# Why the circle can't be homeomorphic to a real interval

1. Oct 8, 2011

### camel_jockey

Hi guys!

Excuse the spam, but I would like to ask something which I read in Armstrongs Basic Topology which I am just not 100% sure about.

He says we wish to define homeomorphism such that a circle cannot be homeomorphic to an interval such as [0,1). A continuous function f : X \mapsto Y is one whose every inverse f^-1(N) (N neighbourhood of a mapped point f(x)) is a neighbourhood in X.

He presents a one-to-one and onto function from [0,1) to the circle

x \mapsto exp ( 2∏i x )

This maps the interval into all of the circle. It has an inverse. Exactly why can one not get a continuous inverse??

I suspect this has something to do with the multi-valued property that the complex LOGARITHM suffers from. But somehow I feel the author does not presume the reader to know this much complex analysis - and that there is a more "primitive" reasons dealing with open sets, neighbourhoods, inverses etc. Is there one? Or is the solution merely that there exists no inverse which is not the logarithm?

2. Oct 8, 2011

### lavinia

the inverse of the function exp(2pix) is not continuous.

3. Oct 8, 2011

### Bacle

Well, [0,1) is not compact, and S1 is.

Also, separating sets are a homeomorphism invariant, i.e., the
sets that, once removed, disconnect your space. And the interval
can be disconnected with a point, but the circle cannot. IOW, if
S1 is homeo. to some interval , then S1-{pt.}
is homeo. to interval-h(pt.), but one is diconnected (after
removing a point.), and the other is not.

4. Oct 8, 2011

### Bacle

Actually, the impossibility of even having a continuou bijection between the circle and the real line is equivalent to the impossibility of defining a continuous argument; an argument is precisely a continuous assignment of real values to the points of the circle.
A continuous bijection between S1 compact, and a (Hausdorff) subset of the real line would be a homeo. I have been getting a lot of mileage from this compact-Hausdorff thing recently.

5. Oct 9, 2011

### camel_jockey

Thank you for your replies! I am understanding better now!

Any particular books or PDFs you recommend which gave you this point of view?

6. Oct 9, 2011

### Bacle

Let me think a bit. I think it has mostly been my obsessive web-surfing
on math websites, where I pick things up, and this one being an excellent
one, with great posts from everyone.

7. Oct 10, 2011

### Bacle

Camel_ Jockey:
I just realized the result you're looking for is part of the
more general result of Borsuk-Ulam theorem:

http://en.wikipedia.org/wiki/Borsuk_ulam

Let me know if you want a proof; it should be possible to find one online; I can't think
of a specific book where one could find a proof.

8. Oct 10, 2011

### camel_jockey

Great scott!! Some fine spotting there....

I recognize the theorem - isnt it in Munkres? I unfortunately dont have that book.

You are probably right but I dont quite understand that the problem would be the antipodal points.

Isnt the problem that the continuous function from the sphere to the line is the logarithm which carries this out

Log z := ln r + iθ = ln | z | + i Arg z.
= i Arg z.

(forget about scaling with 2\pi etc)

Isnt the problem that the point (x,y)=(-1,0) cannot be continuously mapped using the "principal value" of the logarithm?

9. Oct 10, 2011

### lavinia

An open neighborhood of 0 in [0,1) is a half open interval. A half open interval is not homeomorphic to an open interval.

10. Oct 10, 2011

### camel_jockey

Thanks Lavinia - though how does one most simply see that a circle is a "fully open interval" ?

11. Oct 10, 2011

### lavinia

The circle is not a fully open interval but... if there were a homeomorphism from the circle to [0,1) then a small open interval around the point on the circle that is mapped to 0 would be mapped homeomorphically to a half open interval surrounding 0.

12. Oct 10, 2011

### camel_jockey

So I gather that you mean that the circle is not an open set? How would one motivate this (intuitively I know you are correct of course) ?

Does this depend on the topology you CHOOSE, and moreover, does it depend if the circle manifold is chosen to be some SUBSET of R^n ? (we assume that there is no distance-function)

13. Oct 10, 2011

### lavinia

The circle is not an open interval not a half open interval not a closed interval. Intuitively the circle has no end point yet the closed and half closed intervals do. This is enough to differentiate them topologically.

yes the topology matters. But then the circle would not be the circle but just a set. Circle means with its manifold topology.

If you give the circle and the half open interval the discrete topology then they are homeomorphic and the exponential mapping you gave above is a homeomorphism.

Last edited: Oct 10, 2011
14. Oct 10, 2011

### Bacle

My point was that if the circle was homeomorphic to an interval, then S1 could be embedded in the real line, and Borsuk-Ulam shows it cannot. I just thought to give a more general result about Sn and Rn, so an obstruction is that in any candidate for an embedding, (at least) two points will map to the same point.

15. Oct 10, 2011

### lavinia

Borsuk Ulam works.

One could also use the first homotopy groups of the circle and the interval.

16. Oct 11, 2011

### lavinia

It seems to me that this problem boils down to the Intermediate Value Theorem.

Let g be a continuous surjective map from the circle to [0,1).
Parameterize the circle by a map,f, from the closed interval ,[0,1], so that f(0) is mapped to 0 by g. Since f(1) = f(0) and g is surjective, gof must return to zero and thus can not be 1-1.

17. Oct 11, 2011

### Bacle

But there cannot be a continuous surjection , because the circle is compact, and the interval [0,1) is not, so f(S1)=[a,b], since the compact subsets of the real line are the closed+bounded intervals, and f continuous maps compact to compact.

18. Oct 11, 2011

### lavinia

yeah but it doesn't matter. The argument works for a non-constant continuous map. Use the same construction for any point in the image of g.