x$ such that $|f(x) - f(y)| < \epsilon/3$ for all $y \in U_x, f \in \mathscr{F}$. By compactness, there exist points $x_1,\ldots,x_n \in X$ such that the corresponding open neighbourhoods $U_1,\ldots, U_n$ cover $X$. Now, suppose $f,g \in \mathscr{F}$ have $|f(x_i) - g(x_i)| < \epsilon/3$ for all of the values $i = 1 ,\ldots, n$. Fix an arbitrary $x \in X$. Since the $U_i$ cover $X$ there is a particular value $i$ for which $x \in U_i$. We have then $$|f(x) - g(x)| \leq |f(x) - f(x_i)| + |f(x_i) - g(x_i)| + |g(x_i) - g(x)| < \epsilon,$$ as desired.

Now we easily get:

Claim: Let $X$ be a compact Hausdorff space and let $(f_\lambda)_{\lambda \in \Lambda}$ be an equicontinuous net in $C(X)$ converging pointwise to $f \in C(X)$. Then, $f_\lambda \to f$ uniformly.

Proof: Let $\mathscr{F} = \{f_\lambda : \lambda \in \Lambda\} \cup \{f\}$, an equicontinuous set in $C(X)$. Given an $\epsilon > 0$, find $x_1, \ldots,x_n$ as in the above claim. Since $f_\lambda \to f$ pointwise, there exist $\lambda_1,\ldots,\lambda_n \in \Lambda$ such that $|f_\lambda(x_i) - f(x_i)| < \epsilon/3$ when $\lambda \geq \lambda_i$. Since $\Lambda$ is directed, there is a $\lambda_0 \geq \lambda_1,\ldots,\lambda_n$. When $\lambda \geq \lambda_0$ we have $|f_\lambda(x_i) - f(x_i)| < \epsilon/3$ for $i=1,\ldots,n$ so that, by the above claim, $\|f_\lambda - f\|_\text{uniform} < \epsilon$. Thus, $f_\lambda \to f$ uniformly as desired.

• Thank you for this perspective. I added a bit to your answer, I hope you don't mind. – Mike F Dec 16 '13 at 9:40
• No, I don't mind, that's okay. – Daniel Fischer Dec 16 '13 at 9:44