Path-Connected iff Path-Connected to Point

Theorem

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

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

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

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

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

$\Box$

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.

$\blacksquare$

One might think that this gives an equivalent definition for path-connectedness.

However, the definition given incorporates the empty set by vacuous truth, whilst the condition given on this page does not.

When desired, this may be overcome by handling the empty set separately.