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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?

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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, –  Matt E Jul 25 '11 at 15:56
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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 $M = A \cup B$. –  uupsklick Jul 25 '11 at 16:00
    
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? –  Dylan Moreland Jul 25 '11 at 16:01
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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 to show it for other spaces. –  Dylan Moreland Jul 25 '11 at 16:16
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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! –  Mariano Suárez-Alvarez Jul 25 '11 at 17:02

1 Answer 1

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 $\mathbb{R}^3-\{0\} \to S^2$.

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