I really need your help...

How could I show that any countably infinite set has uncountably many infinite subsets of which any two have only a finite number of elements in common?

Thanks, Tania

Hence take the set Q of rationals and for any irrational real number a, consider the sequence S_a of rationals r₁, r₂, by taking the first few digits of the decimal expansion of a. For example, if a = p, we would have r₁ = 3, r₂ = 31/10, r₃ = 314/100, r₄ = 3141/1000 and so on.

Now, for any two irrationals a and b, the sets S_a and S_b will have at most finitely many elements in common. Since we have uncountably many rationals to choose from, this is our required uncountable family.

David

Since your initial set is countably infinite, we can suppose that it is the natural numbers - since it is certainly injective into the naturals, we may associate every element uniquely with a natural number. Furthermore, the injection can be assumed to be surjective by reordering.

So, all you have to do is prove that the natural numbers have this property. You need to form an uncountable collection of subsets of the naturals - can you see any way to do this? Think about which index set to use.

When you've come up with a way to find uncountably many subsets (with an index set) you can modify this to the collection you describe.

Let me know how you get on with this and then we'll investigate how to find this extra property.

-Dave

Thanks David, that's the way I am thinking... Thanks Dave, but David's explanation is more understandable, isn't it?

Tania

A variant of David's method which is immediately obvious is to consider the interval [0.1,1). Take each number in the interval in decimal expansion (excluding recurring 9 if there the decimal terminates) as 0.a₁ a₂ a₃ ... and map the number to the set {a₁ , a₁ a₂ , ...}. This set is clearly a subset of the natural numbers N and [0.1,1) is uncountable.

Kerwin