## Homeomorphisms with the discrete topology

Surely sets with the same cardinality are homeomorphic if we assign both of them the discrete topology. What's preventing us from doing that?

For example, (0,1) and (2,3) \cup (4,5) have the same cardinality. With the natural subspace topology they are not homeomorphic - as one is connected and the other isn't. However I could say that they are homeomorphic by assigning the discrete topology on both. Why can't we do that? Is it because that we would lose our intuition of 'continuous deformation' if we did this?

This may sound stupid but I can't seem to get my head around this.

Thanks.
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 Recognitions: Science Advisor Well, I would say: Sometimes , from the context/situation, there is a natural topology that is assumed, and this topology is not (equivalent to ) the discrete topology. You may think of a subspace as being the restriction of the Euclidean topology. Sometimes you want to make your topologies bigger or smaller to allow, e.g., for more compact sets-- which happens when your topology has fewer open sets-- or allowing more sequences to converge (compactness has a lot of nice consequences ). So you adjust your choice of topology to the needs of the situation. This happens, e.g., in functional analysis; see weak topology, weak* topology, strong topology, etc. Also, if all your subspaces are pairwise- homeomorphic , then your theory becomes too general to be interesting. Think that if you applied this idea to spaces in general, Let me see if I can think of better examples/explanations. Maybe this can explain better: http://mathforum.org/kb/message.jspa...6258&tstart=90 and follow the links.

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 Quote by Flying_Goat Surely sets with the same cardinality are homeomorphic if we assign both of them the discrete topology. What's preventing us from doing that? For example, (0,1) and (2,3) \cup (4,5) have the same cardinality. With the natural subspace topology they are not homeomorphic - as one is connected and the other isn't. However I could say that they are homeomorphic by assigning the discrete topology on both. Why can't we do that? Is it because that we would lose our intuition of 'continuous deformation' if we did this? This may sound stupid but I can't seem to get my head around this. Thanks.
Yeah, everything is right. The two sets with the discrete topology are indeed homeomorphic.

Do you find this unintuitive?

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## Homeomorphisms with the discrete topology

I think the OP wants to know why one doesn't just choose to use the discrete

topology on all subspaces, which would make all subspaces pairwise homeomorphic.

Maybe the OP could clarify this ; I hope I did not misrepresent your question, OP,

and let me know otherwise.

 Quote by micromass Do you find this unintuitive?
Surely the disjoint interval (0,1) with the discrete topology is 'disconnected' by defintion, but intuitively it shouldn't be. What I am confused about is that the topological properties of a space changes as you change the topology - it is not an inherent property of the space. With the usual topologies, (0,1) and (2,3)\cup (4,5) are not homeomorphic, but they are with the discrete topology. By changing the topolgies, I can(probably) make their topological properties the same. So what is the point of finding topological invariants in the first place?

If X and Y are homeomorphic, it means that you can 'continuously deform' X into Y. Under what topology(s) would this 'continuous deformation' actually coincide with our intuitive notion of continuous deformation? Does it only happen when we assign the spaces with the normal euclidean subspace topology, as stated by Bacle?
 The point is that you usually have fixed topologies on your spaces. So a topological invariant depends on both the space and the topology you give it. The same space with a different topology could very well have different invariants.

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