In Connected Smooth Manifold Any Two Points can be Joined by Admissible Curve

Theorem
Let $M$ be a connected smooth manifold with or without a boundary.

Let $p, q \in M$ be points.

Let $\gamma : \closedint a b \to M$ be an admissible curve.

Then:


 * $\forall p, q \in M : \exists \gamma \subset M : \paren {\map \gamma a = p} \land \paren {\map \gamma b = q}$

Proof
For $p, q \in M$, we write:
 * $p \sim q$

there exists an admissible curve $\gamma : \closedint a b \to M$ such that $\map \gamma a = p$ and $\map \gamma b = q$

Let $p \in M$.

Let:
 * $N := \set {q \in M : p \sim q}$

Now we need to show that:
 * $M = N$

Note that $N \ne \O$, as $p \in N$.

Thus it suffices to show that $N$ is clopen, since $M$ is connected.

We claim that each $q \in M$ has a neighborhood $V$ such that either:
 * $V \subseteq N$

or:
 * $V \subseteq M \setminus N$

This means that bot $N$ and $M \setminus N$ are open.

Thus the theorem follows.

To this end, let $q \in M$.

Let $\struct {U, \phi}$ be a chart such that $q \in U$.

As $U$ is open, there is an open ball such that:
 * $\map {B_\epsilon} {\map \phi q} \subseteq U$

Let:
 * $V := {\phi ^{-1} } \sqbrk {\map {B_\epsilon} {\map \phi q} }$

We need to show that if $p \sim q$, then:
 * $\forall r \in V : p \sim r$

Observe that for each $r \in V$, we have a curve segment:
 * $\ell_r : \closedint 0 1 \to M$

defined by:
 * $\ds \map {\ell_r} t := \map {\phi^{-1} } { \map \phi q + t \paren {\map \phi r - \map \phi q} }$

Thus, for each admissible curve
 * $\gamma : \closedint a b \to M$

such that:
 * $\map \gamma a = p$
 * $\map \gamma b = q$

the concatenation of $\gamma$ and $\ell_{\map \gamma b}$:
 * $\tilde \gamma : \closedint a {b'} \to M$

is again an admissible curve.

By construction:
 * $\map {\tilde \gamma} a = p$
 * $\map {\tilde \gamma} {b'} = r$