Proving the Intermediate Value Theorem in Higher Dimensions using Homeomorphisms

[mansi]

hi!
would like to know what a homeomorphism means ( how do you geometrically visualize it?)
AND is the symbol 8 homeomorphic to the symbol X? Why or why not?
( from whatever little i know intuitively about homeomorphisms, i think it is not...)

 

[mathwonk]

well one way to detect non homeomorphic objects is by examining their connectivity. an X can be disconnected by removing one point (the center), but an 8 cannot. so they are not homeomorphic.

 

[Dogtanian]

From my topology course we were given the definition that 2 spaces were homeomorphic if there exists a continuous bijective map with a continuous inverse btween the two spaces.
So yuo need a continuous, reversable transfomation from on space to the other, that is, if you can "convert one space to the other" by doing reversable maps then the two spaces are homeomorphic.

So as you can map the surface of a sphere with apoint missing onto a plane, the two spaces are homeomorphic.

It is often easier to see when two spaces are not homeomorphic, and as mathwonk uses in your example of 8 and X, conectivness is a good way to see this.

 

[Hurkyl]

One "visualization" is that two spaces are homeomorphic if, when they're made of rubber, you can deform one into the other.

And, of course, there's the classic joke: a topologist is someone who can't tell the difference between a doughnut and a coffee cup! (Exercise: imagine how to deform a doughnut into a coffee cup)

 

[mathwonk]

to tell that a line is not homeomorphic to a plane, one again can note that the line is disconnected by removing one point, while the plane is not. can you guess a way to tell a plane is not homeomorphic to euclidean three space?

 

[Palindrom]

Remove a line.
In 3 dimensional space, you have to remove a plane to lose connectedness. So your left with showing that a plane can't be the homeomorphic picture (is that the right term in English?) of a line.
But we already did that...

I guess the famous joke about the coffee pot is not entirely untrue...


Just had the final exam in topology this morning. Went pretty well.
By the way, one of the questions there where to show that 0,1,4,6,8 are not homeomorphic to one another. (That is, when drawn like my lecturer drew them).

 

[mathwonk]

good idea on discnnecting the plane and three sp[ace. that is correct of course, in essence, but not in detail. i.e. if a line disconnects the plane then you have to prove no homeomorphic image of a line can disconnect three space. it is not sufficient to prove a line is not homeomorphic to a plane for this.

the problem is with your assertion that it requies a plane to disconnect three space. this is not true. a sphere disconnects three space. what you have to prove rather is that nothing homeomorphic toa line can disconnect three space. this is easier in the previous case because the only thing homeomorphic toa point is a point, but there ise a huge variety of things homeomorphic to a line.

by the way i do not know an elementary proof of this. if one knows the construction of "homology" groups, and the basic tools for calculating them, it can be shown using them that the 2 sphere and the three sphere are not homeomorphic. it follows that a plane and three space are not homeomorphic, since their one point compactifications are those spheres.

 

[Palindrom]

That's right. I "shot" my post too quickly.
I'd like to take the more advanced topo. course when someone decides to give it here. My T.A. told me some of the things you just mentioned, but in a very "wavy" way... I don't have the proper tools yet.

 

[mathwonk]

topology is not so easy technically. even to check that every curve homeomorphic to a circle separates the plane into exactly two components is highly non trivial, called the jordan curve theorem.

 

[Palindrom]

Heard of it. We mentioned it in the course. But only mentioned.

 

[mathwonk]

here is the first interesting question in topology: prove that a continuous map from the unit disc to the plane, that sends the unit circle identically onto itself, must send some point of the disc to the origin.

Hint: use polar cordinates. This can also be proved using calculus, if you give yourself a smooth, i.e. continuously differentiable map. Then it all follows from the existence of the angle form, "dtheta", and integration i.e. polar coordinates again.

this generalizes to 2 dimensions, the old intermediate value theorem, that a continuous map from the closed unit interval to the line, which sends the end points 0 and 1, to numbers of opposite sign, must send some point of the open unit interval to zero.