Equivalence of Definitions of Irreducible Space/3 iff 7
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:
$\blacksquare$