Are Q and N Homeomorphic in Terms of Topology?

  • Thread starter Scousergirl
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In summary, the conversation discusses whether Q (the set of rational numbers) and N (the set of natural numbers) are homeomorphic. The discussion mentions the existence of a bijection between the two sets, but questions its continuity and whether it is preserved under different topologies. The claim is made that they are not homeomorphic and a proof is provided. However, the other person questions the proof and suggests considering the topologies of the two sets instead. The conversation ends with the suggestion to find a set that is not open in Q to show that they are not homeomorphic.
  • #1
Scousergirl
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Is Q homeomorphic to N?

I understand that there exists a bijection from Q to N but I cannot figure out how this function is continuous and it's inverse is also continuous.
 
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  • #2
With what topology? The subspace topology from being a subspace of the reals with their normal topology, or the discrete topology. They're not homeomorphic in the former, but are in the latter.
 
  • #3
The normal topology...this is what i came up with...not sure if its right:

Claim: they are not homeomorphic
Proof: Assume they are. Then There exists a continuous function f from Q to N. Therefore all of the sequences in Q are mapped to a sequence in N preserving limits. But since sequences in N converge constantly, this cannot be a bijection. therefore they are not homeomorphic.

Does this make sense?
 
  • #4
Why do sequences in N converge constantly? The sequence (1, 2, 3, ...) certainly doesn't converge. Did you mean to say that the only sequences in N that converge are the ones that are eventually constant? But then why is this a contradiction? Maybe your homeomorphism f maps convergent sequences to sequences that are eventually constant. Can this happen?

But honestly I wouldn't bother thinking about it this way. Just think of what the topologies are. N will get the trivial topology, i.e. every set is going to be open (why?). Will Q get the trivial topology? No, it won't. Try to find a set that's not open in Q. This will be enough to show that they're not homeomorphic.
 

1. Is Q homeomorphic to N?

Yes, Q (the set of rational numbers) is homeomorphic to N (the set of natural numbers). This means that there exists a continuous and bijective function between the two sets, with a continuous inverse function. In simple terms, the two sets have the same topological structure.

2. What does it mean for two sets to be homeomorphic?

Two sets are homeomorphic if there exists a continuous and bijective function between them, with a continuous inverse function. This means that the two sets have the same topological structure, and can be mapped onto each other without any tearing or gluing of points.

3. How can we prove that Q is homeomorphic to N?

To prove that Q is homeomorphic to N, we can show that there exists a continuous and bijective function between the two sets. One way to do this is by constructing a function that maps each element of Q to a unique element in N, and vice versa. We can then show that this function has a continuous inverse, thus establishing the homeomorphism between the two sets.

4. What is the significance of Q being homeomorphic to N?

The fact that Q is homeomorphic to N has significant implications in mathematics. It allows us to study the properties of rational numbers by using the more familiar and easier to work with natural numbers. It also helps us to understand the topological structure of Q and N, and how they are related to each other.

5. Are there any other sets that are homeomorphic to Q?

Yes, there are many other sets that are homeomorphic to Q. Some examples include the set of real numbers (R), the set of complex numbers (C), and the set of algebraic numbers. In general, any set that can be mapped onto Q in a continuous and bijective manner is homeomorphic to Q.

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