Topology question; examples of non-homeomorphic metric spaces

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[matt grime]

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.

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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[nocturnal]

oops, I misread the question.

 

[aygunoglu]

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