Convergence in $\overline A $ of a limit of a sequence of functions

Question

Let $X$ be a metric space, $A\subset X$ and $\phi_n$ a sequence of continuous bounded functions in $X$ converging pointwise to $f$, uniformly in $A$. My question is: how can I show that they converge uniformly to $f$ also in $\overline A$ ?

Answer 1

Because the $\phi_n$ are all continuous, we have

$$\sup \{ \lvert \phi_n(x) - \phi_k(x)\rvert : x \in \overline{A}\} = \underbrace{\sup \{ \lvert \phi_n(x) - \phi_k(x)\rvert : x \in A\}}_s,$$

since

$$A \subset (\lvert \phi_n - \phi_k\rvert)^{-1}([0,s]),$$

and the set on the right is closed by continuity of $x \mapsto \lvert \phi_n(x) - \phi_k(x)\rvert$.

Thus the sequence $(\phi_n)$ converges uniformly to some limit $g$ on $\overline{A}$. Since $\phi_n$ by assumption converges to $f$ pointwise on $X$, it follows that $g = f\lvert_{\overline{A}}$.