Locally Path-Connected Space is not necessarily Path-Connected

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is locally path-connected.


Then it is not necessarily the case that $T$ is also path-connected.


Proof

Let $T := \left({\R, \tau_d}\right)$ be the real number line $\R$ under the usual (Euclidean) topology $\tau_d$.

Let $a, b, c \in \R$ where $a < b < c$.

Let $A$ be the union of the two open intervals:

$A := \left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)$

Let $T' := \left({A, \tau_A}\right)$ be the subspace composed of $A$ with the subspace topology induced on $\tau_d$ by $A$.


From Union of Adjacent Open Intervals is Locally Path-Connected, $T'$ is a locally path-connected space.

From Union of Adjacent Open Intervals is not Path-Connected, $T$ is not a path-connected space.

Hence the result.

$\blacksquare$


Sources