Question about continuous functions on two metric spaces Given two metric spaces $X$ and $Y$ and given that $f$and $g$ are continuous mappings from $X \to Y$ and $f(x)=g(x)$ for every $x\in A$ where $A$ is a subset of $X$. Prove that $f(x)=g(x)$ holds true for every $x\in \bar{A}$
What I thought is that if $x\in \bar{A}$ then x is either in A or A'. So the case for A is very easy But I can't think about something for A'

Suppose by contradiction there is a $x\in A'$ with $f(x)\neq g(x)$. As $Y$ is metric and so Hausdorff we have disjoint open sets $U$ and $V$ with $f(x)\in U$ and $g(x)\in V$. As $f$ and $g$ are continuous $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ and so is their intersection. Notice $x\in f^{-1}(U)\cap g^{-1}(V)$, and as $x$ is a limit point of $A$ there must be some $y\in A$ such that $y\in f^{-1}(U)\cap g^{-1}(V)$. The image of $y$ through $f$ is an element of $U$, and the image of $y$ through $g$ is an element of $V$, as these sets are disjoint we must have $f(y)\neq g(y)$ which contradicts our hypothesis as $y\in A$.
