If there exists a uniformly continuous homeomorphism from a metric space $X$ to a complete metric space $Y$ then is $X$ complete?

Question

Let $X,Y$ be metric spaces , $Y$ be complete and $f:X \to Y$ be a homeomorphism which is also uniformly continuous , then is it true that $X$ is complete ?

Answer 1

Yes. If $(x_n)$ is Cauchy in $X$ then by uniform continuity, $(f(x_n))$ is Cauchy in $Y$ and so converges to some $y \in Y$ and then by continuity of $f^{-1}$, the sequence $x_n = f^{-1}(f(x_n))$ converges to $f^{-1}(y)$.

(Given $\varepsilon > 0$, by uniform continuity we can find $\delta > 0$ such that if $d(x,y) < \delta$ then $d(f(x),f(y)) < \varepsilon$. Since $x_n$ is Cauchy, we can find $N$ such that if $n,m > N$ then $d(x_n,x_m) < \delta$ and then $d(f(x_n),f(x_m)) < \varepsilon$ for all $n,m > N$ showing that $(f(x_n))$ is Cauchy).