47: Connected Topological Manifold Is Path-Connected

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description/proof of that connected topological manifold is path-connected

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Topics

About: topological manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof
• 4: Note

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Starting Context

• The reader knows a definition of topological manifold.
• The reader knows a definition of connected topological space.
• The reader knows a definition of path-connected topological space.
• The reader admits the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
• The reader admits the local criterion for openness.

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Target Context

• The reader will have a description and a proof of the proposition that any connected topological manifold is path-connected.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the connected topological manifolds }\}$
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Statements:
$M$: $\in \{\text{ the path-connected topological manifolds }\}$
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2: Natural Language Description

Any connected topological manifold, $M$, is path-connected.

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3: Proof

Whole Strategy: take any point, $p\in M$, and the subset, $S\subseteq M$, as all the points path-connected with $p$, and see that $S=M$; Step 1: take any point, $p\in M$, and the subset, $S\subseteq M$, as all the points path-connected with $p$; Step 2: see that $S$ is open; Step 3: see that $M\setminus S$ is open; Step 4: conclude the proposition.

Step 1:

For any point, $p\in M$, take the subset, $S\subseteq M$, as all the points path-connected with $p$.

Step 2:

Let us prove that $S$ is open.

Let $p_1\in S$ be any. There are an open neighborhood of $p_1$, $U_{p_1}\subseteq M$, and a homeomorphism, ${\varphi }_{p_1}:U_{p_1}\to U_{{\varphi }_{p_1}(p_1)}$, where $U_{{\varphi }_{p_1}(p_1)}\subseteq {\mathbb{R}}^n$ is open on ${\mathbb{R}}^n$.

There is an open ball, $B_{{\varphi }_{p_1}(p_1),ϵ}\subseteq {\mathbb{R}}^n$, such that $B_{{\varphi }_{p_1}(p_1),ϵ}\subseteq U_{{\varphi }_{p_1}(p_1)}$. $B_{{\varphi }_{p_1}(p_1),ϵ}$ is open on $U_{{\varphi }_{p_1}(p_1)}$, and ${\varphi }_{p_1}^{-1}(B_{{\varphi }_{p_1}(p_1),ϵ})\subseteq U_{p_1}$ is open on $U_{p_1}$ and on $M$.

Any $p_2\in {\varphi }_{p_1}^{-1}(B_{{\varphi }_{p_1}(p_1),ϵ})$ is path-connected to $p_1$ on ${\varphi }_{p_1}^{-1}(B_{{\varphi }_{p_1}(p_1),ϵ})$ and on $M$, because ${\varphi }_{p_1}(p_2)$ is path-connected to ${\varphi }_{p_1}(p_1)$ on $B_{{\varphi }_{p_1}(p_1),ϵ}$ and the path can be mapped into ${\varphi }_{p_1}^{-1}(B_{{\varphi }_{p_1}(p_1),ϵ})$ by ${\varphi }_{p_1}^{-1}$: for the path, $\lambda :J\subseteq \mathbb{R}\to B_{{\varphi }_{p_1}(p_1),ϵ}$, take ${\varphi }_{p_1}^{-1}\circ \lambda :J\to {\varphi }_{p_1}^{-1}(B_{{\varphi }_{p_1}(p_1),ϵ})$, which is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.

$p_2$ is path-connected to $p$ on $M$, because $p_2$ can be path-connected to $p$ via $p_1$. So, ${\varphi }_{p_1}^{-1}(B_{{\varphi }_{p_1}(p_1),ϵ})\subseteq S$.

By the local criterion for openness, $S$ is open on $M$.

Step 3:

Let us prove that $M\setminus S\subseteq M$ is open.

Let $p_3\in M\setminus S$ be any. There are an open neighborhood of $p_3$, $U_{p_3}\subseteq M$, and a homeomorphism, ${\varphi }_{p_3}:U_{p_3}\to U_{\varphi (p_3)}$, where $U_{{\varphi }_{p_3}(p_3)}\subseteq {\mathbb{R}}^n$ is open on ${\mathbb{R}}^n$. There is an open ball, $B_{{\varphi }_{p_3}(p_3),ϵ}\subseteq {\mathbb{R}}^n$, such that $B_{{\varphi }_{p_3}(p_3),ϵ}\subseteq U_{{\varphi }_{p_3}(p_3)}$. $B_{{\varphi }_{p_3}(p_3),ϵ}$ is open on $U_{{\varphi }_{p_3}(p_3)}$, and ${\varphi }_{p_3}^{-1}(B_{{\varphi }_{p_3}(p_3),ϵ})\subseteq U_{p_3}$ is open on $U_{p_3}$ and on $M$.

Any $p_4\in {\varphi }_{p_3}^{-1}(B_{{\varphi }_{p_3}(p_3),ϵ})$ is path-connected to $p_3$ on ${\varphi }_{p_3}^{-1}(B_{{\varphi }_{p_3}(p_3),ϵ})$ and on $M$, because ${\varphi }_{p_3}(p_4)$ is path-connected to ${\varphi }_{p_3}(p_3)$ on $B_{{\varphi }_{p_3}(p_3),ϵ}$ and the path can be mapped into ${\varphi }_{p_3}^{-1}(B_{{\varphi }_{p_3}(p_3),ϵ})$ by ${\varphi }_{p_3}^{-1}$, as before.

$p_4$ is not path-connected to $p$ on $M$, because otherwise, $p_3$ would be path-connected to $p$ via $p_4$, a contradiction. So, ${\varphi }_{p_3}^{-1}(B_{{\varphi }_{p_3}(p_3),ϵ})\subseteq M\setminus S$.

By the local criterion for openness, $M\setminus S$ is open on $M$.

Step 4:

Now $S\cap (M\setminus S)=\mathrm{\varnothing }$ and $M=S\cup (M\setminus S)$. As $M$ is connected, one of the 2 subsets has to be empty, but $S$ is not empty containing at least $p$, so, $M\setminus S=\mathrm{\varnothing }$, and so, $S=M$, which means that $M$ is path-connected.

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4: Note

This proposition is not about topological space in general, but about topological manifold.

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References

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