I am trying to prove that the one-point compactification of the interval $(0,1)$ with the Euclidean Topology is $S_{1} = \{(x,y)\in \mathbb{R}^{2}|x^{2}+y^{2}=1\}$ and I am having some trouble getting there.
The process I am going by has two steps:
- Show that $(0,1)$ is homeomorphic to $S_{1} – \{ (1,0)\}$
- Show that the one-point compactification of $S_{1} – \{(1,0)\}$ is $S_{1}$ itself.
Right now, I am stuck on Step 1:
What I did was I set up a mapping $g: (0,1) \to S_{1} – \{ (1,0) \}$ defined by $g: x \mapsto (\cos 2\pi x, \sin 2 \pi x)$. Now, I need to show that $g$ is a homeomorphism.
To show that $g$ is onto, I said that if we let $X:= \cos 2 \pi x$ and $Y:= \sin 2 \pi x $, $X^{2}+Y^{2} = (\cos 2\pi x)^{2} + (\sin 2 \pi x)^{2} = \cos ^{2} 2\pi x + \sin^{2} 2 \pi x = 1$, so every point on the unit circle is given by $(X,Y)$. The only point on the unit circle we are not including is $X=1$, $Y=0$, but the preimage of this point under $g$ would be the point $x = 0$, which is not included in the domain. Therefore, the range consists solely of the image set of $g$, so $g$ is onto.
To show that $g$ is one-to-one, literally what I did was, I drew graphs of the sine and cosine functions from $0$ to $2\pi$ side-by-side on the same plot, and pointed out that as the parameter $x$ ranges from $0$ to $1$, $X = \cos 2\pi x$ ranges from $\cos (0) $ to $\cos (2 \pi)$ and $Y = \sin 2 \pi x$ ranges from $\sin (0)$ to $\sin (2 \pi)$. Looking at the graphs, we see that for no $x_{1} \neq x_{2}$ do we obtain $(\cos (2\pi x_{1}), \sin(2 \pi x_{1})) = (\cos (2 \pi x_{2}), \sin(2 \pi x_{2}))$, so clearly $g$ is one-to-one as well.
I feel as though these two proofs are too handwavey – could anyone provide me with more rigorous looking way of proving these things?
Really, though, what I wanted to ask about is showing that $g$ is continuous and an open map. If I can show these two things, I can show that $g$ is a homeomorphism and thus I will have part 1 completed. However, normally, to show that $g$ is continuous, I would take an arbitrary open set in the range, so an arbitrary open set in $S_{1}-\{ (1, 0) \}$ and show that its preimage is also open in the interval $(0,1)$. The problem here is that I don't know what an arbitrary open set in $S_{1}-\{ (1,0) \}$ looks like! Can somebody please show me how to prove this part? I imagine that once I've seen what this proof is supposed to look like, showing that an open set in $(0,1)$ maps to an open set in $S_{1} – \{(1,0)\}$ won't be too hard. Although, if you want to provide me with a proof of that too, I won't complain!
Thank you ahead of time for your help and patience!
Best Answer
Let $X$ be a compact Hausdoff space, and let $x_0$ be any point of $X$. Then $X$ is the one-point compactification of $X-\{x_0\}$. So to show $S^1$ is the one-point compactification of $(0,1)$ all you need is a homeomorphism from $(0,1)$ to $S^1-\{(1,0)\}$. As you say $f:t\to (\cos2\pi t,\sin2\pi t)$ is a continuous bijection. So you want the reverse to be continuous.
Define
$g_1:\{(x,y)\in S_1:y>0\}\to(0,1)$ by $g_1(x,y)=(2\pi)^{-1}\cos^{-1} x$,
$g_2:\{(x,y)\in S_1:x<0\}\to(0,1)$ by $g_2(x,y)=1/2-(2\pi)^{-1}\sin^{-1} y$,
$g_3:\{(x,y)\in S_1:y<0\}\to(0,1)$ by $g_3(x,y)=1-(2\pi)^{-1}\cos^{-1} x$.
These are continuous, defined on open sets, match up on where each is defined, and so glue together to form a continuous map $g:S^1-\{(1,0)\}\to(0,1)$. This is inverse to $f$.