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.

Is it possible to find a continuous bijection $f: \mathbb{R} \rightarrow \textrm{Z}$ where $\textrm{Z}$ is a compact Hausdorff space? ($\mathbb{R}$ endowed with usual topology)

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Yes. For example, you can take $Z$ to be a wedge of two circles, and let $f$ wrap first around one circle, then the other.

share|improve this answer
2  
Nice counterexample. If I understand your construction correctly, the same topology could be obtained be taking the usual topology on $\mathbb R$ and change local basis at 0 to be the usual neighborhoods plus complement of a bounded closed interval. (Or, equivalently, take the one-point compactfication of reals and make a quotient space by identifying $\infty$ and 0. –  Martin Sleziak Aug 9 '11 at 11:11

Your Answer

 
discard

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.