Let $(X,d)$ and $(Y,d')$ be metric spaces, $A \subset X$ and $f: \overline{A} \to Y$ continuous that is uniformly continuous in $A$. Show that $f$ is uniformly continuous in $\overline{A}$.
I was given an hint that from uniform continuity I can find $\delta$ and use that, but I didn't quite understand this. From uniform continuity I have $$d(x,y) < \delta \implies d'(f(x),f(y)) < \varepsilon$$ so I already have a $\delta$ that works for uniform continuity in $A$?
On
Fix $\varepsilon>0$ and pick $\delta>0$ such that, for $x,y \in A$, $$d(x,y) < \delta \implies d'(f(x),f(y)) < \varepsilon.$$ Now, suppose that $x,y \in \overline{A}$ satisfy $d(x,y)<\delta$. There exist $x_n,y_n \in A$ such that $x_n \to x \text{ and } y_n \to y$. For $n$ sufficiently large, we will have $d(x_n,y_n) < \delta$. Then, for such $n$, $$d'(f(x),f(y)) < d'(f(x),f(x_n)) + \varepsilon + d'(f(y),f(y_n)).$$ Send $n \to +\infty$.
The question was edited after I posted the answer but I have included a solution to the edited version also.
This is fasle. Let $A =(0,1)$ and $f(x)=0$ for $0<x\leq 1$ , $f(0)=1$. Then $f$ is not even continuous on $\overline A$. However, if you also assume that $f$ is continuous on $\overline A$ then we can show that it is uniformly continuous there. If $x, y \in \overline A$ and $d(x,y) <\delta$ then there exist seqeuences $(x_n), (y_n)$ in $A$ such that $x_n \to x, y_n \to y$. Note that $d(x_n,y_n) <\delta$ for $n$ sufficiently large. This gives $d(f(x), f(y))=\lim d(f(x_n), f(y_n)) \leq \epsilon$.