6.5 Equicontinuity
Definition 6.5.1 (Equicontinuous).label Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^{X}$, and $x \in X$, then $\cf$ is equicontinuous at $x$ if for every $U \in \fU$, there exists $V \in \cn_{X}(x)$ such that $(f(x), f(y)) \in U$ for all $y \in V$ and $f \in \cf$.
The set $\cf \subset C(X; Y)$ is equicontinuous if it is equicontinuous at every point in $x$.
Proposition 6.5.2.label Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, $\cf \subset Y^{X}$, and $x \in X$, then the following are equivalent:
- (1)
$\cf$ is equicontinuous at $x$.
- (2)
For $\angles{x_\alpha}_{\alpha \in A}\subset X$ with $x_{\alpha} \to x$, $\angles{f_\alpha}_{\alpha \in A}\subset \cf$, and $U \in \fU$, there exists $\alpha_{0} \in A$ such that $(f_{\alpha}(x_{\alpha}), f_{\alpha}(x)) \in U$ for all $\alpha \ge \alpha_{0}$.
- (3)
There exists a fundamental system of neighbourhoods $\fB \subset \cn_{X}(x)$ at $x$ such that for any $\angles{x_V}_{V \in \fB}\subset X$ with $x_{\alpha} \to x$, $\angles{f_V}_{V \in \fB}\subset \cf$, and $U \in \fU$, there exists $V_{0} \in \fB$ such that $(f_{V}(x_{V}), f_{V}(x)) \in U$ for all $V \subset V_{0}$.
Proof. (1) $\Rightarrow$ (2): Since $\cf$ is equicontinuous at $x$, there exists $V \in \cn_{X}(x)$ such that $(f_{\alpha}(y), f_{\alpha}(x)) \in U$ for all $y \in V$ and $\alpha \in A$. Given that $x_{\alpha} \to x$, there exists $\alpha_{0} \in A$ such that $x_{\alpha} \in V$ for all $\alpha \ge \alpha_{0}$, so $(f_{\alpha}(x_{\alpha}), f_{\alpha}(x)) \in U$ for all $\alpha \ge \alpha_{0}$.
$\neg (1) \Rightarrow \neg (3)$: If $\cf$ is not equicontinuous at $x$, then there exists $U \in \fU$ such that for every $V \in \fB$, there exists $f_{V} \in \cf$ and $x_{V} \in V$ with $(f_{V}(x_{V}), f_{V}(x)) \not\in U$. In which case, $x_{V} \to x$ but $(f_{V}(x_{V}), f_{V}(x)) \not\in U$ for all $V \in \cn_{X}(x)$.$\square$
Definition 6.5.3 (Uniformly Equicontinuous).label Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is uniformly equicontinuous if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$.
Theorem 6.5.4 (Arzelà-Ascoli).label Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, and $\cf \subset C(X; Y)$. If
- (E1)
$\cf$ is equicontinuous.
then
- (C1)
The uniform structures of pointwise and compact convergence on $\cf$ coincide.
- (C2)
The closure of $\cf$ in $Y^{X}$ with respect to the product topology is equicontinuous.
In addition, if $\cf$ satisfies (E1) and
- (E2)
For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$.
then
- (C3)
$\cf$ is a precompact subset of $C(X; Y)$ with respect to the uniform structure of compact convergence.
Conversely, if $X$ is a LCH space, then (C3) implies (E1) + (E2).
Proof. (E1) $\Rightarrow$ (C1): By Proposition 7.2.6, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$ and thus $\cf$.
Let $K \subset X$ be compact, and $U \in \fU$. Since $\cf$ is equicontinuous, for each $x \in K$, there exists $V_{x} \in \cn_{X}(x)$ such that $g(V_{x}) \subset U(g(x))$ for all $g \in \cf$. By compactness of $K$, there exists $\seqf{x_j}\subset K$ such that $K \subset \bigcup_{j = 1}^{n} V_{x_j}$. Let $f, g \in \cf$ such that $(f(x_{j}), g(x_{j})) \in E$ for all $1 \le j \le n$. For any $x \in K$, there exists $1 \le j \le n$ such that $x \in V_{x_j}$. In which case,
so $(f(x), g(x)) \in U \circ U \circ U$. Therefore
so the uniform structures of pointwise and compact convergence coincide.
(E1) $\Rightarrow$ (C2): Let $\cf'$ be the closure of $\cf$ in $Y^{X}$ with respect to the product topology. Let $x \in X$ and entourage $U$ of $Y$. Using Proposition 6.1.14, assume without loss of generality that $U$ is closed. Since $\cf$ is equicontinuous, there exists $V \in \cn_{X}(x)$ such that $(f(x), f(y)) \in U$ for all $f \in \cf$ and $y \in V$. For any element $g \in \cf'$, $(g(x), g(y)) \in \ol U = U$ for all $y \in V$. Therefore $\cf'$ is also equicontinuous.
(E1) + (E2) $\Rightarrow$ (C3): Using (C2), assume without loss of generality that $\cf$ is closed in $Y^{X}$ with respect to the product topology. In which case, $\cf$ is a closed subset of $\prod_{x \in X}\ol{\cf(x)}$ with respect to the product topology. By Tychonoff’s Theorem and Proposition 5.16.2, $\cf$ is compact in the product topology. By (C1), $\cf$ is also compact in the compact uniform topology.
(C3) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_{X}(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j}\subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_{j} \times g)(V) \subset U$. For each $1 \le j \le n$, $f_{j} \in C(X; Y)$, so there exists $V_{j} \in \cn_{X}(x)$ with $V_{j} \subset V$ such that for any $y \in V_{j}$, $(f_{j}(x), f_{j}(y)) \in U$. Let $W = \bigcap_{j = 1}^{n} V_{j}$, then for any $g \in \cf$ with $(f_{j} \times g)(V) \subset U$ and $y \in W$,
Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous.
(C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the uniform structure of compact convergence, $\cf(x)$ is totally bounded for all $x \in X$ by Proposition 6.4.2.$\square$