759: Product of Hausdorff Topological Spaces Is Hausdorff

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description/proof of that product of Hausdorff topological spaces 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 1
• 2: Natural Language Description 1
• 3: Proof 1
• 4: Structured Description 2
• 5: Natural Language Description 2
• 6: Proof 2

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Starting Context

• The reader knows a definition of Hausdorff topological space.
• The reader knows a definition of product topology.

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Target Context

• The reader will have a description and a proof of the proposition that the product of any possibly uncountable number of Hausdorff topological spaces 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 1

Here is the rules of Structured Description.

Entities:
$A$: $\in \{\text{ the possibly uncountable infinite index sets }\}$
$\{T_{\alpha }|\alpha \in A\}$: $T_{\alpha }\in \{\text{ the Hausdorff topological spaces }\}$
$T$: $={\times }_{\alpha \in A}{T}_{\alpha }$, $=\text{ the product topological space }$
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Statements:
$T\in \{\text{ the Hausdorff topological spaces }\}$
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2: Natural Language Description 1

For any possibly uncountable infinite index set, $A$, any Hausdorff topological spaces, $\{T_{\alpha }|\alpha \in A\}$, and the product topological space, $T:={\times }_{\alpha \in A}{T}_{\alpha }$, $T$ is a Hausdorff topological space.

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3: Proof 1

Whole Strategy: Step 1: take any distinct $t_1,t_2\in T$ and see that $t_1({\alpha }^{\prime })\ne t_2({\alpha }^{\prime })$ for an ${\alpha }^{\prime }\in A$; Step 2: take a nonempty open neighborhood of $t_j({\alpha }^{\prime })$, $U_{{\alpha }^{\prime },t_j({\alpha }^{\prime })}\subseteq T_{{\alpha }^{\prime }}$, such that $U_{{\alpha }^{\prime },t_1({\alpha }^{\prime })}\cap U_{{\alpha }^{\prime },t_2({\alpha }^{\prime })}=\mathrm{\varnothing }$; Step 3: construct a nonempty open neighborhood of $t_j$, $U_{t_j}$, using $U_{{\alpha }^{\prime },t_j({\alpha }^{\prime })}$ such that $U_{t_1}\cap U_{t_2}=\mathrm{\varnothing }$.

Step 1:

Let $t_1,t_2\in T$ be any such that $t_1\ne t_2$. $t_1({\alpha }^{\prime })\ne t_2({\alpha }^{\prime })\in T_{{\alpha }^{\prime }}$ for an ${\alpha }^{\prime }\in A$.

Step 2:

As $T_{{\alpha }^{\prime }}$ is Hausdorff, there are some nonempty open neighborhoods, $U_{{\alpha }^{\prime },t_1({\alpha }^{\prime })}\subseteq T_{{\alpha }^{\prime }}$ and $U_{{\alpha }^{\prime },t_2({\alpha }^{\prime })}\subseteq T_{{\alpha }^{\prime }}$, of $t_1({\alpha }^{\prime })$ and $t_2({\alpha }^{\prime })$, such that $U_{{\alpha }^{\prime },t_1({\alpha }^{\prime })}\cap U_{{\alpha }^{\prime },t_2({\alpha }^{\prime })}=\mathrm{\varnothing }$.

Step 3:

Let us define $U_{t_j}:={\times }_{\alpha \in A}{U}_{\alpha ,t_j(\alpha )}$ where $U_{\alpha ,t_j(\alpha )}=T_{\alpha }$ for $\alpha \ne {\alpha }^{\prime }$, which is open on $T$, by the definition of product topology, and is nonempty.

$U_{t_1}\cap U_{t_2}=\mathrm{\varnothing }$, because for any $t\in U_{t_1}$, $t({\alpha }^{\prime })\in U_{{\alpha }^{\prime },t_1({\alpha }^{\prime })}$, and so, $t({\alpha }^{\prime })\notin U_{{\alpha }^{\prime },t_2({\alpha }^{\prime })}$, and so, $t\notin U_{t_2}$.

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4: Structured Description 2

Here is the rules of Structured Description.

Entities:
$\{T_1,...,T_n\}$: $T_j\in \{\text{ the Hausdorff topological spaces }\}$
$T$: $=T_1\times ...\times T_n$, $=\text{ the product topological space }$
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Statements:
$T\in \{\text{ the Hausdorff topological spaces }\}$
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5: Natural Language Description 2

For any Hausdorff topological spaces, $\{T_1,...,T_n\}$, and the product topological space, $T:=T_1\times ...\times T_n$, $T$ is a Hausdorff topological space.

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6: Proof 2

Whole Strategy: Step 1: take any distinct $t_1,t_2\in T$ and see that $t_1^{j^{\prime }}\ne t_2^{j^{\prime }}$ for a $j^{\prime }\in \{1,...,n\}$; Step 2: take a nonempty open neighborhood of $t_j^{j^{\prime }}$, $U_{j^{\prime },t_j^{j^{\prime }}}\subseteq T_{j^{\prime }}$, such that $U_{j^{\prime },t_1^{j^{\prime }}}\cap U_{j^{\prime },t_2^{j^{\prime }}}=\mathrm{\varnothing }$; Step 3: construct a nonempty open neighborhood of $t_j$, $U_{t_j}$, using $U_{j^{\prime },t_j^{j^{\prime }}}$ such that $U_{t_1}\cap U_{t_2}=\mathrm{\varnothing }$.

Step 1:

Let $t_1,t_2\in T$ be any such that $t_1\ne t_2$. $t_1^{j^{\prime }}\ne t_2^{j^{\prime }}\in T_{j^{\prime }}$ for an $j^{\prime }\in \{1,...,n\}$.

Step 2:

As $T_{j^{\prime }}$ is Hausdorff, there are some nonempty open neighborhoods, $U_{j^{\prime },t_1^{j^{\prime }}}\subseteq T_{j^{\prime }}$ and $U_{j^{\prime },t_2^{j^{\prime }}}\subseteq T_{j^{\prime }}$, of $t_1^{j^{\prime }}$ and $t_2^{j^{\prime }}$, such that $U_{j^{\prime },t_1^{j^{\prime }}}\cap U_{j^{\prime },t_2^{j^{\prime }}}=\mathrm{\varnothing }$.

Step 3:

Let us define $U_{t_j}:=U_{1,t_j^1}\times ...\times U_{n,t_j^n}$ where $U_{k,t_j^k}=T_k$ for $k\ne j^{\prime }$, which is open on $T$, by the definition of product topology, and is nonempty.

$U_{t_1}\cap U_{t_2}=\mathrm{\varnothing }$, because for any $t\in U_{t_1}$, $t^{j^{\prime }}\in U_{j^{\prime },t_1^{j^{\prime }}}$, and so, $t^{j^{\prime }}\notin U_{j^{\prime },t_2^{j^{\prime }}}$, and so, $t\notin U_{t_2}$.

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References

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