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.$\square$