3
$\begingroup$

I found a theorem that the $S^2$ is connected however I cannot find a proof via Google. Is there any hint how to proof that the $S^2$ sphere is connected?

  • 2
    Dear uupsklick, There are lot's of ways to prove this. You should probably provide a little more information about your background in topology (e.g. what texts you have studied from) so that people know at what level to aim their hints. Regards,2011-07-25
  • 1
    I just started viewing topological spaces. The definition I'm trying to understand is that there are no non-intersecting, non-empty open sets $A$ and $B$ such that $M = A \cup B$.2011-07-25
  • 0
    One way, which I think will make sense if you know what "connected" means, would be to exhibit a continuous map from a connected space onto $S^2$. Can you think of such a map?2011-07-25
  • 0
    $R^2$ is connected and I can create a map such as the arcustangens to $S^2$. Am I on the correct way?2011-07-25
  • 1
    With $\arctan$? Below, Bruno gives a map that would work. Of course, you have to prove that $\mathbf R^3 \setminus \{0\}$ is path connected, but that's a good exercise too. Do you know this theorem yet, by the way? That the continuous image of a connected space is connected, I mean. Proving connectedness without using these auxiliary lemmas is usually painful: a book will usually prove directly that intervals in $\mathbf R$ are connected, and then try to make use of various [theorems](http://en.wikipedia.org/wiki/Connected_space#Theorems) to show it for other spaces.2011-07-25
  • 0
    What are you trying to do with $\arctan$? Are you trying to show that the open unit disk in $\mathbf{R}^2$ is connected?2011-07-25
  • 2
    Also: do you know what arc-connectedness is and that it implies connectedness? otherwise, all of the answers so far won't help much... This is precisely why it is good to explain your background!2011-07-25

1 Answers 1

10

Probably the easiest way is to notice that $S^2$ path-connected, which implies connectedness. You can join any two points on $S^2$ with a segment of a great circle.

Or you can notice that $\mathbb{R}^3-\{0\}$ is path connected, and that there exists a continuous surjective map $\mathbb{R}^3-\{0\} \to S^2$.