# Locally Connected iff Components of Open Subsets are Open

This article has been proposed for deletion. In particular: This has been subsumed by Equivalence of Definitions of Locally Connected Space, which subsumes this: the relevant section is $(3)$ implies $(4)$. |

## Theorem

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

Then $T$ is locally connected if and only if the connected components of open sets of $T$ are also open in $T$.

## Proof

### Necessary Condition

Follows directly from:

- Open Subset of Locally Connected Space is Locally Connected
- Component of Locally Connected Space is Open
- Open Set in Open Subspace

$\blacksquare$

### Sufficient Condition

## Also see

- Locally Path-Connected iff Path Components of Open Subsets are Open, an analogous result for path components