TITLE: Prove that Image of a closed set is closed in isometry
QUESTION [9 upvotes]: Suppose that $(X, d_X)$ and $(Y, d_Y )$ are complete metric spaces. Let $i \colon X → Y$ be an isometry. Suppose that $F ⊂ X$ is closed. How can I prove that $i(F)$ is a closed subset of $Y$?

REPLY [10 votes]: Take a sequence $(i(x_n))_{n\in\mathbb{N}}$ in $i(F)$ which converge to $l\in Y.$ You have that the sequence is convergent so Cauchy, and 
$$d_Y(i(x_n),i(x_m))=d_X(x_n,x_m)\underset{m,n\to+\infty}{\longrightarrow} 0$$ which implies that $(x_n)_{n\in\mathbb{N}}$ is Cauchy so converge to $x_*\in F$ because $F$ is complete. By continuity of $i,$ you finally have $l=\lim\limits_{n\to+\infty}i(x_n)=i(x_*)\in i(F).$