I am trying to prove the following claim: If a set contains no nonempty open sets, its complement intersects every nonempty open set (the general setting is a complete metric space, $X$). I have the following so far:
Suppose $A$ contains no nonempty open sets. Suppose there is a nonempty open set, $B$, with which $A's$ complement, $X$ \ $A$, does not intersect. Then $A$ must intersect with $B$.
From this point I am not sure how to derive a contradiction.
Best Answer
Let $O$ be a non-empty open set. $A$ contains no non-empty open sets by assumption so $$O \nsubseteq A$$ This logically (by definition of inclusion) means there is some $p\in O$ that is not in $A$. So $$p \in O\cap (X\setminus A) \neq \emptyset$$ It follows that $O$ intersects the complement of $A$. QED