761: Euclidean Topological Space Is Homeomorphic to Product of Lower-Dimensional Euclidean Spaces

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description/proof of that Euclidean topological space is homeomorphic to product of lower-dimensional Euclidean 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: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of Euclidean 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 $d$-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals $d$.

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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:
${\mathbb{R}}^d$: $=\text{ the Euclidean topological space }$, where $d\in \mathbb{N}\setminus \{0\}$
$({\mathbb{R}}^{d_1},...,{\mathbb{R}}^{d_n})$: $\in \{\text{ the sequences of Euclidean topological spaces }\}$, where $d_j\in \mathbb{N}\setminus \{0\}$, such that $\sum _{j\in \{1,...,n\}}{d}_j=d$
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Statements:
${\mathbb{R}}^d\cong {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$, where $\cong$ denotes being homeomorphic.
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2: Natural Language Description

For the $d$-dimensional Euclidean topological space, ${\mathbb{R}}^d$, where $d\in \mathbb{N}\setminus \{0\}$, and any sequence of Euclidean topological spaces, $({\mathbb{R}}^{d_1},...,{\mathbb{R}}^{d_n})$, where $d_j\in \mathbb{N}\setminus \{0\}$, such that $\sum _{j\in \{1,...,n\}}{d}_j=d$, ${\mathbb{R}}^d$ is homeomorphic to ${\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$.

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3: Note

Prevalently, ${\mathbb{R}}^d$ will be sloppily said to equal ${\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$, but they are not exactly the same sets-wise. For example, a point on ${\mathbb{R}}^3$ is $(r_1,r_2,r_3)$ while a point on ${\mathbb{R}}^1\times {\mathbb{R}}^2$ is $(r_1,(r_2,r_3))$, and the 2 points are different, strictly speaking. As they are not the same sets-wise, they cannot be the same 'topological spaces'-wise. So, this proposition says that the 2 topological spaces are homeomorphic, not the same.

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4: Proof

Whole Strategy: Step 1: define the canonical 'sets - map morphisms' isomorphism, $f:{\mathbb{R}}^d\to {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$; Step 2: see that $f$ is continuous at each point, $p\in {\mathbb{R}}^d$; Step 3: see that $f^{-1}$ is continuous at each point, $q\in {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$.

Step 1:

Let us denote any point, $p\in {\mathbb{R}}^d$, as $p=(p^1,...,p^d)$.

Let us denote any point, $q\in {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$, as $q=(q^1,...,q^n)$.

There is the canonical 'sets - map morphisms' isomorphism, $f:{\mathbb{R}}^d\to {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n},(p^1,...,p^d)\mapsto (q^1=(p^1,...,p^{d_1}),...,q^n=(p^{d_1+...+d_{n-1}+1},...,p^d))$, which is indeed bijective, because for each $p,p^{\prime }\in {\mathbb{R}}^d$ such that $p\ne p^{\prime }$, $p^j\ne p^{\prime j}$ for a $j\in \{1,...,d\}$, and there is a $k\in \{1,...,n\}$ such that $p^j$ is in $q^k$ and $p^{\prime j}$ is in $q^{\prime k}$, which means that $q^k\ne q^{\prime k}$, which means that $f(p)\ne f(p^{\prime })$; for each $q=(q^1=(p^1,...,p^{d_1}),...,q^n=(p^{d_1+...+d_{n-1}+1},...,p^d))\in {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$, there is the $p=(p^1,...,p^d)\in {\mathbb{R}}^d$, which means that $f(p)=q$.

Step 2:

Let us see that $f$ is continuous at any point, $p=(p^1,...,p^d)\in {\mathbb{R}}^d$.

Let $q:=f(p)=(q^1,...,q^n)$.

Let $N_q\subseteq {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$ be any neighborhood of $q$. There is an open neighborhood of $q$, $U_q\subseteq {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$, such that $U_q\subseteq N_q$. There is an open $B_{q^1,{ϵ}_1}\times ...\times B_{q^n,{ϵ}_n}\subseteq {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$ such that $B_{q^1,{ϵ}_1}\times ...\times B_{q^n,{ϵ}_1}\subseteq U_q$, where $B_{q^j,{ϵ}_j}\subseteq {\mathbb{R}}^{d_j}$ is the open ball around $q^j$, by the definition of product topology.

Let $ϵ:=min({ϵ}_1,...,{ϵ}_n)$.

