# Two topological spaces are homeomorphic

1. Sep 21, 2009

### symbol0

I had the following thought/conjecture:
Two topological spaces are homeomorphic iff the two topologies are isomorphic.

When I say that the two topologies are isomorphic, I mean that they are both monoids (the operation is union) and there is a bijective mapping f such that f(A) U f(B) = f(A U B) for all A,B in one of the topologies.

Does that make sense? am I on the right track?

I'll appreciate any feedback.

2. Sep 22, 2009

### joeboo

Re: conjecture

Consider any two sets of different size. Now give them both the indiscrete, or trivial topology (only open sets are the empty set, and the entire set itself). How does this affect the argument?

Suppose you also have an underlying isomorphism of sets (ie a bijection). How does this change things?

3. Sep 22, 2009

### symbol0

Re: conjecture

I see joeboo. So your example shows that having isomorphic topologies does not imply homeomorphic topological spaces.
But the other implication is true, right?
That is, If two topological spaces are homeomorphic, then the two topologies are isomorphic.

right?

4. Sep 22, 2009

### joeboo

Re: conjecture

I believe so, yes, but how would you show it? If you have a homeomorphism between two spaces, could you then construct (using said homeomorphism) a isomorphism their respective topologies?

Also, what do you think of the alternative I suggested? It may help you understand the situation better.

5. Sep 22, 2009

### g_edgar

Re: conjecture

Of course. You can probably figure out how to do it.

6. Sep 22, 2009

### symbol0

Re: conjecture

If the two spaces are homeomorphic, there is a bijective correspondence f such that f(V) is open iff V is open. That is, we have a bijective correspondence between the two topologies.
And since f is bijective, it is easy to show that for any sets A,B in a topology,
f(A U B)= f(A) U f(B). So the topologies are isomorphic.

Joeboo, you ask about the alternative you suggested.
I thought that was just a counterexample of the converse implication.
What else is there to understand?

7. Sep 23, 2009

### joeboo

Re: conjecture

I was suggesting you consider the scenario where you have a bijection of sets in addition to an isomorphism of the topologies, and see if this is equivalent to the spaces being homeomorphic.

8. Sep 23, 2009

### symbol0

Re: conjecture

sure, with that scenario, you would actually have more than what you need to get homeomorphic spaces. By just having a bijection f of sets where f is also a bijection between the topologies (they don't need to be isomorphic), then you have the definition of homeomorphism.

9. Sep 23, 2009

### joeboo

Re: conjecture

Careful here; the map between topologies is a correspondence between elements of the topologies, or open sets. The bijection between the spaces is a correspondence between elements in the spaces, or points. They are not the same functions.
However, your above argument is valid IF the isomorphism of the topologies is induced by the bijection between sets.
What if it isn't?

10. Sep 24, 2009

### symbol0

Re: conjecture

I see joeboo,
I'll think about it and reply later. I have a pretty hectic week ahead.
So far I can tell you that if you start with the isomorphism between the topologies, the empty set has to be mapped to the empty set and the full set has to be mapped to the other full set.

11. Oct 9, 2009

### symbol0

Re: conjecture

Hi Joeboo,
I kind of stopped thinking about this, but without thinking too much, I would say that if you start with the isomorphism between the topologies, and then you also have a bijection between the sets, then you would have an homeomorphism, (where the open sets are not necessarily the same open sets you started with).

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