# Path Components are Open iff Union of Open Path-Connected Sets/Lemma 1

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## Theorem

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

Let $C$ be a path component of $T$.

Let $U$ be a path-connected set of $T$.

Then:

- $U \cap C \neq \O$ if and only if $U \neq \O$ and $U \subseteq C$

## Proof

### Necessary Condition

Let $U \cap C \neq \O$.

From Union of Path-Connected Sets with Common Point is Path-Connected, $U \cup C$ is a path-connected set of $T$.

From Set is Subset of Union, $C \subseteq U \cup C$.

By definition of a path component, $C$ is a maximal path-connected set.

Hence $C = U \cup C$.

From Union with Superset is Superset, $U \subseteq C$.

Then:

\(\displaystyle U\) | \(=\) | \(\displaystyle U \cap C\) | Intersection with Subset is Subset | ||||||||||

\(\displaystyle \) | \(\neq\) | \(\displaystyle \O\) | Assumption |

The result follows.

$\Box$

### Sufficient Condition

Let $U \neq \O$ and $U \subseteq C$.

Then:

\(\displaystyle U \cap C\) | \(=\) | \(\displaystyle U\) | Intersection with Subset is Subset | ||||||||||

\(\displaystyle \) | \(\neq\) | \(\displaystyle \O\) | Assumption |

The result follows.

$\blacksquare$