# Do uniformly continuous functions map complete sets to complete sets?

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.

• You should start with a Cauchy sequence on $f(M)$ and try to prove that it converges. You did the opposite. You took a Cauchy sequence on $M$. – Sigur May 15 '13 at 22:37
• I think $f(M)$ is not necessarily complete. because $\Bbb R$ can be homeomorphically embedded in unit circle. the image is open and so not complete. just a raw idea! – user59671 May 15 '13 at 22:48
• Let $d$ be the discrete metric on $M=(0,1)$ and $\rho$ be the standard metric on $N=M$. Consider the identity map from $M$ to $N$. – David Mitra May 15 '13 at 23:05
• @DavidMitra. I think this even demonstrates Lipschitz continuity doesn't, on its own, preserve completeness. – Tom Collinge Jan 22 '17 at 14:31

Take $f:\mathbb R\longrightarrow \mathbb R$, $f(x)=\arctan x$. Then $$f(\mathbb R)=(-\frac\pi2,\frac\pi2)$$ and $f'(x)=\dfrac{1}{1+x^2}\leq 1$ which implies that $f$ is uniformly continuous.
However, $\mathbb R$ is complete, while $f(\mathbb R)=(-\frac\pi2,\frac\pi2)$ is not complete
Let $M:=\Bbb R$ and $N:=(-\pi/2,\pi/2)$ and $f:=\arctan$. This is uniformly continuous (and, also a homeomorphism), but $N$ is not complete.