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

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## Theorem

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

Let the path components of open sets of $T$ be also open in $T$.

Then

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

## Proof

Let $\mathcal B = \set{U \in \tau : U \text{ is path-connected in } T}$.

Let $U$ be open in $T$.

By assumption, the path components of $U$ are open in $T$.

From Path-Connected Set in Subspace, the path components of $U$ are path-connected in $T$.

By the definition of the path components of a topological space, $U$ is the union of its path components.

Hence $U$ is the union of open path-connected sets in $T$.

By definition, $\mathcal B$ is an basis for $T$.

$\blacksquare$