Compactification of a straight line

Question

Like in the case of mapping a infinite-plane to a sphere (Riemann Sphere), I can understand, that I can map the infinite line ($-\infty,\infty$) to a circle.

Secondly, I can also map a finite line segment (of length $L$) onto a circle (of radius $r$) which is even more straightforward by setting say, $2\pi r = L$. But this kind of mapping can't compactify the line segment in the case when $\lim L \rightarrow \infty $.

I would like to know if there can be a map to compactify the semi-infinite line $(0,\infty)$. Or if not is there a way of proving that there can exist no such maps ?

Answer 1

I'm going to construct a homeomorphism between $(0,\infty)$ and $S^1 - \{(0,1)\}$ which can easily be extended to the compactification $(0,\infty)\cup \{p\}$ and $S^1$.

1. Let $f:(0,\infty) \to (-1,1)$ be the map $f(x) = \frac{2}{x+1} -1$. $f$ is clearly injective, continuous and if $r \in (-1,1)$ then $x = \frac{2}{r+1}-1 \in (0,\infty)$ will map to $r$.
2. Now define the map $g:(-1,1) \to S^1/ \{(0,1)\}$ given by $g(x) = (2x\sqrt{1-x^2}, 2x^2 - 1)$.
3. Take their composition $h = g\circ f$.

We can therefore use the map $h:(0,\infty)\cup \{p\} \to S^1$

$$ h(x) \;\; =\;\; \begin{cases} g(f(x)), & \text{if} \; x \in (0,\infty) \\ (0,1), & \text{if} \; x=p \end{cases}. $$