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Is this statement always true?

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

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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.

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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 $\varnothing$

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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.

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1  
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

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