Equivalence of Definitions of Irreducible Space/3 iff 7

Theorem
Let $X$ be a topological space.


 * $(1):\quad$ $X$ is irreducible
 * $(2):\quad$ Every open subset of $X$ is connected

1 implies 2
Let $X$ be irreducible.

Let $U \subseteq X$ be an open subset.

Support $U$ not connected.

Then there exist nonempty open subsets $V,W$ of $U$ that are disjoint (and whose union is $U$).

By Open Set in Open Subspace, $V$ and $W$ are open subsets of $X$.

Because $V\cap W = \varnothing$, $X$ is not irreducible.

This is a contradiction.

Thus $U$ is connected.

2 implies 1
Let $V$ and $W$ be open subsets of $X$.

Then their union $V\cup W$ is open in $X$.

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

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

Because $V\cup W$ is connected, $V\cap W$ is nonempty.

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

Also see

 * Open Subset of Irreducible Space is Irreducible
 * Space is Ultraconnected iff Closed Subsets are Connected, whose proof is almost the same