202: Connected Topological Component Is Exactly Connected Topological Subspace That Cannot Be Made Larger

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A description/proof of that connected topological component is exactly 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 connected topological component.

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Target Context

• The reader will have a description and a proof of the proposition that any connected topological component is exactly any 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 connected topological component, $T_1\subseteq T$, is exactly any connected topological subspace that cannot be made larger.

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2: Proof

Is $T_1$ connected? Suppose that $T_1$ was not connected. $T_1=U_1\cup U_2$, $U_1\cap U_2=\mathrm{\varnothing }$ where $U_i$ would be a non-empty open set on $T_1$. $U_i={U_i}^{\prime }\cap T_1$ where ${U_i}^{\prime }$ would be an open set on $T$, by the definition of subspace topology. There would be 2 points, $p_1,p_2\in T_1$, such that $p_1\in {U_1}^{\prime }$ and $p_2\in {U_2}^{\prime }$. As $p_1$ and $p_2$ would be connected, there would be a connected subspace, $p_1,p_2\in T_2\subseteq T$, but in fact, $T_2\subseteq T_1$, because all the points on $T_2$ belong to the equivalence class. $T_2=T_2\cap ({U_1}^{\prime }\cup {U_2}^{\prime })=(T_2\cap {U_1}^{\prime })\cup (T_2\cap {U_2}^{\prime })$, because $U_1\cup U_2$ would contain $T_1$, so, ${U_1}^{\prime }\cup {U_2}^{\prime }$ would contain $T_2$. ${U_1}^{\prime }$ and ${U_2}^{\prime }$ would not share any point on $T_2$, because otherwise, $U_1$ and $U_2$ would share the point. So, $T_2$ would not be connected, a contradiction, so, $T_1$ is connected.

Adding any point to $T_1$ makes the result not a connected topological subspace, because the added point does not belong to the equivalence class, which means that there is no 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 connected.

Suppose that $T_3$ is any 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 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 connected topological component is connected, because connected topological component is defined based on connected-ness of any pair of points in the component, which is about the existence of a connected topological subspace, which is not necessarily the component; the component is certainly the union of such connected subspaces, but there is no guarantee that such a union is connected, although the union of any sequence of connected subspaces that share a point pair-wise sequentially is connected.

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References

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