216: In Nest of Topological Subspaces, Connected-ness of Subspace Does Not Depend on Superspace

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A description/proof of that in nest of topological subspaces, connected-ness of subspace does not depend on superspace

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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.
• The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.

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Target Context

• The reader will have a description and a proof of the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a 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 nest of topological subspaces, $T_1,T_2$, such that $T_2\subseteq T_1\subseteq T$, if $T_2$ is connected or disconnected as a subspace of $T_1$, $T_2$ is connected or disconnected respectively as a subspace of $T$; if $T_2$ is connected or disconnected as a subspace of $T$, $T_2$ is connected or disconnected respectively as a subspace of $T_1$.

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2: Proof

Suppose that $T_2$ is disconnected as a subspace of $T_1$. $T_2=U_1\cup U_2$, $U_1\cap U_2=\mathrm{\varnothing }$ where $U_i$ is non-empty open on $T_2$. By the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, $U_i$ is open with $T_2$ regarded to be a subspace of $T$. So, $T_2$ is disconnected as a subspace of $T$.

Suppose that $T_2$ is disconnected as a subspace of $T$. Likewise, $T_2$ is disconnected as a subspace of $T_1$.

As [$T_2$ is disconnected as a subspace of $T_1$] $⟹$ [$T_2$ is disconnected as a subspace of $T$], [$T_2$ is not disconnected as a subspace of $T$] $⟹$ [$T_2$ is not disconnected as a subspace of $T_1$], but as "not disconnected" is nothing but 'connected', [$T_2$ is connected as a subspace of $T$] $⟹$ [$T_2$ is connected as a subspace of $T_1$].

Likewise, [$T_2$ is connected as a subspace of $T_1$] $⟹$ [$T_2$ is connected as a subspace of $T$].

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References

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