1. ## 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.

2. You're making it too complicated I'm afraid. Can you think of two subsets of $\mathbb{R}$, with the usual topology, which would work in this case?

3. 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?

4. What about $[0,1]$ and $(0,1)$?

5. Of course.

6. 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?