Properties of Homeomorphisms between topological spaces

[Goldbeetle]

Dear all,
a homomorphism is a continuous 1-1 function between two topological spaces, that is invertible with continuous inverse. My question is as follows. Let's take the topologies of two topological spaces. Is there a 1-1 function between the two collections of open sets defining the topologies of each of the topological spaces?

Thanks,
Goldbeetle

 

[Landau]

Let's take the topologies of two topological spaces. Is there a 1-1 function between the two collections of open sets defining the topologies of each of the topological spaces?

In this generality, the answer is obviously NO. There are usually many, many topologies on a set. An extreme example: take X with the trivial topology and X with the discrete topology. These topologies have cardinality |X| and 2^X, respectively.

Or were you assuming that the two spaces were homeomorphic?

 

[Goldbeetle]

Landau,
thanks. You're right. I forgot to add that that the two spaces are homomorphic. Does in this case the 1-1 association among open sets of the two topologies exists?

 

[quasar987]

Then, the answer is yes... the association being given by the homeomorphism. And it is because of this property of heomeomorphic space that people say that homeomorphic spaces are "essentially the same" topological spaces.

 

[micromass]

If the spaces are homeomorphic, then the answer is yes. In fact, the function

\mathcal{T}\rightarrow \mathcal{T}^\prime:G\rightarrow f(G)

is a bijection.

An interesting follow-up question is the following: given that there is an (order-preserving) bijection between two topologies, are the spaces homeomorphic?

The answer turns out to be positive under very weak assumptions. We just need the spaces to be sober (a condition much weaker then Hausdorff), and then an order-preserving bijection between the topologies induces a homeomorphism. This question and more is studied in what is called "point-free topology".

 

[Goldbeetle]

Thanks. Where can I find a proof of the result of my question?

 

[micromass]

I don't think you'll find a proof anywhere. It's very easy, so try proving yourself that

\mathcal{T}\rightarrow \mathcal{T}^\prime:G\rightarrow f(G)

is a bijection with inverse

\mathcal{T}^\prime\rightarrow \mathcal{T}:G\rightarrow f^{-1}(G).

 

[Goldbeetle]

It is indeed trivial: (1) the fact that there's a 1-1 association between the power sets of the sets on which each topology is defined is derivable by the fact that there's a bijection between sets, (2) continuity guarantees that the associated sets are open in the respective topologies.

Right?
Goldbeetle

 

[micromass]

Correct!

 

[Goldbeetle]

Thanks!