Answer to Question #217077 in Real Analysis for Prathibha Rose

Question

Let f and g be continuous functions from a metric space X to metric space Y such that they agree on a dense subset of a X .then prove that f=g

Answer 1

Let $M=\{x\in X: f(x)=g(x)\}\subset X$. Since f and g are continuous functions from a metric space X to metric space Y, M is a closed subset (see below), i.e. $\overline{M}=M$. By condition, M is dense, i.e. $\overline{M}=X$. Therefore, $M=X$ and $f=g$ everywhere. Q.E.D.

Let's show that M is a closed subset. Let $x_n\in M$ be any sequence in M, converging to $x\in X$.

That is, $x=\lim\limits_{n\to\infty}x_n$ . Since f and g are continuous, then $f(x)=\lim\limits_{n\to\infty}f(x_n)=\lim\limits_{n\to\infty}g(x_n)=g(x)$. This means that $x\in M$ and that M is closed.