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Prove or disprove: If $M$ is complete and $f:(M, d )\to (N, p)$ is continuous, then $f(M)$ is complete.

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Asymptotes and such things kill this: consider the function $x\mapsto e^{-x^2}$, for example. –  Mariano Suárez-Alvarez May 20 '12 at 20:18

4 Answers 4

Let $M=N=\mathbb{R}$ and $f(x)={x \over 1+|x|}$. $f$ is of course continuous, $\mathbb{R}$ is complete but its image $(-1,1)$ isn't.

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In the case of acknowledged homework it seems preferable to give a hint $-$ even a pretty broad one, as Asaf and I did $-$ rather than a complete solution. –  Brian M. Scott May 20 '12 at 9:32
Oh. I'm sorry, I missed the homework tag. Should I rewrite/delete the answer? –  Najib Idrissi May 20 '12 at 10:32
It happens; I’ve done it myself a time or two. Up to you, though there’s no real reason to delete it: Asaf’s hint is almost as specific. (Mine’s probably the most cryptic, since I didn’t actually say what metric to use on $N$.) –  Brian M. Scott May 20 '12 at 10:36
OK. I think I'll rewrite it to not include the complete answer. –  Najib Idrissi May 20 '12 at 10:37

In general this is not true as pointed out in several counter-examples/hints on other posts: I'm going to add something else here. Continuity is just too weak to do the job. What you need from $f$ in order for the completeness of $M$ to be transmitted to $f(M)$ is some control over the sequences in $M$ that map to Cauchy-sequences in $f(M)$.

One sufficient property is $f$ being Bi-Lipschitz. One says that a function $f:(X,d)\to (Y,e)$ is Bi-Lipschitz if there exists $M\geq 1$ so that $\frac{1}{M}d(x,y)\leq e(f(x),f(y))\leq Md(x,y)$ for all $x,y\in X$. It is quite straight-forward to show that such property is indeed enough for $f(M)$ to be complete if $M$ is. And in fact, there doesn't seem to be many properties that you could drop out and make it still work: maybe this is close to necessary condition as well.

Added: For completness sake, I will add a counter-example of a Lipschitz function that fails to do the job. Since $[1,\infty[$ is a closed subset $\mathbb{R}$ it is thus complete and $\frac{1}{x}:[1,\infty[\to]0,1]$, denoted by $f$, is a continuous bijection, yet $]0,1]$ is not complete. Choose e.g. the standard Cauchy-sequence $(\frac{1}{n})_{n=1}^{\infty}$ which does not converge in $]0,1]$.

In addition $f\in C^{1}([1,\infty[)$ and $|f\,'(x)|=|-\frac{1}{x^{2}}|=\frac{1}{x^{2}}\leq 1$ so $f$ is $1$-Lipschitz by the intermediate value theorem. So even a Lipschitz function is not enough for the result! Maybe you can figure out something closer to Bi-Lipschitz (but not quite) that also fails.

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BROAD HINT: Let $M=N=\Bbb Q$, and let $d$ be a discrete metric on $M$, e.g., $$d(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\;.\end{cases}$$ Put the right non-discrete topology on $N$, and use the identity function $f:M\to N:x\mapsto x$. The discrete metric on $M$ ensures that $f$ is continuous no matter what metric you put on $N$.

It will help to ask yourself what are the Cauchy sequences in $\langle M,d\rangle$.

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Beautiful Counter example :) –  CSA May 16 at 19:41

Hint consider both spaces as the real numbers with he standard metric and $f$ to be a continuous function which is a bijection with an open interval, e.g. $\arctan$.

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