Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 4, Problem 4

Problem

Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p)=f(p)$ for all $p \in E$, prove that $g(p)=f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.)

Answer

For $y \in f(X)$, there exists $x \in X$ such that $y=f(x)$. For $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon$. Since $E$ is dense in $X$, there exists $z \in E$ such that $z \in N_\delta(x)$, which implies $f(x) \in N_\epsilon(f(z))$. This means $f(E)$ is dense in $f(X)$.

For $p \in X$ and $\epsilon >0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon/2, d_Y(g(x),g(x'))<\epsilon/2$. For $q \in E$ satisfying $q \in N_\delta(p)$, $d_Y(g(p),f(p)) \leq d_Y(g(p),g(q)) + d_Y(g(q),f(q)) + d_Y(f(q),f(q)) < \epsilon$. Since $\epsilon$ is arbitrary, $g(p)=f(p)$.

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