Showing two countably infinite sets have a 1-1 correspondence

[k3k3]

Homework Statement

Suppose A and B are both countably infinite sets. Prove there is a 1-1 correspondence between A and B.

Homework Equations

The Attempt at a Solution

Since A is countably infinite, there exists a mapping f such that f maps ℕto A that is 1-1 and onto.

Similarly for B, there exists a mapping g such that g maps ℕto B that is 1-1 and onto.

Since g is 1-1 and onto, g^-1 exists and g^-1 maps B into ℕ

Hence g^-1(b)=n for all b in B.

Since f(n)=a for all n in ℕ, then f(g^-1(b))=a for all b in B since g^-1(b) yields its corresponding number from ℕ.

Since f and g are bijective, we can define h as a mapping from B to A by h(b)=a for all b in B by using the ordering created by the mapping g.

Does this make sense?

 

[jbunniii]

Homework Statement

Suppose A and B are both countably infinite sets. Prove there is a 1-1 correspondence between A and B.

Homework Equations

The Attempt at a Solution

Since A is countably infinite, there exists a mapping f such that f maps ℕto A that is 1-1 and onto.

Similarly for B, there exists a mapping g such that g maps ℕto B that is 1-1 and onto.

Since g is 1-1 and onto, g^-1 exists and g^-1 maps B into ℕ

Everything is fine up to this point.

Hence g^-1(b)=n for all b in B.

What is n?

Since f(n)=a for all n in ℕ,

What? This means that f is a constant function, certainly not a bijection.

You have the right idea, but you aren't expressing it correctly.

f is a 1-1 mapping of N onto A
g is a 1-1 mapping of N onto B

You seek a 1-1 mapping of A onto B. What is that mapping, and how do you know that it is 1-1 and onto?

 

[SammyS]

Homework Statement

Suppose A and B are both countably infinite sets. Prove there is a 1-1 correspondence between A and B.

Homework Equations

The Attempt at a Solution

Since A is countably infinite, there exists a mapping f such that f maps ℕto A that is 1-1 and onto.

Similarly for B, there exists a mapping g such that g maps ℕto B that is 1-1 and onto.

Since g is 1-1 and onto, g^-1 exists and g^-1 maps B into ℕ

Hence g^-1(b)=n for all b in B.

Since f(n)=a for all n in ℕ, then f(g^-1(b))=a for all b in B since g^-1(b) yields its corresponding number from ℕ.

Since f and g are bijective, we can define h as a mapping from B to A by h(b)=a for all b in B by using the ordering created by the mapping g.

Does this make sense?

Not really.

At least three of your statements say something other than what you probably intended.

"Hence g^-1(b)=n for all b in B." says every b in B gets mapped to the same n in ℕ.

Similarly, "Since f(n)=a for all n in ℕ" says that all of the elements of ℕ get mapped to the same element of A.

Also, "then f(g^-1(b))=a for all b in B" says that all of the elements of B get mapped to one single element of A.

It should be enough to note that since g is a bijection, then so is g^-1. Then since f and g^-1 are bijections, so is f○g^-1.

 

[k3k3]

n is supposed to be some n in ℕ. What ever it is mapped to.

I think the mapping is the composite function f(g^-1(n)) and since f and g are 1-1 and onto, the composite function is too.

 

[k3k3]

I am trying to say that f(some n) = some unique a when I say "f(n)=a". Should I say that for all a in A and n in ℕ?

 

[SammyS]

I am trying to say that f(some n) = some unique a when I say "f(n)=a". Should I say that for all a in A and n in ℕ?

What you want to say is covered by the composition, of f with g^-1 being a bijection.

Otherwise:

Each element of ℕ is the image of exactly one element of B.

etc.​

would be OK, I suppose.

 

[k3k3]

Here is my revised argument:

Since A is countably infinite, there exists a mapping f such that f maps ℕto A that is 1-1 and onto.
Similarly for B, there exists a mapping g such that g maps ℕto B that is 1-1 and onto.

Since g is 1-1 and onto, g^-1 exists and g^-1 is 1-1 and onto and maps B into ℕ.

Then for every element of ℕ, there exists exactly one b that it is the image of.
Hence there exists a b in B such that g^-1(b)=n for some n in ℕ

Using g^-1 in f creates the 1-1 correspondence from A to B.

Therefore f○g^-1 is a bijective mapping from A to B.