Let $A$ be a subset of a topological space $X$. It can be shown that $x\in \overline{A}$ if and only if every open set $U$ containing $x$ intersects $A$.
I can prove this by proving the contrapositive:
$x\notin \overline{A}$ if and only if there exists an open set $U$ containing $x$ that does not intersect $A$.
But is this the only way to prove this theorem? That is, can it only be shown using the contrapositive statement, or is there a more 'direct' way to prove it?
Best Answer
Defining $\overline{A} =\bigcap\{C \subseteq X \text {closed}: A \subseteq C\}$ (the smallest closed set containing $A$ as a subset):
Let $x \in \overline{A}$ and let $U$ be an open neighbourhood of $x$. If $U \cap A = \emptyset$ then $A \subseteq X\setminus U$, and the latter set is closed, so $\overline{A} \subseteq X\setminus U$, in this definition. Contradiction, as witnessed by $x$. So $A \cap U \neq \emptyset$.
Suppose $x$ has the neighbourhood intersection property. Let $C$ be a closed set that contains $A$. If $x \notin C$, then $X \setminus C$ is a neighbourhood of $x$ that misses $A$, which cannot be. So $x \in C$, and as $C$ was arbitrary closed around $A$, $x \in \overline{A}$.
Pretty direct I'd say. We need small proofs by contradiction, but this is inherent to the statement, I think. To show $U \cap A \neq \emptyset$, we either have to find a concrete point that witnesses it or assume the opposit to make progress.