# Equivalent Definition for Locally Connected

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## Theorem

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

Then these definitions for local connectedness are logically equivalent:

- $T$ is locally connected if and only if the components of open subsets of $S$ are also open in $S$.

- $T$ is locally connected if and only if there is a base consisting entirely of connected sets.

## Proof

Assume that $T$ is such that it has a basis $\mathcal B$ which consists entirely of connected sets.

For each $x \in S$ we define $\mathcal B_x = \left\{{B \in \mathcal B: x \in B}\right\}$.

This is a local basis.

As all the elements of $\mathcal B_x$ is also an element of $\mathcal B$, it follows that $\mathcal B_x$ is also formed of connected sets.

Thus, for each point $x \in S$, there is a local basis which consists entirely of connected sets.

Thus, $T$ is locally connected by definition.

$\Box$

Assume that $T$ is locally connected.

Then by definition, for each point $x \in S$, there exists a local basis $\mathcal D_x$ which consists entirely of open sets, each of which is connected.

Consider the set $\displaystyle \mathcal D = \bigcup_{x \in S} \mathcal D_x$.

From Union of Local Bases is Basis, $\mathcal D$ is a basis for the topology $\tau$.

Since each $\mathcal D_x$ consists entirely of connected sets, $\mathcal D$, the set of all this sets by definition, also consists entirely of connected sets.

$\blacksquare$