350: Euclidean Topological Space Nested in Euclidean Topological Space Is Topological Subspace

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A description/proof of that Euclidean topological space nested in Euclidean topological space is topological subspace

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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 topological space.
• The reader knows a definition of subspace topology.

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Target Context

• The reader will have a description and a proof of the proposition that any Euclidean topological space nested in any Euclidean topological space is a topological subspace of the nesting Euclidean topological 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

For any Euclidean topological spaces, ${\mathbb{R}}^{d_1}$ and ${\mathbb{R}}^{d_2}$, such that ${\mathbb{R}}^{d_2}\subseteq {\mathbb{R}}^{d_1}$ where inevitably $d_2\le d_1$, where it does not mean that ${\mathbb{R}}^{d_1}={\mathbb{R}}^{d_2}\times {\mathbb{R}}^{d_1-d_2}$, because ${\mathbb{R}}^{d_2}$ may have be turned and/or translated against ${\mathbb{R}}^{d_1}$, the topology of ${\mathbb{R}}^{d_2}$ is the subspace topology of ${\mathbb{R}}^{d_1}$.

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2: Proof

It suffices to show that any open ball, $B_1$, on ${\mathbb{R}}^{d_1}$ is an open ball, $B_2=B_1\cap {\mathbb{R}}^{d_2}$, or the empty set on ${\mathbb{R}}^{d_2}$, and for any open ball, $B_2$, on ${\mathbb{R}}^{d_2}$, there is an open ball, $B_1$, on ${\mathbb{R}}^{d_1}$ such that $B_2=B_1\cap {\mathbb{R}}^{d_2}$, because then, any open set, $U_2$, on ${\mathbb{R}}^{d_2}$ is a union of some open balls, ${\cup }_{\alpha }{B}_{2_{\alpha }}$, on ${\mathbb{R}}^{d_2}$, and each of the open balls will be open in the subspace topology, so, $U_2$ will be open in the subspace topology, while any subset, $S_2$, on ${\mathbb{R}}^{d_2}$ that is open in the subspace topology is $S_2=U_1\cap {\mathbb{R}}^{d_2}$ where $U_1$ is open on ${\mathbb{R}}^{d_1}$, but $U_1$ is a union of some open balls, ${\cup }_{\alpha }{B}_{1_{\alpha }}$, on ${\mathbb{R}}^{d_1}$, and $S_2=({\cup }_{\alpha }{B}_{1_{\alpha }})\cap {\mathbb{R}}^{d_2}={\cup }_{\alpha }(B_{1_{\alpha }}\cap {\mathbb{R}}^{d_2})$, but as $B_{1_{\alpha }}\cap {\mathbb{R}}^{d_2}$ will be an open ball or empty set on ${\mathbb{R}}^{d_2}$, $S_2$ will be open on ${\mathbb{R}}^{d_2}$.

Now, ${\mathbb{R}}^{d_2}$ is the subset of ${\mathbb{R}}^{d_1}$ where the $d_2+1,...,d_1$ components are 0, which (the subset) is afterward turned around the origin and then translated. So, we can take the global chart on ${\mathbb{R}}^{d_1}$ that is the standard chart turned around the origin and then translated in the same way with for ${\mathbb{R}}^{d_2}$. In the new chart, ${\mathbb{R}}^{d_2}$ is the subset of ${\mathbb{R}}^{d_1}$ where the $d_2+1,...,d_1$ components are 0, and any point on ${\mathbb{R}}^{d_2}$ has the same $1,...,d_2$ components with the ${\mathbb{R}}^{d_2}$ standard chart and with the ${\mathbb{R}}^{d_1}$ new chart. Let us always use the new chart from now on.

$B_1$ is $\{x\in {\mathbb{R}}^{d_1}|\sum _{i=1,...,d_1}(x_i-p_i{)}^2<{ϵ}^2\}$ where $p$ is the center of the ball. $B_1\cap {\mathbb{R}}^{d_2}$ is $\{x\in {\mathbb{R}}^{d_2}|\sum _{i=1,...,d_2}(x_i-p_i{)}^2+\sum _{i=d_2+1,...,d_1}(p_i{)}^2<{ϵ}^2\}=\{x\in {\mathbb{R}}^{d_2}|\sum _{i=1,...,d_2}(x_i-p_i{)}^2<{ϵ}^2-\sum _{i=d_2+1,...,d_1}(p_i{)}^2\}$, which is an open ball or the empty set on ${\mathbb{R}}^{d_2}$.

$B_2$ is $\{x\in {\mathbb{R}}^{d_2}|\sum _{i=1,...,d_2}(x_i-p_i{)}^2<{ϵ}^2\}$ where $p$ is the center of the ball. There is a subset on ${\mathbb{R}}^{d_1}$, $\{x\in {\mathbb{R}}^{d_1}|\sum _{i=1,...,d_2}(x_i-p_i{)}^2+\sum _{i=d_2+1,...,d_1}(x_i-0{)}^2<{ϵ}^2\}$, which is an open ball, named $B_1$, on ${\mathbb{R}}^{d_1}$ with the center $(p_1,...,p_{d_2},0,...,0)$, and $B_1\cap {\mathbb{R}}^{d_2}$ is $B_2$.

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References

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