# Homework Help: Is H^n homeo. to R^n?

1. Feb 21, 2006

### Palindrom

I'm trying to see why H^n is not homeo. to R^n.

I'm now having doubts on whether this is at all true.

Is it? If not, why not?

*By H^n I mean the closed "upper" space, i.e. x_n>=0.

2. Feb 21, 2006

### HallsofIvy

For one obvious point, Hn has a boundary and Rn doesn't. That is, neighborhoods of points (x, y, 0) are different from neighborhoods of (x, y, z) for z> 0. There is no way to homeomorphically map Rn to Hn that will respect that difference.

3. Feb 21, 2006

### Palindrom

But H^n has no boundary as a topological space. For any topological space X, the boundary of X is empty.

And why are the nhbds you mentioned different? A nhbd of a point with z>0 is homeo. to R^n, whereas a nhbd of a point with z=0 is homeo. to H^n. So that comes down to the same question: Why is H^n not homeo. to R^n?

You see my problem?

4. Feb 21, 2006

### Palindrom

Oh, come on, don't give up, help me out here...

5. Feb 21, 2006

### topsquark

Consider the space $$X = (1,2) \cup (3,4)$$ in the standard order topology. BdX={1,2,3,4} is not empty.

-Dan

6. Feb 21, 2006

### AKG

topsquark, regarding (1,2) U (3,4) as a subset of R, it does have that boundary. But regarding X as a space itself, what Palindrom said is correct: For any topological space X, the boundary of X is empty.

The boundary of a set S in a topological space X is defined as:

$$Bd(S) = \overline{S} \cap \overline{X - S}$$

So

$$Bd(X) = \overline{X} \cap \overline{X - X} = X \cap \overline{\emptyset} = X \cap \emptyset = \emptyset$$

7. Feb 21, 2006

### AKG

The one-point compactification of R is homeomorphic to S1, and you can remove any point from S1 and still have a path-connected space (you will have something homeomorphic to (0,1), which is in fact homeomorphic to R). The one-point compactification of H is homeomorphic to, I believe, the interval I (i.e. the compact space [0, 1]). However, you can't just remove any point from I and be left with a path-connected space; in fact, if you remove any point except the extreme points, you won't even be left with a connected space.

Now you know that Rn = R x ... x R (n times), and you know that Hn = R x ... x R x H (where there are n-1 copies of R). So Rn and Hn differ by their nth factor, and we know that the nth factors are not homeomorphic, so can we use this to say that the product spaces are not homeomorphic? That is, can you argue that if A and B are not homeomorphic, then for any non-empty space X (for us, X = Rn-1), X x A and X x B are not homeomorphic?

8. Feb 21, 2006

### Hurkyl

Staff Emeritus
I don't think we need to go quite that far: removing any point from Rn changes some of its qualitative topological properties -- but you can remove points from Hn without said change.

I wanted to suggest something about removing a whole hyperplane, when is intuitively obvious, but it wasn't clear to me how the details would pan out.

By the way, XxA and XxB can be homeomorphic without A and B being homeomorphic.

As an example, let X be the product of infinitely many copies of R. Then, simply let A=R and B=R².

Or, as a simpler example, let X, A, and B all be discrete, A and B be nonempty with different cardinality, and X have infinite cardinality at least as much as both A and B.

Last edited: Feb 21, 2006
9. Feb 21, 2006

### matt grime

In general you cannot use Krull-Schmidt (cancellation) properties like that unless you prove they work. If we assumed they worked even in finite dimensional cases then there would be smooth vector fields on the sphere.

10. Feb 21, 2006

### topsquark

:uhh: (Ahem!) I knew that.

-Dan

11. Feb 21, 2006

### AKG

For every point in Euclidean space p, and for every open ball B about p, the boundary of B does not contain p. The same is not true for points in half-spaces, in particular, any point with nth coordinate 0 in a half-space is contained in the boundary of every ball containing it.

