278: Subset of Product Topological Space Is Closed iff It Is Intersection of Finite Unions of Products of Closed Subsets Only Finite of Which Are Not Whole Spaces

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A description/proof of that subset of product topological space is closed iff it is intersection of finite unions of products of closed subsets only finite of which are not whole 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 1
• 2: Proof 1
• 3: Description 2
• 4: Proof 2
• 5: Note 1
• 6: Note 2

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Starting Context

• The reader knows a definition of product topology.
• The reader knows a definition of closed set.
• The reader admits the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets.
• The reader admits the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets.
• The reader admits the proposition that any possibly-infinite-wise product topological space for which the indices set is finite is homeomorphic to the corresponding finite product topological space.

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Target Context

• The reader will have a description and a proof of the proposition that any subset of any product topological space is closed if and only if it is the intersection of any possibly uncountable number of finite unions of the products of any closed subsets only finite number of which (the subsets) are not the whole spaces, and especially only if only 1 of the subsets is not the whole space.

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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 1

For any product topological space, $T={\times }_{\alpha \in A}{T}_{\alpha }$ where $A$ is any possibly uncountable indices set, any subset, $S\subseteq T$, is closed if and only if $S$ is the intersection of any possibly uncountable number of finite unions of the products of any closed subsets only finite of which (the subsets) are not the whole spaces, which is $S={\cap }_{\beta \in B}{\cup }_{j\in J_{\beta }}{\times }_{\alpha \in A}{C}_{\beta -j-\alpha }$ where $B$ is any possible uncountable indices set, $J_{\beta }$ is any finite indices set for each $\beta$, and $C_{\beta -j-\alpha }\ne T_{\alpha }$ for only finite number of $C_{\beta -j-\alpha }$s for each $(\beta ,j)$, and especially only if $C_{\beta -j-\alpha }\ne T_{\alpha }$ for only 1 of $C_{\beta -j-\alpha }$s for each $(\beta ,j)$.

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2: Proof 1

1st, let us suppose that $S={\times }_{\alpha \in A}{C}_{\alpha }$, a simplified case in which $B=\{1\}$ and $J_1=\{1\}$. $S=T\setminus ({\cup }_i({\times }_{\alpha }{U}_{i-\alpha }))$ where $i$ corresponds to each $\alpha$ such that $C_{\alpha }\ne T_{\alpha }$ and $U_{i-\alpha }=T_{\alpha }\setminus C_{\alpha }$ when $\alpha =i$ and $U_{i-\alpha }=T_{\alpha }$ otherwise, because for any point, $p\in S$, $p_i\in C_i$ for each $i$, $p_i\notin U_{i-i}$ for each $i$, so, $p\notin {\times }_{\alpha }{U}_{i-\alpha }$ for each $i$, so, $p\notin {\cup }_i({\times }_{\alpha }{U}_{i-\alpha })$; for any point, $p\in T\setminus ({\cup }_i({\times }_{\alpha }{U}_{i-\alpha }))$, $p\notin {\cup }_i({\times }_{\alpha }{U}_{i-\alpha })$, $p\notin {\times }_{\alpha }{U}_{i-\alpha }$ for each $i$, $p_i\notin U_{i-i}$ for each $i$, $p_i\in C_i$ for each $i$, so, $p\in {\times }_{\alpha }{C}_{\alpha }=S$. As ${\cup }_i({\times }_{\alpha }{U}_{i-\alpha })$ is open by the definition of product topology, $S$ is closed.

Now, let us suppose that $S={\cap }_{\beta \in B}{\cup }_{j\in J_{\beta }}{\times }_{\alpha \in A}{C}_{\beta -j-\alpha }$. By the previous paragraph, ${\times }_{\alpha \in A}{C}_{\beta -j-\alpha }$ is closed, so, $S$ is closed as an intersection of finite unions of closed sets.

Let us suppose that $S$ is any closed set, $S\subseteq T$. $S=T\setminus {\cup }_{\beta \in B}{\times }_{\alpha \in A}{U}_{\beta -\alpha }$ where $U_{\beta -\alpha }$ is open but $U_{\beta -\alpha }\ne T_{\alpha }$ for only finite number of $\alpha$s, by the definition of product topology. $S={\cap }_{\beta \in B}(T\setminus {\times }_{\alpha \in A}{U}_{\beta -\alpha })$, by the proposition for any set, the intersection of the compliments of any possibly uncountable number of subsets is the complement of the union of the subsets. ${\times }_{\alpha \in A}{U}_{\beta -\alpha }={\cap }_i({\times }_{\alpha \in A}{U}_{\beta -i-\alpha })$ where $i$ corresponds to $\alpha$ such that $U_{\beta -\alpha }\ne T_{\alpha }$ and $U_{\beta -i-\alpha }=U_{\beta -\alpha }$ when $\alpha =i$ and $U_{\beta -i-\alpha }=T_{\alpha }$ otherwise, because for any $p\in {\times }_{\alpha }{U}_{\beta -\alpha }$, $p_{\alpha }\in U_{\beta -i-\alpha }$ for each $i$ and $\alpha$, because if $\alpha =i$, $p_i\in U_{\beta -i}=U_{\beta -i-i}$, and otherwise, $p_{\alpha }\in T_{\alpha }=U_{\beta -i-\alpha }$, so, $p\in {\times }_{\alpha \in A}{U}_{\beta -i-\alpha }$ for each $i$; for any $p\in {\cap }_i({\times }_{\alpha \in A}{U}_{\beta -i-\alpha })$, $p\in {\times }_{\alpha \in A}{U}_{\beta -i-\alpha }$ for each $i$, $p_{\alpha }\in U_{\beta -i-\alpha }$ for each $i$ and $\alpha$, so, if $\alpha$ is an $i$, $p_i\in U_{\beta -i-i}=U_{\beta -\alpha }$, and otherwise, $p_{\alpha }\in T_{\alpha }=U_{\beta -\alpha }$, so, $p\in {\times }_{\alpha }{U}_{\beta -\alpha }$.

