Equivalence of Definitions of Irreducible Space

$(1)$ implies $(2)$
Let $T$ be an irreducible space by Definition 1.

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

$T$ 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.

By Union is Associative:
 * $S = F_1 \cup \cdots \cup \paren {F_{n - 1} \cup F_n}$

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.

$(2)$ 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.

$(1)$ 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 $T$.

$U$ 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.

$(3)$ implies $(1)$
Let $T$ be an irreducible space by Definition 3.

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

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.

$(3)$ iff $(4)$
By definition of everywhere dense subset, an open set is dense it has non-empty intersection with any other open set.

$(4)$ iff $(5)$
Follows from Everywhere Dense iff Interior of Complement is Empty.