271: Finite Product of Topological Spaces Equals Sequential Products of Topological Spaces

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A description/proof of that finite product of topological spaces equals sequential products of topological spaces

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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 product topology.
• The reader knows a definition of product of sets.

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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 topological spaces equals the sequential products of the topological spaces.

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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 spaces, $T_1,T_2,...,T_n$, the product topological space, $T_1\times T_2\times ...\times T_n$, equals the sequential products, $(...(T_1\times T_2)...)\times T_n$.

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2: Proof

Sets-product-wise, $T_1\times T_2\times ...\times T_n$ is nothing but $(...(T_1\times T_2)...)\times T_n$.

1st, let us think of the case, $n=3$. Any open set, $U$, on $T_1\times T_2\times T_3$ is $U={\cup }_{\alpha }(U_{1-\alpha }\times U_{2-\alpha }\times U_{3-\alpha })$, by the definition of product topology. Any open set, $U^{\prime }$, on $(T_1\times T_2)\times T_n$ is $U^{\prime }={\cup }_{\beta }({\cup }_{\gamma }(U_{1-\beta -\gamma }\times U_{2-\beta -\gamma })\times U_{3-\beta })$, by the definition of product topology.

In order to realize $U$ by $U^{\prime }$, $\{i-\beta -\gamma \}$ can be taken to be 1 element for each $\beta$, then, $U^{\prime }={\cup }_{\beta }((U_{1-\beta -1}\times U_{2-\beta -1})\times U_{3-\beta })={\cup }_{\beta }(U_{1-\beta -1}\times U_{2-\beta -1}\times U_{3-\beta })$, and $\{U_{i-\beta -1}\}$ and $\{U_{3-\beta }\}$ can be taken to be $\{U_{i-\alpha }\}$.

In order to realize $U^{\prime }$ by $U$, some $U_{3-\alpha }$s can be chosen to be the same for some multiple $\alpha$s, then, the new indexes set, $\{\beta -\gamma \}$, can be chosen such that $\{U_{i-\beta -\gamma }\}=\{U_{i-\alpha }\}$ where $U_{3-\beta -\gamma }$s are the same for the same $\beta$, then, $U={\cup }_{\alpha }(U_{1-\alpha }\times U_{2-\alpha }\times U_{3-\alpha })={\cup }_{\beta -\gamma }(U_{1-\beta -\gamma }\times U_{2-\beta -\gamma }\times U_{3-\beta -\gamma })={\cup }_{\beta }({\cup }_{\gamma }(U_{1-\beta -\gamma }\times U_{2-\beta -\gamma })\times U_{3-\beta })$ where $U_{3-\beta -\gamma }$ is written as $U_{3-\beta }$ as it does not really depend on $\gamma$.

So, the 2 topologies are the same.

Obviously, the case for any $n$ can be proven likewise.

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References

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