Take the 2-minute tour ×
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.

enter image description here

enter image description here

enter image description here

Is my work correct?

Could they give me another counterexample?

share|improve this question
What exactly is the problem? I think you are missing something here. –  copper.hat Jun 1 '12 at 5:51
A counterexample to what? –  Chris Eagle Jun 1 '12 at 10:05
The question Is my work correct? is sweet since the amount of mathematical work shown by the OP is null. –  Did Jun 3 '12 at 8:29

1 Answer 1

up vote 1 down vote accepted

If the missing conclusion of the theorem at the top is that $h$ is continuous iff $f$ is continuous, then yes, the example in (c) shows that compactness of $Y$ cannot be omitted from the hypothesis. There is a small error, though: $g\circ f$ is the identity on $[0,2\pi)$, not $[2\pi,0)$.

There are many similar examples; here is what is perhaps the simplest. Let $$X=Z=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$ and $Y=\Bbb N$, all with the usual Euclidean metric given by $d(x,y)=|x-y|$. $X$ is compact, but $Y$ isn’t. Let $$f:X\to Y:x\mapsto\begin{cases}0,&\text{if }x=0\\\\\frac1x,&\text{if }x\ne0\;,\end{cases}$$ and let $$g=f^{-1}:Y\to X:n\mapsto\begin{cases}0,&\text{if }n=0\\\\\frac1n,&\text{if }n\in\Bbb Z^+\;.\end{cases}$$

Then $g$ is one-to-one and continuous, and $g\circ f$ is the identity map on $X$, but $f$ is not continuous at $0$.

share|improve this answer
Small correction/typo: $f$ should be $\frac{1}{n} \mapsto n$. –  Tom Cooney Jun 1 '12 at 9:31
@Tom: Thanks; fixed. (Actually, I reversed $f$ and $g$.) –  Brian M. Scott Jun 1 '12 at 9:36

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.