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I'm trying to prove that a manifold $M$, that is connected, is pathwise connected. I know the standard proof of this theorem: just use that the set of points that can be joined to a point $x \in M$ is clopen. However, I'm trying to find a more direct proof. I want to find explicitly a path that connect any two given points $x, y \in M$. Of course, I can create a sequence of points $(z_\alpha)_{\alpha \in A}$ indexed by some ordinal and consequently a sequence of sets $(U_\alpha)_{\alpha \in A}$ homeomorphic to the euclidean space, such that each open set $U_{z_{\alpha}}$ intersects with its successor and, then, concatenate each path to the other inside the intersection. However I don't know how to find a finite sequence (whether it's possible) nor to proof that this concatenation of paths will be continuous. Is there some way to finish it or an analogous proof?

Thanks in advance.

user40276
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Here is my partial attempt, which eventually gives a constructive proof that connected and compact manifolds are path-connected.

Lemma 1: Any connected set that admits a finite open cover $\{U_i\}_{i = 0}^n$ of path-connected sets is itself path-connected.

Proof:

$U_0$ is path-connected.

For $n \geq 1$, assume by induction that $\bigcup_{i \in I}U_i$ is path-connected for any $I \subset \{0, 1, \dotsc, n\}$ (proper subset) such that $\bigcup_{i \in I}U_i$ is connected.

Suppose that $\bigcup_{i = 0}^nU_i$ is connected; let $\Gamma$ be the set of components of $(\bigcup_{i = 0}^nU_i) - U_0$; if $\Gamma = \emptyset$, then $\bigcup_{i = 0}^nU_i \subseteq U_0$ which is path-connected. Otherwise, choose some $C \in \Gamma$. Note that $$ V = \bigcup\{U_i\ |\ i \in \{0, 1, \dotsc, n\}, U_i \cap C \neq \emptyset\} $$

is connected, and thus path-connected by the induction hypothesis ($U_0 \nsubseteq V$). By a theorem of Kuratowski's, $W = (\bigcup_{i = 0}^nU_i) - C$ is connected. Since $V$ and $W$ form an open cover of the connected $\bigcup_{i = 0}^nU_i$, they must intersect. It follows immediately that $\bigcup_{i = 0}^nU_i = V \cup W$ is path-connected.

In particular, since $V \cap W$ contains some $x$, we may concatenate paths with endpoint $x$ to explicitly construct paths, which is what the question wishes for. This construction can be inductively built up from cases with lower $n$. $\Box$

Corollary 1: Any connected manifold $M$ that admits a finitary atlas $\{(U_i, \phi_i\}_{i \in I}$ is path-connected.

Proof:

Apply Lemma 1 by noting that $\{U_i\}_{i \in I}$ is a finite open cover of the connected $M$, where each $U_i$ is homeomorphic to Euclidean space, which is path-connected. $\Box$

Proposition 1: Any connected and compact manifold $M$ is path-connected.

Proof:

Consider the atlas $\{(U_x, \phi_x)\}_{x \in M}$, where $U_x$ is some open neighborhood of $x$ that admits a homeomorphism $\phi_x$ to Euclidean space. $\{U_x\}_{x \in M}$ is an open cover of $M$, thus there exists a finite $I \subseteq M$ such that $\bigcup_{x \in I}U_x = M$. Apply Corollary 1 to the finitary atlas $\{(U_x, \phi_x)\}_{x \in I}$. $\Box$

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