Equivalence of Definitions of Locally Connected Space/Definition 4 implies Definition 3
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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 $\BB = \set {U \in \tau : U \text{ is connected in } T}$.
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, $\BB$ is an basis for $T$.
$\blacksquare$