Equicontinuity on a topological domain and pointwise relative compactness

Let $X$ be a topological space, let $(Y,d)$ be a metric space, and let $\mathcal{F}\subseteq C(X,Y)$.

The family $\mathcal{F}$ is equicontinuous at $x\in X$ if, for every $\epsilon >0$, there is a neighbourhood $U$ of $x$ such that

$d(f(y),f(x))<\epsilon$

for every $f\in \mathcal{F}$ and every $y\in U$. It is equicontinuous if it is equicontinuous at every $x\in X$. The same neighbourhood must serve every member of the family. The empty family is equicontinuous, and when $X=\mathrm{\varnothing }$ the pointwise condition is vacuous.

For $x\in X$, write $\mathcal{F}(x):=\{f(x):f\in \mathcal{F}\}$. The family is pointwise relatively compact if the closure $\overline{\mathcal{F}(x)}$ is a compact subset of $Y$ for every $x\in X$. Thus an empty family is pointwise relatively compact because the empty set is compact, and an empty domain again makes the condition vacuous.