# Topological counterexample

• Sep 1st 2010, 01:33 AM
ihateyouall
Topological counterexample
How do I show, I am assuming by example, that X can be homeomorphic to a subspace of Y and Y is homeomorphic to a subspace of X but X and Y are not homeomorphic?

I have tried things involving the indiscrete and the discrete topologies as these seemed simple to work with, however I have been unable to come up with anything.

Edit: I have added an extra question later on in the thread that I am stuck on.
• Sep 1st 2010, 01:57 AM
Defunkt
You're making it too complicated I'm afraid. Can you think of two subsets of \$\displaystyle \mathbb{R}\$, with the usual topology, which would work in this case?
• Sep 1st 2010, 02:28 AM
ihateyouall
I thought about this before. I was finding difficulty because I was considering using something like rationals as one subspace however this does not seem so easy to get to be homeomorphic to a subset of something. If one set is countable and the other is uncountable then it would seem that I will have problems getting the larger to be homeomorphic to a subset of the smaller; so I think I might need to have two uncountable sets, or maybe some countable ones. I don't think that I can have two intervals as these would be homeomorphic. Any ideas?
• Sep 1st 2010, 02:42 AM
Defunkt
What about \$\displaystyle [0,1]\$ and \$\displaystyle (0,1)\$?
• Sep 1st 2010, 02:48 AM
ihateyouall
Of course.
• Sep 2nd 2010, 01:16 AM
ihateyouall
I am sorry to ask again but I have a related problem. I have to find X, Y which are not homeomorphic but for whom X*I and Y*I are, where I is the unit interval with the normal topology. I am thinking that I must have to make some move in the added 'dimension' to the space X*I that gets to something that can be called Y*I and the move does not correspond to something valid for homeomorphism in the non-extended state. I have the hint to consider a disk with two arcs attached. Any ideas?