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.

$\blacksquare$


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