217: Union of 2 Connected Subspaces Is Connected if Each Neighborhood of Point on Subspace Contains Point of Other Subspace

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A description/proof of that union of 2 connected subspaces is connected if each neighborhood of point on subspace contains point of other subspace

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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 subspace topology.
• The reader knows a definition of connected topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace.

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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 subspaces, $T_1,T_2\subseteq T$, $T_1\cup T_2$ is connected if there is a point, $p\in T_1$, whose each neighborhood, $U_p\subseteq T$, contains a point on $T_2$.

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2: Proof

Suppose that there is such a point, $p$. Suppose that $T_1\cup T_2$ was not connected. $T_1\cup T_2=U_1\cup U_2$, $U_1\cap U_2=\mathrm{\varnothing }$ where $U_i$ would be nonempty open on $T_1\cup T_2$, and $U_i={U_i}^{\prime }\cap (T_1\cup T_2)$ where ${U_i}^{\prime }$ would be open on $T$. Without loss of generality, $p\in U_1$. ${U_1}^{\prime }$ would be a neighborhood of $p$, so, would contain a point, $p^{\prime }\in T_2$. $T_2=(({U_1}^{\prime }\cup {U_2}^{\prime })\cap T_2)$, as $U_1\cup U_2$ would contain $T_2$ and ${U_1}^{\prime }\cup {U_2}^{\prime }$ would contain $U_1\cup U_2$. $T_2=({U_1}^{\prime }\cap T_2)\cup ({U_2}^{\prime }\cap T_2)$. ${U_1}^{\prime }$ and ${U_2}^{\prime }$ would not share any point on $T_2$, because otherwise, $U_1$ and $U_2$ would share the point. As $T_2$ would be connected, ${U_2}^{\prime }\cap T_2$ would have to be empty as $p^{\prime }\in {U_1}^{\prime }$. So, ${U_2}^{\prime }$ would have to contain a point of $T_1$, but then, $T_1=({U_1}^{\prime }\cup {U_2}^{\prime })\cap T_1=({U_1}^{\prime }\cap T_1)\cup ({U_2}^{\prime }\cap T_1)$, a union of disjoint nonempty open sets, because ${U_1}^{\prime }$ and ${U_2}^{\prime }$ would not share any point on $T_1$, a contradiction. So, $T_1\cup T_2$ is connected.

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References

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