Path-Connected iff Path-Connected to Point

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

Let $U \subseteq S$ be a non-empty subset of $T$.

Then $U$ is path-connected iff:


 * $\exists p \in U: \forall q \in U: \exists f: \left[{0 .. 1}\right] \to U: f $ continuous, $f \left({0}\right) = p$ and $f \left({1}\right) = q$

Necessary Condition
When $U$ is path-connected, any $p \in U$ satisfies the assertion.

Because $U$ is non-empty, such a $p$ exists.

Sufficient Condition
Suppose $p$ is such a point.

For $q \in U$, denote $p \sim q$ when $p$ and $q$ are path-connected.

To show that $U$ is path-connected, we have to show:


 * $\forall q,r \in U: q \sim r$

We already know that $p \sim q$ and $p \sim r$.

From Path-Connectedness is Equivalence Relation, we find that hence $q \sim p$ by symmetry of $\sim$.

Subsequently, we conclude $q \sim r$ from transitivity of $\sim$.

Hence the result.