1160: Subspace of Hausdorff Topological Space Is Hausdorff

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description/proof of that subspace of Hausdorff topological space is Hausdorff

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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: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of topological subspace.
• The reader knows a definition of Hausdorff topological space.

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Target Context

• The reader will have a description and a proof of the proposition that any subspace of any Hausdorff topological space is Hausdorff.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$T^{\prime }$: $\in \{\text{ the Hausdorff topological spaces }\}$
$T$: $\subseteq T^{\prime }$ with the subspace topology
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Statements:
$T\in \{\text{ the Hausdorff topological spaces }\}$
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2: Proof

Whole Strategy: Step 1: take any $t_1,t_2\in T$ such that $t_1\ne t_2$ and some open neighborhoods of $t_1,t_2$, $U_{t_1}^{\prime },U_{t_2}^{\prime }\subseteq T^{\prime }$, such that $U_{t_1}^{\prime }\cap U_{t_2}^{\prime }=\mathrm{\varnothing }$; Step 2: see that $U_{t_1}^{\prime }\cap T,U_{t_2}^{\prime }\cap T\subseteq T$ are some open neighborhoods of $t_1$ and $t_2$ and $(U_{t_1}^{\prime }\cap T)\cap (U_{t_2}^{\prime }\cap T)=\mathrm{\varnothing }$.

Step 1:

Let $t_1,t_2\in T$ be any such that $t_1\ne t_2$.

There are some open neighborhood of $t_1$ and $t_2$, $U_{t_1}^{\prime },U_{t_2}^{\prime }\subseteq T^{\prime }$, such that $U_{t_1}^{\prime }\cap U_{t_2}^{\prime }=\mathrm{\varnothing }$.

Step 2:

$U_{t_1}^{\prime }\cap T\subseteq T$ is an open neighborhood of $t_1$, because it is open on $T$ and $t_1\in U_{t_1}^{\prime }\cap T$.

$U_{t_2}^{\prime }\cap T\subseteq T$ is an open neighborhood of $t_2$, likewise.

$(U_{t_1}^{\prime }\cap T)\cap (U_{t_2}^{\prime }\cap T)=\mathrm{\varnothing }$, because $(U_{t_1}^{\prime }\cap T)\cap (U_{t_2}^{\prime }\cap T)\subseteq U_{t_1}^{\prime }\cap U_{t_2}^{\prime }=\mathrm{\varnothing }$.

So, $T$ is Hausdorff.

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References

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