Is H^n Homeomorphic to R^n?

[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.

 

[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.

 

[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?

 

[Palindrom]

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

 

[topsquark]

For any topological space X, the boundary of X is empty.

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

-Dan

 

[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

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

 

[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 n^th factor, and we know that the n^th 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 = R^n-1), X x A and X x B are not homeomorphic?

 

[Hurkyl]

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.

 

[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.

 

[topsquark]

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

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

(Ahem!) I knew that.

-Dan

 

[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 R^n-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".

 

[Palindrom]

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.

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?)?

 

[Hurkyl]

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!)

 

[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.

 

[AKG]

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

 

[Hurkyl]

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.

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)

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

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

 

[AKG]

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

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?

 

[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.

 

[Hurkyl]

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.

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.

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!

 

[AKG]

So, it's actually somewhat tricky to show R^n - {P} isn't contractible!

I would think so. Thinking of homotopies of surfaces instead of functions now (or equivalently I guess, looking at functions from S^n-1 to Rⁿ now instead of functions from Rⁿ) then it would be easy to show that every imbedding of S^n-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.

 

[matt grime]

Thinking back several years now I seem to recall that De Rham (co)homology with compact support will distinguish between these two spaces.

 

[Palindrom]

I think I've got it, using a (famous) theorem we proved in class, that S^n isn't contractible. Tell me what you think:

Assume H^n and R^n are homeo. This induces a natural homeo. between their respective one point compactifications.
However, R^n's one point compactification is homeo. to S^n (by the stereo. projection), wheras by the same homeo., it is clear that H^n's one point compactification is homeo. to the subset of S^n for which x_n>=0.
Now, that last space is clearly homeo. to B^(n-1) (the n-1- ball), simply by the projection x_n=0, which is itself contractible! And this is a contradiction to the theorem I mentioned above, that S^n isn't contractible.

Seems O.K?

 

[AKG]

Hⁿ's one-point compactification is homeomorphic to the top half of Sⁿ, which is the subset of Sⁿ with x_n+1 > 0, not x_n > 0. The projection onto the first n co-ordinates (not n-1) gives the closed n-ball (not the n-1 ball) which is contractible.

 

[Palindrom]

Hⁿ's one-point compactification is homeomorphic to the top half of Sⁿ, which is the subset of Sⁿ with x_n+1 > 0, not x_n > 0. The projection onto the first n co-ordinates (not n-1) gives the closed n-ball (not the n-1 ball) which is contractible.

The subset of Sⁿ with x_n+1 > 0 is homeo. to the subset of Sⁿ with x_n > 0... And anyway, the contradiction still holds, as we've got two homeo. spaces, one is contractible and the other isn't, right?

 

[AKG]

The subset of Sⁿ with x_n+1 > 0 is homeo. to the subset of Sⁿ with x_n > 0...

Right, but that's an unnatural way to look at it. You were probably thinking that x_n is the last co-ordinate, which is why you though the projection x_n = 0 gives the n-1 ball. If you want to regard the one-point compactification of the half space as the subset of the n sphere (which is subset of n+1 dimensional Euclidean space) with x_n > 0 instead of the last coordinate x_n+1 > 0, you have to still remember that the projection x_n = 0 gives the (closed) n ball, not the n-1 ball. Of course, both of these are contractible.

And anyway, the contradiction still holds, as we've got two homeo. spaces, one is contractible and the other isn't, right?

Yes, the idea of the proof still works, there's just that small technical detail.

 

[Palindrom]

Now I see what you meant, sorry!
And thanks!

 

[Palindrom]

O.K, now I've been thinking about that one for a while, this is a generalization:

Can an open subset of R^n be homeomorphic to a non-open subset of R^n?

 

[AKG]

Maybe try a simpler problem: can a path-connected non-open subset of R² be homeomorphic to the open unit disk? Can you argue that if A and B are homeomorphic, then so are A^C and B^C? If so, the comeplement of the open disk, A, is closed, and every sequence has a limit point in A^C. The complement of a non-open subset B is non-closed, so it has some sequence that doesn't have a limit point in B^C. Yeah, I don't know if this is a good way to go about it, but it may be a start.

 

[Palindrom]

It's not true that if A and B are homeomorphic, then so are their complements; take [0,infty) and [0,1)- they are homeomorphic, but the first complement is connected while the other isn't...

And it would satisfy me to prove the simpler problem that you presented.

 

[Palindrom]

Ah wait, you said R^2. I really want to work with R^n, R^2 won't satisfy me.