description/proof of that product of Hausdorff topological spaces is Hausdorff
Topics
About: topological space
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
Starting Context
- The reader knows a definition of Hausdorff topological space.
- The reader knows a definition of product topology.
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.
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.
Main Body
1: Structured Description 1
Here is the rules of Structured Description.
Entities:
\(A\): \(\in \{\text{ the possibly uncountable infinite index sets }\}\)
\(\{T_\alpha \vert \alpha \in A\}\): \(T_\alpha \in \{\text{ the Hausdorff topological spaces }\}\)
\(T\): \(= \times_{\alpha \in A} T_\alpha\), \(= \text{ the product topological space }\)
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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, \(\{T_\alpha \vert \alpha \in A\}\), and the product topological space, \(T := \times_{\alpha \in A} T_\alpha\), \(T\) is a Hausdorff topological space.
3: Proof 1
Whole Strategy: Step 1: take any distinct \(t_1, t_2 \in T\) and see that \(t_1 (\alpha') \neq t_2 (\alpha')\) for an \(\alpha' \in A\); Step 2: take a nonempty open neighborhood of \(t_j (\alpha')\), \(U_{\alpha', t_j (\alpha')} \subseteq T_{\alpha'}\), such that \(U_{\alpha', t_1 (\alpha')} \cap U_{\alpha', t_2 (\alpha')} = \emptyset\); Step 3: construct a nonempty open neighborhood of \(t_j\), \(U_{t_j}\), using \(U_{\alpha', t_j (\alpha')}\) such that \(U_{t_1} \cap U_{t_2} = \emptyset\).
Step 1:
Let \(t_1, t_2 \in T\) be any such that \(t_1 \neq t_2\). \(t_1 (\alpha') \neq t_2 (\alpha') \in T_{\alpha'}\) for an \(\alpha' \in A\).
Step 2:
As \(T_{\alpha'}\) is Hausdorff, there are some nonempty open neighborhoods, \(U_{\alpha', t_1 (\alpha')} \subseteq T_{\alpha'}\) and \(U_{\alpha', t_2 (\alpha')} \subseteq T_{\alpha'}\), of \(t_1 (\alpha')\) and \(t_2 (\alpha')\), such that \(U_{\alpha', t_1 (\alpha')} \cap U_{\alpha', t_2 (\alpha')} = \emptyset\).
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 \(\alpha \neq \alpha'\), which is open on \(T\), by the definition of product topology, and is nonempty.
\(U_{t_1} \cap U_{t_2} = \emptyset\), because for any \(t \in U_{t_1}\), \(t (\alpha') \in U_{\alpha', t_1 (\alpha')}\), and so, \(t (\alpha') \notin U_{\alpha', t_2 (\alpha')}\), and so, \(t \notin U_{t_2}\).
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.
6: Proof 2
Whole Strategy: Step 1: take any distinct \(t_1, t_2 \in T\) and see that \(t_1^{j'} \neq t_2^{j'}\) for a \(j' \in \{1, ..., n\}\); Step 2: take a nonempty open neighborhood of \(t_j^{j'}\), \(U_{j', t_j^{j'}} \subseteq T_{j'}\), such that \(U_{j', t_1^{j'}} \cap U_{j', t_2^{j'}} = \emptyset\); Step 3: construct a nonempty open neighborhood of \(t_j\), \(U_{t_j}\), using \(U_{j', t_j^{j'}}\) such that \(U_{t_1} \cap U_{t_2} = \emptyset\).
Step 1:
Let \(t_1, t_2 \in T\) be any such that \(t_1 \neq t_2\). \(t_1^{j'} \neq t_2^{j'} \in T_{j'}\) for an \(j' \in \{1, ..., n\}\).
Step 2:
As \(T_{j'}\) is Hausdorff, there are some nonempty open neighborhoods, \(U_{j', t_1^{j'}} \subseteq T_{j'}\) and \(U_{j', t_2^{j'}} \subseteq T_{j'}\), of \(t_1^{j'}\) and \(t_2^{j'}\), such that \(U_{j', t_1^{j'}} \cap U_{j', t_2^{j'}} = \emptyset\).
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 \(k \neq j'\), which is open on \(T\), by the definition of product topology, and is nonempty.
\(U_{t_1} \cap U_{t_2} = \emptyset\), because for any \(t \in U_{t_1}\), \(t^{j'} \in U_{j', t_1^{j'}}\), and so, \(t^{j'} \notin U_{j', t_2^{j'}}\), and so, \(t \notin U_{t_2}\).
