Let $(X,d)$ be a metric space and $F \subseteq C(X)$ be a finite set. Prove that $F$ is equicontinuous on $X$. Let $(X,d)$ be a metric space and $F \subseteq C(X)$ be a finite set. Prove that $F$ is equicontinuous on $X$.
Since $F$ is in $C(X)$, then $F$ is continuous and so for all $\epsilon > 0$ there exists some $\delta >0$ such that $d_x(x,y) < \delta$ implies $d_y(f(x),f(y)) < \epsilon$.
How is this different from the definition of equicontinuous?
 A: But $F$ here is a family (set) of functions, not just a single function. Since each single function $f$ in $F$ is continuous, yes, for a given $\epsilon$, there exists a $\delta$ such that $d(x, y) < \delta \implies d(f(x), f(y)) < \epsilon$. However, for the family to be equicontinuous, there must be a single $\delta$ which works for all the functions; this is a similar distinction to the distinction of "continuity" and "uniform continuity" for functions from $\mathbb{R} \to \mathbb{R}$.
Fortunately, in this problem the family $F$ is finite; so for a given $\epsilon$, you can just list all the $\delta$'s and pick the smallest one. 
A: You need for $\delta$ to work for every element of the family.
Let the family be $F=\{f_1,f_2,\cdots, f_n\}.$ Fix $\epsilon>0$ and $x\in X.$ For each $k$, the continuity of $f_k$ guarantees the existence of $\delta_k>0$ so that $d_y(f_k(x),f_k(y))<\epsilon$ whenever $y\in X$ and $d_x(x,y)<\delta_k.$  Set $\delta=\min_k\delta_k$ (we can take the minimum  there are finitely many $\delta_k$'s). Now, if $y\in X$ and $d_x(x,y)<\delta,$ it follows from the choice of $\delta$ that $d_y(f(x),f(y))<\epsilon$ for any $f\in F,$ and hence we have (pointwise) equicontinuity.
Note that this follows from the fact that $\delta\leq\delta_k$ for all $k$, and so the equicontinuity comes from the continuity of each element of the family. The key is that the finiteness of $F$ allows us to pick a minimal $\delta$ that works for every element of $F$.
