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

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 $T = \struct{S, \tau}$ be locally path-connected by Definition 4:
 * the components of the open sets of $T$ are also open in $T$.

That is, $T = \struct{S, \tau}$ is locally path-connected by Definition 3.