287: Subset of R^{d-k} Is Open If the Product of R^k and Subset Is Open

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A description/proof of that subset of $R^{d-k}$ is open if the product of $R^k$ and subset is open

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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 Euclidean topological space.

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Target Context

• The reader will have a description and a proof of the proposition that any subset of $R^{d-k}$ is open if the product of $R^k$ and the subset is open.

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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 the Euclidean topological spaces, ${\mathbb{R}}^k$ and ${\mathbb{R}}^{d-k}$, any set, $U\subseteq {\mathbb{R}}^{d-k}$, is open on ${\mathbb{R}}^{d-k}$ if ${\mathbb{R}}^k\times U$ is open on ${\mathbb{R}}^d$.

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2: Proof

For any point, $p\in {\mathbb{R}}^k\times U$, there is an open ball, $B_p:=\{\mathrm{\forall }(x^1,...,x^d)|\sum _{i=1,...,d}(x^i-p^i{)}^2<{\epsilon }^2\}$, contained in ${\mathbb{R}}^k\times U$. Then, $\{\mathrm{\forall }(p^1,...,p^k,x^{k+1},...,x^d)|\sum _{i=k+1,...,d}(x^i-p^i{)}^2<{\epsilon }^2\}$ is contained in ${\mathbb{R}}^k\times U$ because it is contained in $B_p$. For any $p^{\prime }\in U$, $(p^{k+1},...,p^d)$, there is a $p\in {\mathbb{R}}^k\times U$, $(p^1,...,p^k,p^{k+1},...,p^d)$, and $\{\mathrm{\forall }(p^1,...,p^k,x^{k+1},...,x^d)|\sum _{i=k+1,...,d}(x^i-p^i{)}^2<{\epsilon }^2\}$ is contained in ${\mathbb{R}}^k\times U$, and that $(x^{k+1},...,x^d)$ is an open ball contained in U.

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References

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