EDIT: No wait, that's wrong.

EDIT: Okay, I don't know how to make a full proof out of this, but I think that every isometric embedding of Rn-1 into Rn separates Rn in the sense that if Z denotes the image of this embedding, then Rn - Z is disconnected. On the other hand, the same does not hold for the half-space, in particular, then the image Z is the "bottom" of the space, i.e. Z is the set of points with nth co-ordinate 0. I think this is what Hurkyl might have been trying to get at with his idea about "removing a whole hyperplane".

Last edited: Feb 21, 2006
12. Feb 21, 2006

### Palindrom

Please, just give me one property that is always changed by removing a point from R^n, but that there exists a point in H^n that when removed, doesn't change that property.

I also thought about the hyperplane thing for a minute, but then i saw I couldn't formalize it.

So then are they homeomorphic or not (clearly not, but why?)?

13. Feb 21, 2006

### Hurkyl

Staff Emeritus
R^n is contractible. R^n - {P} is not. (This was the first one that came to mind)

(By the way, you can't talk about an isometric embedding of R^{n-1} into R^n, because we haven't picked a metric!)

Last edited: Feb 21, 2006
14. Feb 21, 2006

### AKG

Well we must have a topology, right? And that topology is bound to be the standard topology. But the standard topology is precisely the topology induced by the standard metric.

15. Feb 21, 2006

### AKG

How is $\mathbb{H}^n - \{p\}$ contractible? Wouldn't it be noncontractible for the same reasons $\mathbb{R}^n - \{p\}$ would be?

16. Feb 21, 2006

### Hurkyl

Staff Emeritus
But the standard topology doesn't force the metric. Also, we're mainly interested in homeomorphisms, and they don't respect the metric.

Incidentally, I can pick metrics on the two spaces so that I can isometrically embed R^{n-1}-->R^n without disconnecting R^n.

e.g. on R^{n-1}, I can take the metric d(P,Q) = arctan(||P-Q||), where ||.|| is the usual Euclidean norm. Any isometric embedding of this into R^n would simply map it to an open ball of codimension 1 and radius pi/2. (e.g. if n=3, to a disk)

Remove a "boundary" point of H^n. The resulting space is contractible.

17. Feb 21, 2006

### AKG

I just read this definition today. So it means that the identity map is homotopic to a constant function. So you'd remove a point p from the "boundary". Then pick your constant function to be any point c "above" the "bottom" and have your homotopy to just map (x, t) to the point a distance of t||x-c|| from the point c in the direction of x, right? This works because from the point of view of c, there is nothing behind p, whereas a homotopy like this in Rn would end up trying to send points behind p to p itself?

18. Feb 21, 2006

### AKG

Actually, I was originally trying to think of something like this, but I was thinking of using the idea that some loops are not homotopic to a point in R² - p but they are in H² - p, but I couldn't think of how to generalize it to dimensions higher than 2.

19. Feb 21, 2006

### Hurkyl

Staff Emeritus
Right -- that explicit example proves R^n and H^n and H^n - {P} (with P on the boundary) are all contractible.

But contractions can be much more interesting -- for example, if X is three-quarters of an annulus, then X is contractible.

So, it's actually somewhat tricky to show R^n - {P} isn't contractible! But it is the approach I would take.

And it would be based on this idea, by looking at a copy of S^{n-1} embedded in R^n such that P is in its interior.

Surely this problem can be easily done with homology. But, alas, there's only one homology theory with which I have even a rudimentary understanding, and I don't know if it can be applied to non-compact spaces! I suppose you could simply apply the one-point compactification, though. I was trying to avoid that!

20. Feb 21, 2006

### AKG

I would think so. Thinking of homotopies of surfaces instead of functions now (or equivalently I guess, looking at functions from Sn-1 to Rn now instead of functions from Rn) then it would be easy to show that every imbedding of Sn-1 into the half-space minus a well-chosen point is null-homotopic but the same is not true of Euclidean space minus any point.