S^1 (the circle) is not homeomorphic to any interval of the the real line.

I can sort of see how this works.

We can wrap the interval around itself to produce a circle. (continuous)

But if we take a circle and "rip" it and make it a line, that is no longer continuous.

But is there a way to state this more formally? Also this question is listed under connectedness, is there some sort of argument that can be made that way? But both spaces seem to be connected though.

Thank you for your help.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

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Originally Posted by

**Jame** I can sort of see how this works.

We can wrap the interval around itself to produce a circle. (continuous)

But if we take a circle and "rip" it and make it a line, that is no longer continuous.

But is there a way to state this more formally? Also this question is listed under connectedness, is there some sort of argument that can be made that way? But both spaces seem to be connected though.

Thank you for your help.

This is annoying to do this way (by connectedness) haha. We have three choices, the interval is open, half-open, or closed. If it's either of the first two then they can't be homeomorphic to $\displaystyle \mathbb{S}^$ by a compactness argument, a closed interval $\displaystyle [a,b]$ is not homeomorphic to $\displaystyle \mathbb{S}^1$ because $\displaystyle [a,b]$ possesses non-cut-points and $\displaystyle \mathbb{S}^1$ does. Said differently, assume that $\displaystyle f:[a,b]\to\mathbb{S}^1$ was a homeomorphism then $\displaystyle f: (a,b]\to\mathbb{S}^1-\{f(a)\}$ is a homeomorphism, but $\displaystyle (a,b]$ is connected but $\displaystyle \mathbb{S}^1-{f(a)}$ isn't (removing any point of $\displaystyle \mathbb{S}^1$ diconnects it).

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

If you know something about loops or the fundamental group, then you can do it pretty easily. If not, the method needs a bit of explaining:

The idea is to consider closed paths / curves in each space. (A path is a continuous function $\displaystyle \gamma:[0,1]\to X$. If it is closed, then its beginning point is the same as its end point; that is, $\displaystyle \gamma(0)=\gamma(1)$.) Then you can try to play around with the loops by deforming them continuously. It then turns out that you can put different kinds of loops into different equivalence classes - we say that two loops are equivalent if one can be deformed into the other (and vice versa). (For example, a loop that wraps around a circle once cannot be deformed into a loop that wraps around the circle twice.) Then, if two spaces are homeomorphic, those equivalence classes of loops should be the same, in some sense. It is possible to show that $\displaystyle S^1$ has infinitely many classes of loops and a line segment has only one. Therefore, the two cannot be homeomorphic.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

Thank you for your replies.

I am confused though. If we remove a single point from the circle doesn't it remain connected?

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

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Originally Posted by

**Jame** Thank you for your replies.

I am confused though. If we remove a single point from the circle doesn't it remain connected?

Lol, of course. I meant two points. $\displaystyle f: (a,b)\to\mathbb{S}^1-\{f(a),f(b)\}$ would be a homeo, and we know that $\displaystyle f(a)\ne f(b)$.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

Could we also assume

$\displaystyle f:[a,b] \rightarrow S^1$ is a homeomorphism,

then $\displaystyle f:[a,b] - \left\{ a \right\} \rightarrow S^1 - \left\{f(a)\right\}$ must be a homeomorphism as well.

But $\displaystyle [a,b] - \left\{ a \right\}$ is not connected, where $\displaystyle S^1-\left\{f(a)\right\}$ is connected?

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

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Originally Posted by

**Jame** Could we also assume

$\displaystyle f:[a,b] \rightarrow S^1$ is a homeomorphism,

then $\displaystyle f:[a,b] - \left\{ a \right\} \rightarrow S^1 - \left\{f(a)\right\}$ must be a homeomorphism as well.

But $\displaystyle [a,b] - \left\{ a \right\}$ is not connected, where $\displaystyle S^1-\left\{f(a)\right\}$ is connected?

No, because $\displaystyle [a,b]-\{a\}=(a,b]$ is connected.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

you see, i would assume it would be better to take this approach:

if the circle and an interval were homeomorphic, then they are still homeomorphic if we remove one point from each. (using the appropriate induced relative topologies).

but if x is an interior point of an interval I, then I - {x} is not connected, whereas $\displaystyle S^1 - \{f^{-1}(x)\}$ is.

of course, it's possible I has no interior points (I = [a,a] for some real a). but then I - {a} is null, and certainly not homeomorphic to $\displaystyle S^1 - \{f^{-1}(a)\}$.

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Originally Posted by

**roninpro** If you know something about loops or the fundamental group, then you can do it pretty easily. If not, the method needs a bit of explaining:

The idea is to consider closed paths / curves in each space. (A path is a continuous function $\displaystyle \gamma:[0,1]\to X$. If it is closed, then its beginning point is the same as its end point; that is, $\displaystyle \gamma(0)=\gamma(1)$.) Then you can try to play around with the loops by deforming them continuously. It then turns out that you can put different kinds of loops into different equivalence classes - we say that two loops are equivalent if one can be deformed into the other (and vice versa). (For example, a loop that wraps around a circle once cannot be deformed into a loop that wraps around the circle twice.) Then, if two spaces are homeomorphic, those equivalence classes of loops should be the same, in some sense. It is possible to show that $\displaystyle S^1$ has infinitely many classes of loops and a line segment has only one. Therefore, the two cannot be homeomorphic.

$\displaystyle \pi_1(I) \cong \{0\}, \pi_1(S^1) \cong \mathbb{Z}$, right? that is, an interval is contractible, the circle is not.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

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Originally Posted by

**Deveno** you see, i would assume it would be better to take this approach:

if the circle and an interval were homeomorphic, then they are still homeomorphic if we remove one point from each. (using the appropriate induced relative topologies).

but if x is an interior point of an interval I, then I - {x} is not connected, whereas $\displaystyle S^1 - \{f^{-1}(x)\}$ is.

of course, it's possible I has no interior points (I = [a,a] for some real a). but then I - {a} is null, and certainly not homeomorphic to $\displaystyle S^1 - \{f^{-1}(a)\}$.

Said basically the same thing I did, no?

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$\displaystyle \pi_1(I) \cong \{0\}, \pi_1(S^1) \cong \mathbb{Z}$, right? that is, an interval is contractible, the circle is not.

Yeah, and so any homeomorphism would induce an isomorphism $\displaystyle 0\xrightarrow{\approx}\mathbb{Z}$.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

does my suggestion work if the point removed from [a,b] is not either endpoint? i think it would be alright. that would result in a disconnected space. the circle will still be connected though.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

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Originally Posted by

**Jame** does my suggestion work if the point removed from [a,b] is not either endpoint? i think it would be alright. that would result in a disconnected space. the circle will still be connected though.

Yes, that could work.

Re: S^1 (the circle) is not homeomorphic to any interval of the the real line.

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**Drexel28** Said basically the same thing I did, no?

ach, so touchy these days! well, fewer cases, right? my argument dispatches with open, closed, half-open distinctions, we either have a singleton, or an interval with an interior point. and there's no mention of compactness, either. the focus is on connectedness, how many snips of my scissors do i need to make the argument? one.

but, arguably, on some "meta-level" they're the same, since we're using the same inspiration. i like roninpro's idea the best, because "sets" are messy (proving continuity of something can be a chore, especially if you have some wacked-out topology. even metric spaces (which are fairly well-behaved) involve a few dirty inequalities, bounding this and that), whereas groups are nice and clean, you have fewer options,and you get more clear-cut answers.

and all we really care about is homotopy classes anyway, right? functions are just so 20th century....we're modern peeps, lets get functors in da house.