# Equivalence of Definitions of Irreducible Space/3 iff 6

## Theorem

The following definitions of the concept of Irreducible Space are equivalent:

### Definition 3

A topological space $T = \struct {S, \tau}$ is irreducible if and only if every two non-empty open sets of $T$ have non-empty intersection:

$\forall U, V \in \tau: U, V \ne \O \implies U \cap V \ne \O$

### Definition 6

A topological space $T = \struct {S, \tau}$ is irreducible if and only if the closure of every non-empty open set is the whole space:

$\forall U \in \tau: U^- = S$

## Proof

Let $T = \struct {S, \tau}$ 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 $U \ne \O$.

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, $U \cap V \ne \O$.

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.

$\blacksquare$