219: Path-Connected Topological Component Is Open and Closed on Locally Path-Connected Topological Space

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A description/proof of that path-connected topological component is open and closed on locally path-connected topological space

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of path-connected topological component.
• The reader knows a definition of locally path-connected topological space.
• The reader knows a definition of closed set.
• The reader admits the proposition that any 2 points that are path-connected on any topological subspace are path-connected on any larger subspace.
• The reader admits the local criterion for openness.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that any path-connected topological component is open and closed on any locally path-connected topological space.

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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: Description

For any locally path-connected topological space, $T$, any path-connected component, $T_1\subseteq T$, is open and closed.

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2: Proof

Around any point, $p\in T_1$, there are a neighborhood, $p\in N_p\subseteq T$, and a path-connected neighborhood, $p\in {N_p}^{\prime }\subseteq N_p$. ${N_p}^{\prime }\subseteq T_1$, because each point on ${N_p}^{\prime }$ is path-connected with $p$ on ${N_p}^{\prime }$, so, also on $T$, by the proposition that any 2 points that are path-connected on any topological subspace are path-connected on any larger subspace. So, $T_1$ is open by the local criterion for openness.

Suppose that $T_1$ was not closed. While the closure $\overline{T_1}$ would be closed, there would be a point, $p\in \overline{T_1}$ and $p\notin T_1$. By the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset, $p$ would be an accumulation point of $T_1$. For any neighborhood of $p$, $N_p$, there would be a path-connected neighborhood, $p\in {N_p}^{\prime }\subseteq N_p$. As $p$ would be an accumulation point of $T_1$, ${N_p}^{\prime }$ would contain a point of $T_1$, so, $p$ would be path-connected with the point of $T_1$ on ${N_p}^{\prime }$, and would be so on $T$, by the proposition that any 2 points that are path-connected on any topological subspace are path-connected on any larger subspace. So, $p$ would be in $T_1$, a contradiction. So, $T_1$ is closed.

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References

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