# Homework Help: Topology question; examples of non-homeomorphic metric spaces

1. Oct 8, 2006

### notmuch

Hello,

Here's a problem that I'm having trouble with:

Give an example of metric spaces X and Y and continuous maps f: X->Y and g: Y->X such that f and g are both bijective but X and Y are not homeomorphic.

I can find plenty of examples where I can find one such function, but finding the second is always a problem.

2. Oct 8, 2006

### HallsofIvy

Let X be the real numbers with the "usual" topology: d(x,y)= |x-y|.

Let Y be the real numbers with the "discrete" topology: d(x,y)= 0 if x= y, 1 otherwise.

Let f(x)= x.

3. Oct 8, 2006

### notmuch

HallsofIvy,

Thanks for the reply, but that won't work with this definition of continuity:

f:A->B is continuous if
f^(-1)(U) is open in A for every open set U of B.

4. Oct 8, 2006

### matt grime

I suggest you re-examine the definition of homeomorphic. Given any space X and two topologies S and T, (X,S) and (X,T) are never homeomorphic.

5. Oct 8, 2006

### notmuch

Matt,

I understand that. I was not refuting the fact that those two topoligies are not homeomorphic. Finding such examples of non-homeomorphic spaces is easy, but finding the functions satisfying the conditions is the hard part.

6. Oct 8, 2006

### StatusX

You can form a topology on R from the basis of half open intervals whose endpoints are integers. Can you think of anything to do with this? EDIT: I don't know if this will work- in the example I had in mind it turns out the spaces are homeomorphic.

Last edited: Oct 8, 2006
7. Oct 8, 2006

### matt grime

Did you stop to think about which topology we take as that on the domain and the range? The map f(x)=x is a bijection from R to R, with obvious bijective inverse. Give one copy of R the norm topology, the other the discrete. Done.

If you've found one bijective function, then it is invertible, as a function. That is either the definition of bijective, or a trivial consequence of the definition of bijective.

Last edited: Oct 8, 2006
8. Oct 8, 2006

### AKG

I don't think HallsofIvy's example will work. For every y in Y, {y} is open. For any bijection from X to Y, the inverse image of {y} is a one-point set, which is not open in X, so there are no continuous bijections from X to Y.

9. Oct 8, 2006

### matt grime

Ah, see the question properly now.

Remember that any continuous bijection from a compact space to a hausdorff space is a homeomorphism (or perhaps I mean from a hausdorff space to a compact space), thus one will need to find and use a non-hausdorff space, or a non-compact space.

10. Oct 8, 2006

### notmuch

Matt,

Sorry for the confusion. I was a little confused by your response so I didn't reply, I was trying to see if I was understanding you correctly. AKG expressed what I meant to say.

You're right about the Hausdorff/compactness conditions... we are stuck with finding both X and Y being noncompact spaces, since they must also be metric spaces (thus Hausdorff). I still have problems finding such an example.

11. Oct 8, 2006

### nocturnal

Let X = [0,1) with the normal metric on R

Let Y be the uinit square, ie: $Y = \{(x_1,x_2): x_1^2 + x_2^2 = 1\}$, with the normal metric on R^2

Let $f:X \rightarrow Y, \ \ \ \theta \mapsto (cos2 \pi \theta, sin2 \pi \theta)$.

Take U = [0, 1/2). U is open in X but f(U) is not open in Y.

12. Oct 8, 2006

### AKG

nocturnal, do you understand the stated problem? You've found a continuous bijection from X to Y and shown that it isn't a homeomorphism. This is almost entirely pointless. You don't need to show that one particular continuous bijection from X to Y isn't a homeomorphism, you have to show that no continuous bijection from X to Y is a homeomorphism. Although you haven't done this, it can be done. More importantly, you have to find a continuous bijection from Y to X as well. This cannot be done.

13. Oct 9, 2006

### nocturnal

14. May 25, 2007

### aygunoglu

do you really think that problem is true,
I thnk it is wrong, it is a different defination of homemorphizm