Equivalence of Definitions of Irreducible Space/3 iff 6

Proof
Let $T = \left({S, \tau}\right)$ be an irreducible space by Definition 3.

That is, let every non-empty open set in $T$ have a non-empty intersection with every other non-empty open set.

Let $U \in \tau$ be open in $T$ such that $U \ne \varnothing$.

If $U = S$ then $U^- = S$ trivially.

So, let $U \ne S$.

Let $x \in S$ such that $x \ne U$.

Let $V$ be any non-empty open set of $T$ such that $x \in V$.

By hypothesis, $U \cap V \ne \varnothing$.

Therefore $V$ contains some point of $U$ distinct from $x$.

$V$ is arbitrary, so it follows that every open set $V$ of $T$ such that $x \in V$ contains some point of $U$ distinct from $x$.

That is, $x$ is a limit point of $U$.

So every point of $T$ is a limit point of $U$.

Thus, by definition, every point of $T$ is in the closure of $U$.

The argument reverses directly.