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Okay here is the setup to a problem I have. We have $\mathbb{R}^3$ along with the standard topology and then two subspaces that are given the subspace topology. I forget the explicit definitions of the two subspaces $A,B$ but it is essentially a 3d parabola and a kind of cone shape. I can see looking at them that they would be homeomorphic I think but I don't know how to prove it.

So right now I have $A=\{(x,y,z) \in \mathbb{R^3}: \text{Something}\}$ and $B=\{(x,y,z) \in \mathbb{R^3}: \text{Something}\}$ and I want to show they are homeomorphic so what is a good way to do this?

Also I don't really see the significance of $A,B$ be given the subspace topology what difference does that make?

In general is it better to explicitly give such a homeomorphism or is there something else I can do?

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  • $\begingroup$ Giving $A$ and $B$ the subspace topology is just the normal way to turn them into topological spaces. If you don't give them a topology, then they're just sets. Also, if you give $A$ two different topologies, then the two copies are no longer homeomorphic, at least not via the identity map. So it makes a difference as to what $A$ will be homeomorphic to. $\endgroup$
    – user49640
    Feb 20 '17 at 18:28
  • $\begingroup$ As for how to show that $A$ and $B$ are homeomorphic, depending on their specific characteristics, you may be able to either write down a specific homeomorphism between them, or give a more abstract argument along these lines: math.stackexchange.com/questions/165629/… That specific result may not be useful to you, though. $\endgroup$
    – user49640
    Feb 20 '17 at 18:31
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    $\begingroup$ There is no efficient way to show that two spaces are homeomorphic. This task is extremely difficult and in most cases you just construct a homeomorphism directly or indirectly as a composition of homeomorphisms. It is alot easier to show that two spaces are not homeomorphic by looking at topological invariants. $\endgroup$
    – freakish
    Feb 20 '17 at 18:58
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    $\begingroup$ do not vandalise your posts... You may edit your question to clarify it but not delete it. $\endgroup$
    – Surb
    Mar 4 '17 at 22:37
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Every subset of $\mathbb R^3$ is its own special snowflake, and there is no algorithm for deciding when two of them are homeomorphic. It's not simply that no algorithm is known; no algorithm for solving this can possibly exist because there are too many subspaces.

When you are given these problems you can collect tools that will help you solve new ones, but it's possible that none of your known tools will work on a new problem you're presented with and you'll have to come up with a new idea. This is typical for advanced math problems. There are very few patterns, and solving new problems means coming up with new ideas.

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