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Could you please check if this construction makes sense and answer to questions to the parts in bold?

My construction:

The construction can be given explicitly as an inverse stereographic projection. Consider the map

$s: \mathbb{R} \rightarrow S^1$ given by

$x \rightarrow (\frac{1-x^2}{1 + x^2},\frac{2x}{1 + x^2})$

Then $s$ is a homeomorphism between $\mathbb{R}$ and $S^1 \setminus \{(-1,0)\}$. So, since $S^1 \setminus \{(-1,0)\}$ is dense in $S^1$ and $S^1 \setminus\left(S^1 \setminus\{(-1,0)\}\right)$ consists of a single point, $S^1$ is the one-point compactification of $\mathbb{R}$.

Question 1: Why is s a homeomorphism? (How to prove that?) I take this as a known fact from math. analysis, but I am not sure.

Question 2: How do we know that $S^1 \setminus \{(-1,0)\}$ is dense in $S^1$?

Tereza Tizkova
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    The structure of your question is poor. In question 1 you ask something about a mysterious map $s$, but nobody knows what $s$ is. I recommend to define $s$ at the beginning and then to ask your questions. Anyway, your arguments are correct. – Paul Frost Apr 30 '22 at 23:03
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    For $x\in\Bbb R$ let $f(x)=2\arctan x\in (-\pi,\pi).$ Then $s(x)=(\cos f(x),,\sin f(x),)$ is easily seen to be a continuous bijection. And $s$ maps any bounded open real interval $(a,b)$ to ${(\cos u,\sin u): f(a)<u<f(b)}$, which is an open set in $S^1$ \ ${(-1,0)}$. So $s$ is an open map. A continuous open bijection is a homeomorphism.... If $T$ is any subset of $\Bbb R^2$ and if $ \overline T$ is the closure of $T$ in $\Bbb R^2$ then $T$ is dense in the space $\overline T.$ – DanielWainfleet May 01 '22 at 08:35
  • @PaulFrost Fixed, thank you! – Tereza Tizkova May 01 '22 at 09:46

1 Answers1

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Stereographic projection from $p = (-1,0)$ is given by $$\sigma : S^1 \setminus \{p\} \to \mathbb R, \sigma(x,y) = \frac{y}{1+x}.$$

Your map $s$ has the property $p \notin s(\mathbb R)$ (because $s(x)$ has second coordinate $0$ only for $x = 0$, and $s(0) = (1,0)$). Thus we can regard $s$ as a map $s : \mathbb R \to S^1 \setminus \{p\}$. Noting $x^2 + y^2 = 1$ for $(x,y) \in S^1$ we get

$$s(\sigma(x,y)) = s\left(\frac{y}{1+x}\right) = \left(\frac{1- (\frac{y}{1+x})^2}{1+ (\frac{y}{1+x})^2}, \frac{2\frac{y}{1+x}}{1+ (\frac{y}{1+x})^2}\right) = \left(\frac{(1+x)^2 -y^2}{(1+x)^2 +y^2}, \frac{2(1+x)y}{(1+x)^2 +y^2}\right) \\=\left(\frac{(1+2x +x^2 -y^2}{1+2x +x^2 +y^2}, \frac{2(1+x)y}{1+2x +x^2 +y^2}\right) = \left(\frac{(2x +2x^2}{2+2x}, \frac{2(1+x)y}{2+2x}\right) =(x,y), $$

$$\sigma(s(x)) = \sigma\left(\frac{1-x^2}{1+x^2},\frac{2x}{1+x^2} \right) = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{2x}{1+x^2 + 1 -x^2} = x .$$

This shows that $s$ and $\sigma$ are inverse to each other, i.e. both maps are homeomorphisms.

To see that $S^1 \setminus \{p\}$ is dense in $S^1$, observe that the points $p_n = \left(-\sqrt{1 -\frac{1}{n^2}},\frac 1 n \right)$ are in $ S^1 \setminus \{p\}$ and $p_n \to p$.

Paul Frost
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