209: Path-Connected Topological Component Is Exactly Path-Connected Topological Subspace That Cannot Be Made Larger

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A description/proof of that path-connected topological component is exactly path-connected topological subspace that cannot be made larger

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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
• 3: Note

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Starting Context

• The reader knows a definition of path-connected topological component.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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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 exactly any path-connected topological subspace that cannot be made larger.

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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 topological space, $T$, any path-connected topological component, $T_1\subseteq T$, is exactly any path-connected topological subspace that cannot be made larger.

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2: Proof

Is $T_1$ path-connected? For any points, $p_1,p_2\in T_1$, as $p_1$ and $p_2$ are path-connected on $T$, there is a path, $\lambda :[0,1]\to T$. $\lambda ([0,1])\subseteq T_1$, because for any point, $p_3\in \lambda ([0,1])$, $p_3=\lambda (r_3)$, and by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, ${\lambda }^{\prime }:[0,r_3]\to T=\lambda {|}_{[0,r_3]}$ is a path on $T$ that connects $p_1$ and $p_3$, so, $p_3\in T_1$. By the proposition that any restriction of any continuous map on the domain and the codomain is continuous, ${\lambda }^{″}:[0,1]\to T_1$ as the restriction of $\lambda$, is a path on $T_1$. So, yes, $T_1$ is path-connected.

Adding any point to $T_1$ makes the result not a path-connected topological subspace, because the added point does not belong to the equivalence class, which means that there is no path-connected topological subspace that contains the added point and a point of $T_1$, so, the result subspace that contains the both points cannot be path-connected.

Suppose that $T_3$ is any path-connected topological subspace that contains a point of $T_1$ and cannot be made larger. All the points of $T_3$ belong to the equivalence class of the point, so, $T_3\subseteq T_1$, but as $T_1$ is a path-connected subspace, $T_1\subseteq T_3$, so, $T_3=T_1$.

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3: Note

It is not so obvious that any path-connected topological component is path-connected, because path-connected topological component is defined based on path-connected-ness of any pair of points on the component, which is about the existence of a path-connected topological subspace, which is not necessarily the component; the component is certainly the union of such path-connected subspaces, but there is no guarantee that such a union is path-connected, although the union of any sequence of path-connected subspaces that share a point pair-wise sequentially is path-connected.

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References

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