Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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.

share|cite|improve this answer

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.

share|cite|improve this answer

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 $\langle M,d\rangle$.

share|cite|improve this answer
Beautiful Counter example :) – CSA May 16 at 19:41
@Aaron: $\Bbb Q$ with the discrete metric is complete. – Brian M. Scott Sep 30 at 20:25

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. $\arctan$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.