Equivalence of Definitions of Locally Connected Space

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

1 implies 2
Follows directly from Local Basis is Neighborhood Basis.

2 implies 1
Let $x\in S$ and $x \in U \in \tau$.

By definition of local basis, we have to show that there exists a connected open set $V\in\tau$ with $x\in V\subset U$.

Let $V = \operatorname{Comp}_x \left({T}\right)$ denote the component of $x$ in $U$.

By definition of topological subspace, it suffices to show that $V$ is open in $U$.

By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points.

Let $y\in V$.

By assumption, there exists a connected neighborhood $W$ of $y$ in $T$ with $W\subset U$.

By Neighborhood in Topological Subspace iff Intersection of Neighborhood and Subspace, $W$ is a neighborhood of $y$ in $U$.

By definition of component:
 * $W\subset\operatorname{Comp}_y \left({U}\right) = \operatorname{Comp}_x \left({U}\right) = V$

By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$.

Because $y$ was arbitrary, $V$ is open in $U$.

Because $U$ was arbitrary, $x$ has a local basis of connected (open) sets.

1 implies 3
Follows directly from Union of Local Bases is Basis.

3 implies 1
Follows directly from Basis induces Local Basis.

Also see

 * Locally Connected iff Components of Open Subsets are Open
 * Equivalence of Definitions of Locally Path-Connected Space, whose proof is almost the same