4.14 Local Path-Connectedness
Definition 4.14.1 (Locally Path-Connected). Let $X$ be a topological space, then $X$ is locally path-connected if for every $x \in X$, there exists a fundamental system of neighbourhoods consisting of path-connected sets at $x$.
Proposition 4.14.2. Let $X$ be a locally path-connected space, then
The path components of $X$ are open.
The path components of $X$ are equal to its components.
$X$ is connected if and only if it is path-connected.
Every open subset of $X$ is locally path-connected.
Proof. (1): Let $A \subset X$ be a path component and $x \in A$, then there exists $U \in \cn(x)$ connected. By Proposition 4.13.4, $A \cup U \in \cn(x)$ is also connected. Since $A$ is a path-component, $A \cup U \subset A \in \cn(x)$. Thus $A$ is open by Lemma 4.4.3.
(2): Let $P$ be a path component in $X$ and $C \supset P$ be its connected components. If $C$ is not path-connected, then $P \subsetneq C$ there exists path-components $\seqi{P}\subset 2^{C}$ such that $C = \bigsqcup_{i \in I}P_{i}$. In which case, $C \setminus P$ is open by (1), and $C$ is not connected, which is impossible.
(3): By (2) applied to $X$.$\square$