# Homeomorphism of a compact real line to the real line

Any open interval is homeomorphic to the real line, for example $(-1,1)$, correct? But a one-point compactification of the real line is not homeomorphic to the real line because the real line is open and the one-point compactification of the real line has a compact circumference. Correct? Then is the two point compactification of the real line homeomorphic to the real line? The answer would be no, right? If I define an order on the extended real line of $-\infty\leq a \leq \infty$, the extended real line is a compact Hausdorff space that is homeomorphic to the interval $[0,1]$. But there is no way for it to be homeomorphic to an open interval and thus to the real line, right? I just want to make sure I understood this correctly.

A one-point compactification of the real line is homeomorphic to a circle. What is the two-point compactification of the real line homeomorphic to, $[0,1]$? Thanks.

• $[0,1]$ is indeed a compactification of $\mathbb R$ where you've added two points. Note that there isn't a general notion of "two-point compactification" like there is one of "one-point compactification". Some spaces admit a compactification where you add two points (like $\mathbb R$), some don't (like $\mathbb R^2$). There is however, a general compactification which gives the one you're interested in as a special example, the end compactification: en.wikipedia.org/wiki/End_%28topology%29 Sep 9, 2015 at 19:13
• Thanks for the link and the answer. I will go read that. I wasn't sure at all about the two-point compactification. Thanks again. Sep 10, 2015 at 0:46

Every open interval $(a,b)\subseteq \mathbb R$ is indeed homeomorphic to $\mathbb R$. The one-point compactification of $\mathbb R$ is homeomorphic to the unit circle $\mathcal S^1$.

Now, $\mathbb R$ and $\mathcal S^1$ are not homeomorphic because the latter is compact (and the real line isn't). Is that what you mean by "compact circumference"?

The two-point compactification of $\mathbb R$ is homeomorphic to the closed interval $[0,1]$ (or any other closed real interval).

So it looks like you got it right.

[Also, here's a post where you can find a definition of the two-point compactification and also an interesting discussion: Question on compactification ]

• Yes, that is what I meant. I probably should have just said it is compact. Thanks. Thanks for the pointer. I appreciate it. Thanks for the validation. Sep 10, 2015 at 0:45
• If I read that right, then a Riemann sphere is homeomorphic to a one-point compaction of R^2, correct? Thanks. Sep 10, 2015 at 1:07
• Yes, that's correct. Sep 10, 2015 at 16:20