# Equivalence of Definitions of Locally Path-Connected Space/Definition 3 implies Definition 4

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## Theorem

Let $T = \struct{S, \tau}$ be a topological space.

Let $T$ have a basis consisting of path-connected sets in $T$.

Then

- the path components of open sets of $T$ are also open in $T$.

## Proof

Let $U$ be an open subset of $T$.

From Open Set is Union of Elements of Basis, $U$ is a union of open path-connected sets in $T$.

From Open Set in Open Subspace and Path-Connected Set in Subspace, $U$ is a union of open path-connected sets in $U$.

From Path Components are Open iff Union of Open Path-Connected Sets, the path components of $U$ are open in $U$.

From Open Set in Open Subspace then the path components of $U$ are open in $T$.

$\blacksquare$