760: Finite Product of 2nd-Countable Topological Spaces Is 2nd-Countable

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description/proof of that finite product of 2nd-countable topological spaces is 2nd-countable

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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: Natural Language Description
• 3: Proof
• 4: Note

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Starting Context

• The reader knows a definition of 2nd-countable 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 finite number of 2nd-countable topological spaces is 2nd-countable.

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

For any 2nd-countable topological spaces, $\{T_1,...,T_n\}$, and the product topological space, $T:=T_1\times ...\times T_n$, $T$ is a 2nd-countable topological space.

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3: Proof

Whole Strategy: Step 1: construct countable $B:=\{U_{1,k_1}\times ...\times U_{n,k_n}|k_j\in J_j\}$ where $B_j=\{U_{j,k_j}|k_j\in J_j\}$ is a countable basis for $T_j$; Step 2: see that $B$ is a basis for $T$.

Step 1:

Each $T_j$ has a countable basis, $B_j=\{U_{j,k_j}|k_j\in J_j\}$, where $J_j$ is a countable index set. $B:=\{U_{1,k_1}\times ...\times U_{n,k_n}|k_j\in J_j\}$ is countable.

Step 2:

Let us prove that $B$ is a basis for $T$. Each $U_{1,k_1}\times ...\times U_{n,k_n}$ is open on $T$, by the definition of product topology. For any neighborhood of any point, $p=(p^1,...,p^n)\in T$, $N_p\subseteq T$, there is an open neighborhood of $p$, $U_p\subseteq T$, such that $U_p\subseteq N_p$, by the definition of neighborhood of point; $U_p={\cup }_{\alpha \in A}{U}_{1,\alpha }\times ...\times U_{n,\alpha }$ where $A$ is a possibly uncountable index set and $U_{j,\alpha }\subseteq T_j$ is an open subset, by Note of the article on the definition of product topology; $p\in U_{1,\alpha }\times ...\times U_{n,\alpha }$ for an $\alpha$; there is an element, $U_{j,k_j}$, of $B_j$ such that $p^j\in U_{j,k_j}\subseteq U_{j,\alpha }$; $U_{1,k_1}\times ...\times U_{n,k_n}$ is an element of $B$, and $p\in U_{1,k_1}\times ...\times U_{n,k_n}\subseteq U_{1,\alpha }\times ...\times U_{n,\alpha }\subseteq U_p\subseteq N_p$.

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4: Note

The product has to be finite for this proposition, while the product of Hausdorff topological spaces does not need to be finite for the product to be Hausdorff.

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References

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