# Irreducible Space is Locally Connected

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## Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is irreducible.

Then $T$ is locally connected.

## Proof

Let $T = \left({S, \tau}\right)$ 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

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 4$