Let $f: (M, d) \rightarrow (N, \rho)$ be uniformly continuous. Prove or disprove that if M is complete, then $f(M)$ is complete.
If I am asking a previously posted question, please accept my apologies and tell me to bugger off. I saw a similar problem but the solution was dealing with a Bi-Lipschitz function or some such business.
I believe this statement to be true and here is a rough sketch of my reasoning:
Since $f$ is uniformly continuous, then $f$ maps Cauchy to Cauchy. Let $(x_n)$ be a Cauchy sequence in $M$. Since $M$ is complete, $x_n \rightarrow x \in M$. Again, because of $f$'s uniform continuity, we now have $(f(x_n))$ is Cauchy in $N$ and $f(x_n) \rightarrow f(x) \in N$. Thus $N$ is complete.
By the way, I am studying for an exam. This is certainly not homework. I gladly accept your criticisms. Thank you in advance for your help.