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

Theorem
Let $T = \struct {S, \tau}$ be a topological space. Let the components of the open sets of $T$ are also open in $T$.

Then
 * $T$ has a basis consisting of connected sets in $T$.

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

Let $U$ be open in $T$.

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

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

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

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

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