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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As the topics. Proving that $\partial P'\subset \partial P $ if and only in $P'\cap P^0 \subset P'^0$. I am not sure how to start

share|cite|improve this question
Just to check, $\partial$ is boundary, $'$ is derived set, you're using a superscript zero for interior, and the setting is an arbitrary topological space? – Kevin Carlson Oct 13 '12 at 8:03
yup, the notation are defined in that way – Mathematics Oct 13 '12 at 8:04
Pick elements.. – Berci Oct 13 '12 at 10:10

One way to define $\partial P$ is as $P'\setminus P^0$. So if $P'\cap P^0\subset P'^0,$ then we want to show $\partial P' \cap P^0$ is empty. But $\partial P'\subset P'$ since derived sets are closed, so that $\partial P'\cap P^0\subset P'\cap P^0\subset P'^0$, while by the definition of boundary $\partial P'\cap P'^0$ is empty.

Approach the converse via the contrapositive: assume $x\in P'\cap P^0 \setminus P'^0$.Then in particular $x\in P'\setminus P'^0,$ that is $x\in \partial P'$, while on the other hand $x\notin \partial P$ since it's in the interior of $P$. Thus $\partial P'$ is not a subset of $\partial P$.

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.