Let us take the open ball around $p$, $B_{p,ϵ}\subseteq {\mathbb{R}}^d$.

Let us see that $f(B_{p,ϵ})\subseteq B_{q^1,{ϵ}_1}\times ...\times B_{q^n,{ϵ}_n}$.

Let $p^{\prime }=(p^{\prime 1},...,p^{\prime d})\in B_{p,ϵ}$ be any. $f(p^{\prime })=(q^{\prime 1}=(p^{\prime 1},...,p^{\prime d_1}),...,q^{\prime n}=(p^{\prime d_1+...+d_{n-1}+1},...,p^{\prime d}))$. $q^{\prime j}\in B_{q^j,{ϵ}_j}$, because $\sum _{j\in \{d_1+...+d_{j-1}+1,...,d_1+...+d_{j-1}+d_j\}}(p^{\prime j}-p^j{)}^2\le \sum _{j\in \{1,...,d\}}(p^{\prime j}-p^j{)}^2<{ϵ}^2\le {ϵ}_j^2$. So, $f(p^{\prime })\in B_{q^1,{ϵ}_1}\times ...\times B_{q^n,{ϵ}_n}$. So, $f(B_{p,ϵ})\subseteq B_{q^1,{ϵ}_1}\times ...\times B_{q^n,{ϵ}_n}$.

So, $f(B_{p,ϵ})\subseteq B_{q^1,{ϵ}_1}\times ...\times B_{q^n,{ϵ}_n}\subseteq U_q\subseteq N_q$.

So, $f$ is continuous at $p$.

Step 3:

Let us see that $f^{-1}$ is continuous at any point, $q=(q^1=(p^1,...,p^{d_1}),...,q^n=(p^{d_1+...+d_{n-1}+1},...,p^d))\in {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$.

Let $p:=f^{{}^1}(q)=(p^1,...,p^n)$.

Let $N_p\subseteq {\mathbb{R}}^d$ be any neighborhood of $p$. There is an open neighborhood of $p$, $U_p\subseteq {\mathbb{R}}^d$, such that $U_p\subseteq N_p$. There is an open $B_{p,ϵ}\subseteq {\mathbb{R}}^d$ such that $B_{p,ϵ}\subseteq U_p$, where $B_{p,ϵ}\subseteq {\mathbb{R}}^d$ is the open ball around $p$.

Let ${ϵ}^{\prime }:=ϵ/\sqrt{n}$.

Let us take the open ball around $q$, $B_{q^1,{ϵ}^{\prime }}\times ...\times B_{q^n,{ϵ}^{\prime }}\subseteq {\mathbb{R}}^{d_1}\times ...\times {\mathbb{R}}^{d_n}$.

Let us see that $f^{-1}(B_{q^1,{ϵ}^{\prime }}\times ...\times B_{q^n,{ϵ}^{\prime }})\subseteq B_{p,ϵ}$.

Let $q^{\prime }=(q^{\prime 1}=(p^{\prime 1},...,p^{\prime d_1}),...,q^{\prime n}=(p^{\prime d_1+...+d_{n-1}+1},...,p^{\prime d}))\in B_{q^1,{ϵ}^{\prime }}\times ...\times B_{q^n,{ϵ}^{\prime }}$ be any. $p^{\prime }:=f^{-1}(q^{\prime })=(p^{\prime 1},...,p^{\prime d})$. $p^{\prime }\in B_{p,ϵ}$, because $\sum _{j\in \{1,...,d\}}(p^{\prime j}-p^j{)}^2=\sum _{j\in \{1,...,d_1\}}(p^{\prime j}-p^j{)}^2+...+\sum _{j\in \{d_1+...+d_{n-1}+1,...,d\}}(p^{\prime j}-p^j{)}^2<{ϵ}^{\prime 2}+...+{ϵ}^{\prime 2}=n{ϵ}^2/n={ϵ}^2$. So, $f^{-1}(q^{\prime })\in B_{p,ϵ}$. So, $f^{-1}(B_{q^1,{ϵ}^{\prime }}\times ...\times B_{q^n,{ϵ}^{\prime }})\subseteq B_{p,ϵ}$.

So, $f^{-1}(B_{q^1,{ϵ}^{\prime }}\times ...\times B_{q^n,{ϵ}^{\prime }})\subseteq B_{p,ϵ}\subseteq U_p\subseteq N_p$.

So, $f^{-1}$ is continuous at $q$.

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References

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