So, $S={\cap }_{\beta \in B}(T\setminus ({\cap }_i({\times }_{\alpha \in A}{U}_{\beta -i-\alpha })))={\cap }_{\beta \in B}{\cup }_i(T\setminus {\times }_{\alpha \in A}{U}_{\beta -i-\alpha })$, by the proposition for any set, the union of the complements of any possibly uncountable number of subsets is the complement of the intersection of the subsets. But as has been shown above (this is a case of that for ${\times }_{\alpha \in A}{C}_{\alpha }=T\setminus ({\cup }_i({\times }_{\alpha }{U}_{i-\alpha }))$, there is only 1 $C_{\alpha }$ such that $C_{\alpha }\ne T_{\alpha }$), $T\setminus {\times }_{\alpha \in A}{U}_{\beta -i-\alpha }={\times }_{\alpha \in A}{C}_{\beta -i-\alpha }$ where $C_{\beta -i-\alpha }$s are closed sets and $C_{\beta -i-\alpha }\ne T_{\alpha }$ for $\alpha =i$ but $C_{\beta -i-\alpha }=T_{\alpha }$ otherwise. So, $S={\cap }_{\beta \in B}{\cup }_i{\times }_{\alpha \in A}{C}_{\beta -i-\alpha }$.

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3: Description 2

For any product topological space, $T=T_1\times T_2\times ...\times T_n$, any subset, $S\subseteq T$, is closed if and only if $S$ is the intersection of any possibly uncountable number of finite unions of the products of any closed subsets, which is $S={\cap }_{\beta \in B}{\cup }_{j\in J}{C}_{\beta -j-1}\times C_{\beta -j-2}\times ...\times C_{\beta -j-n}$ where $B$ is any possible uncountable indices set, $J_{\beta }$ is any finite indices set for each $\beta$, and especially only if $C_{\beta -j-i}\ne T_i$ for only 1 of $C_{\beta -j-i}$s for each $(\beta ,j)$.

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4: Proof 2

By the proposition that any possibly-infinite-wise product topological space for which the indices set is finite is homeomorphic to the corresponding finite product topological space, $T$ is homeomorphic to $T^{\prime }={\times }_i{T}_i$ where $i\in \{1,2,...,n\}$, and Description 1 applies to $T^{\prime }$.

$S={\cap }_{\beta \in B}{\cup }_{j\in J}{C}_{\beta -j-1}\times C_{\beta -j-2}\times ...\times C_{\beta -j-n}$ corresponds to $S^{\prime }={\cap }_{\beta \in B}{\cup }_{j\in J}{\times }_i{C}_{\beta -j-i}$, which is closed by Description 1, so, $S$ is closed by the homeomorphism.

And for any closed $S\in T$, the corresponding $S^{\prime }$ is $S^{\prime }={\cap }_{\beta \in B}{\cup }_{j\in J_{\beta }}{\times }_{\alpha }{C}_{\beta -j-\alpha }$, which corresponds to $S={\cap }_{\beta \in B}{\cup }_{j\in J_{\beta }}{C}_{\beta -j-1}\times C_{\beta -j-2}\times ...\times C_{\beta -j-n}$.

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5: Note 1

The "finite unions" part of the necessary conditions cannot be omitted. As an example, let us think of the case, $T_1=\{1,2\}$ and $T_2=\{1,2\}$ with the discrete topologies. $S=(\{2\}\times \{1,2\})\cup (\{1,2\}\times \{2\})$ is a closed subset of $T=T_1\times T_2$. But $S$ cannot be expressed as ${\cap }_{\beta \in B}{C}_{\beta -1}\times C_{\beta -2}$, because any such a $C_{\beta -1}\times C_{\beta -2}$ has to contain both $\{2\}\times \{1,2\}$ and $\{1,2\}\times \{2\}$, and the only possibility is $C_{\beta -1}\times C_{\beta -2}=\{1,2\}\times \{1,2\}$, which cannot create $S$.

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6: Note 2

As is stated in the proposition that any possibly-infinite-wise product topological space for which the indices set is finite is homeomorphic to the corresponding finite product topological space, $T=T_1\times T_2\times ...\times T_n$ and $T^{\prime }={\times }_i{T}_i$ are not exactly the same, as any element of $T$ is like a $⟨p_1,p_2,...,p_n⟩$ while any element of $T^{\prime }$ is a function, $f:\{1,2,...,n\}\to {\cup }_i{T}_i$, although some people may sloppily say that they are the same. While Description 2 seems obvious from the homeomorphism anyway, we have bothered to go more explicit.

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References

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