Expanding from this question about $\mathbb{R}$ i want to know if this is also possible for $S^1$. Through the stereographic projection we have a homeomorphism from $\mathbb{R}$ to $S^1\setminus\{p\}$, but i can't quiet finish the proof.
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What do you mean be an open interval in $S^1$? Do you mean the image of an interval (in $\mathbb{R}$) under the map $t\mapsto e^{i t}$? – Hayden May 12 '17 at 10:22
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@Hayden, yes, alternatively because i mentioned it in the question the image of an open interval in $\mathbb{R}$ of the inverse of the stereographic projection – PlatinTato May 12 '17 at 10:25
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Yes, but personally I don't find that to be a satisfying definition of 'interval' in $S^1$; it misses a lot of things that seem like 'intervals', and with that definition not every open subset of $S^1$ can be written as a countable union of disjoint 'open intervals' (take $S^1$ to be your open set; certainly isn't the countable union of disjoint preimages of open intervals of $\mathbb{R}$ under any stereographic projections). That said, any open proper subset of $S^1$ will be the preimage of a countable union of open interval of $\mathbb{R}$ under some stereographic projection. – Hayden May 12 '17 at 10:49