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 this statement always true?

let $U,V$ are disjoin open sets in $X$ and $P\subset \overline{U}\cap \overline{V}$, then $P\cap(U\cup V)=\varnothing$.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

If $x\in P$ then in every open neighborhood of $x$ there are elements from $U$ and $V$. If $x\in U$ then it had an open neighborhood which meets $V$ as well, but that is impossible (can you see why?), similarly if $x\in V$.

Therefore if $x\in P$ then $x\notin U\cup V$, and so the intersection is indeed empty.

share|improve this answer

I think you can choose $P$ maximal so $P=\overline{U}\cap \overline{V}$. Expanding leads to $$ (\partial U \cap V)\cup (\partial V \cup U) \cup (\partial V \cap \partial U)\cap (U \cup V)=$$ Since $U,V$ are open and disjunct the first two terms don't matter (maybe incorrect) so we get $$ (\partial V \cup \partial U) \cap (U \cup V) $$. This is the same as $$((\partial U \cup \partial V)\cap U )\cup ((\partial U \cup \partial V) \cap V)$$, and again we have $$ (\partial U \cap V) \cup (\partial V \cap U)$$. So yeah it its $\varnothing$

share|improve this answer

If by $\bar{A}$ you mean the closure of $A$, then this is obviously not true, as long as $P\neq\emptyset$.

If else you mean the complement of $A$, which is usually denoted by $A^c$, it is always true. You have $(P\cap(U\cup V))^c = P^c\cup(U^c\cap V^c) = X$ by assumption, and by noting that $(A^c)^c=A$, you get your statement.

share|improve this answer
Obviously not true? You have two other answers on this page contradicting your "obvious" observation. Care to elaborate on your answer? –  Asaf Karagila Feb 14 '13 at 15:38
Oh, sorry. I didn't notice the "disjoint" part. My bad. –  Daniel Robert-Nicoud Feb 14 '13 at 15:51

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.