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 $V \in \cn_{X}(x)$.$\square$