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 $\map \gamma b = q$

Lemma 1
The $\sim$ is a equivalent relation on $M$.

Proof of Lemma 1
For each $p \in M$, we define:
 * $N_p := \set {q \in M : p \sim q}$

Now we need to show that:
 * $\forall p \in M : M = N_p$

Lemma 2
For each $p \in N$, $N_p$ is a non-empty open set.

Proof of Lemma 2
Let $p \in M$.

Let $q \in N_p$.

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

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

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

We shall show that:
 * $V \subseteq N_p$

Let $r \in V \setminus \set q$.

Define a regular curve segment:
 * $\ell_r : \closedint 0 1 \to M$

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

In particular, $\ell_r$ is an admissible curve such that:
 * $\map {\ell_r} a = q$
 * $\map {\ell_r} b = r$

Therefore:
 * $q \sim r$

As $p \sim q$, we have by Lemma 2:
 * $p \sim r$

That is:
 * $r \in N_p$

Lemma 3
If $N_p \cap N_q \ne \O$, then $N_p = N_q$.

Proof of Lemma 3
Let $r \in N_p \cap N_q$.

That is, $r \sim p$ and $r \sim q$.

By Lemma 1, then $p \sim q$.

Let $p \in M$.

Recall $N_p$ is an non-empty open set.

Observe:
 * $\ds M \setminus N_p = \bigcup_{q \mathop \in M \setminus N_p} N_q$

is also an open set.

Thus $N_p$ is an non-empty clopen set.

Since $M$ is connected, we have:
 * $M = N_p$