# Proving that $\partial P'\subset \partial P$ if and only in $P'\cap P^0 \subset P'^0$

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

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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 $\partial P$.