I am not asking for a proof, but rather an explanation of what this question is trying to get me to do. The question is:
Let $f$ be a continuous function on a metric space $\left(X, d\right)$ and $B$ be a nonempty set in $X$. Show that if $f$ is uniformly continuous on the closure of $B$, then it is uniformly continuous on $B$.
I don't understand what there is to show. In my book, the definition of the closure of $B$ is the union of $B$ and its accumulation points. Doesn't this mean that $f$ is automatically uniformly continuous on $B$ since it is contained in its closure? What am I missing here?
This is indeed trivial. Given $\epsilon>0$, any $\delta$ that witnesses uniform continuity on $\overline{B}$ will also work for uniform continuity on $B$, since $B\subseteq\overline{B}$. I'm guessing that this problem was intended to be stated the other way around (given uniform continuity on $B$, prove it on $\overline{B}$).