Which of $(-\infty,\infty]$ and $[-\infty,\infty]$ is homeomorphic to $S^1$?

Question

Is it correct to say that $(-\infty,\infty]$ is homeomorphic to $S^1$? or it is $[-\infty,\infty]$? (considering standard topology).

Would you please provide some explanation or better a rigorous mathematical proof for your answer?

Thank you.

Answer 1

Neither is homeomorphic to $S^1$. $(-\infty,\infty]$ is not compact, while $S^1$ is, so they cannot be homeomorphic. $[-\infty,\infty]$ is compact, being homeomorphic to $[0,1]$, but removing any one point of $\Bbb R$ from $[-\infty,\infty]$ disconnects it, while removing one point from $S^1$ does not disconnect $S^1$, so they are not homeomorphic either.

Answer 2

If $(-\infty,+\infty]$ is construed as a mere set rather than as a topological space, then it's not homeomorphic to anything including itself. Now lets add a topology: a basic open neighborhood of a point other than $+\infty$ is an open interval containing that point; a basic open neighborhood of $+\infty$ is a set of the form $(a,+\infty]\cup (-\infty,b)$; and an open set is the union of any collection of these "basic" open sets. That space is homeomorphic to a circle, and one homeomorphism is this: $$ p(t) = \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right). \tag 1 $$ The point $p(t)$ is on the circle $x^2+y^2=1$ and we consider $p(+\infty)$ to be $\lim\limits_{t\to+\infty} p(t) = (-1,0)$.

But that is perhaps not the most usual topology on the set $(-\infty,+\infty]$. In the order topology, one can define basic open sets as above except that a basic open neighborhood of $+\infty$ would be of the form $(a,+\infty]$.

But if you're going to do things according to $(1)$ above, it is better not to distinguish between two objects $\pm\infty$, but rather to have just one $\infty$ that is at both ends of the line. Then one can define, for example, $\tan(\pi/2)$ to be $\infty$ and that makes it a continuous function, and likewise every rational function becomes continuous at points where it has vertical asymptotes and at $\infty$/

Answer 3

Related to some of the comments, you might be interested to know that $(-\infty,\infty]$ in the cyclic order topology is homeomorphic to $S^1$. The cyclic order topology is coarser than the corresponding linear order topology, in this case strictly coarser.