204: Subspace That Contains Connected Subspace and Is Contained in Closure of Connected Subspace Is Connected

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A description/proof of that subspace that contains connected subspace and is contained in closure of connected subspace is connected

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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 space.
• The reader knows a definition of closure of subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space and any connected subspace, any subspace that contains the connected subspace and is contained in the closure of the connected subspace is 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: Description

For any topological space, $T$, and any connected subspace, $T_1\subseteq T$, any subspace, $T_2\subseteq T$, such that $T_1\subseteq T_2\subseteq \overline{T_1}$ is connected.

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2: Proof

Suppose that $T_2$ was not connected. $T_2=U_1\cup U_2$, $U_1\cap U_2=\mathrm{\varnothing }$ where $U_i$ is a non-empty open set on $T_2$ and $U_i={U_i}^{\prime }\cap T_2$ where ${U_i}^{\prime }$ is open on $T$. $U_i={U_i}^{\prime }\cap (U_1\cup U_2)=({U_i}^{\prime }\cap U_1)\cup ({U_i}^{\prime }\cap U_2)$. $U_1=({U_1}^{\prime }\cap U_1)\cup ({U_1}^{\prime }\cap U_2)$ and ${U_1}^{\prime }\cap U_2=\mathrm{\varnothing }$; likewise ${U_2}^{\prime }\cap U_1=\mathrm{\varnothing }$. $T_1$ could not be separated into $U_1$ and $U_2$, because otherwise, $T_1=S_1\cup S_2$ where $S_i\subseteq U_i$, $S_1\cap S_2=\mathrm{\varnothing }$, $S_i={U_i}^{\prime }\cap S_i$ as $U_i={U_i}^{\prime }\cap U_i$, $S_i={U_i}^{\prime }\cap (S_1\cup S_2)={U_i}^{\prime }\cap T_1$, so, $T_1$ would not be connected. Suppose that $T_1\subseteq U_1$ without loss of generality. $T_1\subseteq U_1\subseteq T\setminus {U_2}^{\prime }$, because ${U_2}^{\prime }\cap U_1=\mathrm{\varnothing }$. So, $T\setminus {U_2}^{\prime }$ is a closed set that contains $T_1$, so, $\overline{T_1}\subseteq T\setminus {U_2}^{\prime }$, but as $U_2\subseteq {U_2}^{\prime }$, it would not be that $T_2\subseteq \overline{T_1}$, a contradiction. So, $T_2$ is connected.

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3: Note

As $T_2$ can be taken to be $\overline{T_1}$, $\overline{T_1}$ is connected.

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References

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