Htykuut

I am an undergraduate student doing a first course in topology. I am stuck with a conclusion that I feel like should be relatively straightforward after proving quite some statements. The assignment is relatively long (giving a "recipe") for the final conclusion, so please bear with me.

Let $S^1$ denote the unit circle in $mathbb{R}^2$. Consider for $n geq 2$ the mapping:

$R: S^1 rightarrow M_{n+1}$ given by $(u,v) mapsto$ $ begin{bmatrix}
u & 0 & -v \
0 & I_{n-1} & 0 \
v & 0 & u
end{bmatrix}$

I have proved the following things:

$(textbf{A})$ that $R(cos (phi), sin(phi))$ defines a rotation by an angle $phi$ in the $(x_{1}, x_{n+1})$-plane.

Now, letting $H:={x in mathbb{R}^n vert x_1 >0 }$

$(textbf{B})$ that the map $f:H times S^1 mapsto mathbb{R}^{n+1}$, $(x, u, v) mapsto $
$R(u,v)begin{bmatrix}
x_{1} \
x_{2} \
vdots \
x_{n}\
0
end{bmatrix}$
is a homeomorphism from

$H times$ $S^1$ (equipped with the topology induced from $mathbb{R}^n times mathbb{R}^2$)
onto an open subset of $mathbb{R}^{n+1}$.

$(textbf{C})$ When $A subset H$ and $A times S^1 subset mathbb{R}^n times mathbb{R}^2$ are equipped with the induced topology, then $f|_{A×S^1}$ is an embedding of $A times S^1$ onto $mathbb{R}^{n+1}$.

$(textbf{D})$Finally I showed that if $A$ is bounded in $mathbb{R}^n$ that $f(A times S^1)$ is bounded in $mathbb{R}^{n+1}$.

Using these results I have to conclude the following:

For $m geq$ 1 let the $m$-fold product $X_m := S^1 times ... times S^1 subset (mathbb{R}^2)^m = mathbb{R}^{2m}$ be equipped with induced topology. Show for all $m geq 1$ there exists an embedding of $X_m$ onto a bounded subset of $mathbb{R}^{m+1}$ .

I feel like one can picture the $(m-1)$-fold product of $S^1$ to sit in $H$, as the bounded subset $A$ from part $textbf{D}$. But I do not know how to realize this since $S^1$ is the unit circle and the first coordinate of elements of $H$ is greater than $0$.

Any help would be appreciated!

We prove that $X_m$ embeds into $mathbb{R}^{m+1}$ as a bounded subset by induction on $m$.

The base case $m=1$ is just the statement that $S^1subseteq mathbb{R}^2$ is bounded.

Now assume inductively that $X_m$ embeds into $mathbb{R}^{m+1}$ as a bounded subset. Let $g:X_mrightarrow mathbb{R}^{m+1}$ be such an embedding.

Because $g(X_m)$ is bounded it lies inside of some cube: $g(X_m)subseteq [a_1, b_1]times [a_2,b_2]times...times [a_{m+1}, b_{m+1}]$. Consider a new function $h:X_mrightarrow mathbb{R}^{m+1}$ given by $h(x) = g(x) + (2|a_1|,0,...,0)$. I leave it to you to show that $h$ is also an embedding and that $h(X_m)subseteq H$.

Great. How does this give an embedding of $X_{m+1}$ into $mathbb{R}^{m+2}$? Well, now that we have $h(X_m)subseteq A$, part C can be used. Can you finish from here?