Irreducible Space is Locally Connected
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Theorem
Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is locally connected.
Proof
Let $T = \struct {S, \tau}$ be irreducible.
Then:
- $\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
So trivially there are no two open sets that can form a separation of $T$.
As a basis consists of open sets, this applies to all sets in a basis for $T$.
The result follows from definition of locally connected.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness