Connection between one-point compactification of $\mathbb{R}$ and $S^1$

796 Views Asked by At

Definition I got: A one-point compactification of $X$ is $\hat{X}=(X\cup\{\infty\},\tau).$ The new topology $\tau$ is generated by open subsets of $X$, and $U\cup\{\infty\}$ where $U=K^C$ where $K$ is a compact subset in $X$.

Question 1: What is $\infty$ here? Is it a point?

Question 2: I am trying to prove that $\hat{\mathbb{R}}$ is homeomorphic to $S^1=\{(x,y)\in \mathbb{R}^2: x^2+y^2=1\}$ rigorously. I know that $S^1$ is homeomorphic to the quotient space $[0,1]/\sim$ where we identify $0$ with $1$. (Hopefully I got this part correct). I attempted to establish a homeomorphism $h:[0,1]/\sim \to \hat{\mathbb{R}}$ but I don't know how to do it. I think we would want the homeomorphism to satisfy $h([0])=h([1])=\infty$. But then I don't know how to proceed. Moreover, the topology on $\hat{\mathbb{R}}$ is so unintuitive that even if I have a function, I don't know how to check the continuity...

1

There are 1 best solutions below

4
On BEST ANSWER

1) Yes, $\infty$ is a new point.

2) Choose $[-\frac{\pi}{2},\frac{\pi}{2}]$ and identify the endpoints again, then take your homeomorphism to be $\tan x$ for $x\ne \frac{\pi}{2}$, and $\infty$ at $\frac{\pi}{2}$. This will be a continuous bijection with continuous inverse. Continuity at infinity is fairly easy to check, you just have to show that every basic nhood of $\infty$ (i.e. every set of the form $[\infty,-N)\cup(N,\infty]$ contains the image of some $\delta$ ball around $\frac{\pi}{2}=\frac{-\pi}{2}$.

I hope this helps, I'm not sure if this is explicit enough, or well explained enough.