2023-02-19

206: 2 Points Are Topologically Path-Connected iff There Is Path That Connects 2 Points

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A description/proof of that 2 points are topologically path-connected iff there is path that connects 2 points

Topics


About: topological space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that any 2 points are path-connected on any topological space if and only if there is a path that connects the 2 points on the topological space.

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.


Main Body


1: Description


For any topological space, \(T\), any points, \(p_1, p_2 \in T\), are path-connected if and only if there is a path, \(\lambda: [r_1, r_2] \rightarrow T\), such that \(\lambda (r_1) = p_1\) and \(\lambda (r_2) = p_2\)


2: Proof


Suppose that there is a \(\lambda\). Let us prove that \(T_1 := \lambda ([r_1, r_2])\) is a path-connected topological subspace. For any points, \(p_3, p_4 \in T_1\), \(p_3 = \lambda (r_3)\) and \(p_4 = \lambda (r_4)\), and \({\lambda}': [r_3, r_4] \rightarrow T_1\), which is the restriction of \(\lambda\) on the domain and the codomain, is a path on \(T_1\), by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. So, \(p_3\) and \(p_4\) are path-connected on \(T_1\). So, \(p_1\) and \(p_2\) are path-connected on \(T\), as there is a path-connected topological subspace that contains the both points.

Suppose that \(p_1\) and \(p_2\) are path-connected. There is a path-connected topological subspace, \(p_1, p_2 \in T_1 \subseteq T\). There is a path, \(\lambda': [r_1, r_2] \rightarrow T_1\), that connects the 2 points, but by the proposition that any expansion of any continuous map on the codomain is continuous, \(\lambda: [r_1, r_2] \rightarrow T\) is a path on \(T\).


3: None


This proposition may be a definition for some people, but as I have adopted another definition, this proposition is due.


References


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