# Equivalence of Definitions of Irreducible Space

## Theorem

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

### Definition 1

A topological space $T = \struct {S, \tau}$ is **irreducible** if and only if there exists no cover of $T$ by two proper closed sets of $T$.

### Definition 2

A topological space $T = \struct {S, \tau}$ is **irreducible** if and only if there is no finite cover of $T$ by proper closed sets of $T$.

### 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 4

A topological space $T = \struct {S, \tau}$ is **irreducible** if and only if every non-empty open set of $T$ is (everywhere) dense in $T$.

### Definition 5

A topological space $T = \struct {S, \tau}$ is **irreducible** if and only if the interior of every proper closed set of $T$ is empty.

### 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:

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

### Definition 7

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

## Proof

### $(1)$ implies $(2)$

Let $T$ be an irreducible space by Definition 1.

That is, $T$ is not the union of any two proper closed sets.

Aiming for a contradiction, suppose $T$ admits a finite cover by proper closed sets.

By the Well-Ordering Principle, there exists a minimal natural number $n \in \N$ such that $T$ has a cover by $n$ proper closed sets, say $F_1, \ldots, F_n$.

By definition of proper subset, it must be the case that $n > 1$.

By Finite Union of Closed Sets is Closed in Topological Space, $F_{n - 1} \cup F_n$ is closed.

- $S = F_1 \cup \cdots \cup \paren {F_{n - 1} \cup F_n}$

By minimality of $n$, $F_{n - 1} \cup F_n$ is not a proper subset.

Thus $F_{n - 1} \cup F_n = S$.

This contradicts the definition of an irreducible space by Definition 1.

Thus $T$ has no finite cover by proper closed sets.

That is, $T$ is an irreducible space by Definition 2.

$\Box$

### $(2)$ implies $(1)$

Let $T$ be an irreducible space by Definition 2.

Then $T$ has no finite cover by proper closed sets.

A cover by two sets is a finite cover.

Hence it immediately follows that $T$ has no finite cover by $2$ proper closed sets.

That is, $T$ is an irreducible space by Definition 1.

$\Box$

### $(1)$ implies $(3)$

Let $T$ be an irreducible space by Definition 1.

That is, $T$ is not the union of any two proper closed sets.

Let $U$ and $V$ be two arbitrary non-empty open sets of $T$.

Aiming for a contradiction, suppose $U$ and $V$ have an empty intersection.

By De Morgan's Laws, their complements are proper closed sets whose union is $T$.

This contradicts the definition of an irreducible space by Definition 1.

Thus:

- $U \cap V \ne \O$

As $U$ and $V$ are arbitrary, it follows that $T$ is an irreducible space by Definition 3.

$\Box$

### $(3)$ implies $(1)$

Let $T$ be an irreducible space by Definition 3.

Let $F$ and $G$ be two arbitrary proper closed sets of $T$

Aiming for a contradiction, suppose their union is $T$.

By De Morgan's Laws, their complements are non-empty open sets whose intersection is empty.

This contradicts the definition of an irreducible space by Definition 3.

Thus $F \cup G \ne T$.

As $F$ and $G$ are arbitrary, it follows that $T$ is an irreducible space by Definition 1.

$\Box$

### $(3)$ iff $(4)$

By definition of everywhere dense subset, an open set is dense if and only if it has non-empty intersection with any other open set.

$\Box$

### $(4)$ iff $(5)$

Follows from Everywhere Dense iff Interior of Complement is Empty.

$\Box$

### $(3)$ iff $(6)$

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.

$\Box$

### $(3)$ iff $(7)$

### 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:

$\blacksquare$