# Equivalence of Definitions of Irreducible Space/3 iff 7

## Contents

## Theorem

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

### Open Sets Intersect

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:

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

### Open Sets are Connected

A topological space $T = \struct {S, \tau}$ is **irreducible** if and only if every open set of $T$ is connected.

## Proof

### Open Sets Intersect implies Open Sets are Connected

Let $T = \struct {S, \tau}$ be irreducible in the sense that:

Let $U \subseteq S$ be an open set of $T$.

Aiming for a contradiction, suppose $U$ is not connected.

Then there exist non-empty open sets $V, W$ of $U$ that are disjoint and whose union is $U$.

By Open Set in Open Subspace, $V$ and $W$ are open sets of $T$.

Because $V \cap W = \O$, $T$ is not irreducible.

This is a contradiction.

Thus $U$ is connected.

Thus $T = \struct {S, \tau}$ is irreducible in the sense that:

$\Box$

### Open Sets are Connected implies Open Sets Intersect

Let $T = \struct {S, \tau}$ be irreducible in the sense that:

Let $V$ and $W$ be open sets of $T$.

By definition of topology, their union $V \cup W$ is open in $T$.

By hypothesis, $V \cup W$ is connected.

By Open Set in Open Subspace, $V$ and $W$ are open sets of $V \cup W$.

Because $V \cup W$ is connected, $V \cap W$ is non-empty.

Because $V$ and $W$ were arbitrary, $T$ is irreducible.

Thus $T = \struct {S, \tau}$ is irreducible in the sense that:

$\blacksquare$