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 the following true?

Suppose $A\subset X$ is closed and $H\colon X\to Y$ is a homeomorphism.

Then $\partial H(A) = H(\partial A)$.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Since $H$ is bijective and $A$ is closed $$H(\partial A)=H(\overline A\setminus \operatorname{int}(A)) =H(A\setminus \operatorname{int}(A))=H(A)\setminus H(\operatorname{int}(A))$$ and $H$ maps closed sets to closed one $$\partial H(A)=\overline{H(A)}\setminus\operatorname{int}(H(A)=H(A)\setminus\operatorname{int}(H(A)),$$ we have to show that $H(\operatorname{int}(A))=\operatorname{int}(H(A))$. Note that since $H(\operatorname{int}(A))\subset H(A)$ and $H(\operatorname{int}(A))$ is open, we have $H(\operatorname{int}(A))\subset \operatorname{int}(H(A))$. Now, let $x\in \operatorname{int}(H(A))$. Then $x$ is in an open set $O$ which is contained in $H(A)$. Since $H^{-1}(x)$ is in $H^{-1}(O)$, open set contained in $A$, $H^{-1}(x)$ is in $\operatorname{int}(A)$, and $x=H(H^{-1}(x))$ is in $H(\operatorname{int}(A))$.

share|improve this answer
add comment

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.