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I have a feeling no such uncountable set exists but have no idea how I could formulate a proof to show this. If such an uncountable set did exist I could try and use a form of the diagonalization method to show this, but haven't been able to find one so far... If someone could just start me off and give me an idea where to go with this question that would be great.

Europa
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  • What's T? You should write the text of your problem clearly – Exodd Dec 18 '14 at 13:57
  • Ah sorry T should be S... – Europa Dec 18 '14 at 13:58
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    For each irrational, choose a sequence of rationals that converge to it. Then, "transfer" to $\cal P(\Bbb N)$. – David Mitra Dec 18 '14 at 13:59
  • What do you mean transfer? Form a bijection? Sorry, I don't really understand the significance of considering the irrationals. – Europa Dec 18 '14 at 14:06
  • David is saying: the irrationals are uncountable, and there's a way to associate to each irrational an element of $P(\mathbb{N})$, such that the intersections are finite – Exodd Dec 18 '14 at 14:08
  • The set of all finite sets is countable. (Oh, the other comment has been deleted.) – Heimdall Dec 18 '14 at 14:10
  • The point is (and is not my idea), finding such a family of sets is relatively easy to do in $\Bbb R$. Given two distinct irrationals, the corresponding sequences can have at most finitely many terms in common. – David Mitra Dec 18 '14 at 14:13
  • Thanks David, I think I can see what to do now. – Europa Dec 18 '14 at 14:15
  • Identify (find a bijection) the set of rationals and $\Bbb N$. Use that to do the "transfer". – David Mitra Dec 18 '14 at 14:17

2 Answers2

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For every real $\alpha\in [0,1)$ with $\alpha=0.a_1a_2a_3\ldots$ the digits of $\alpha$, let $A_{\alpha}=\{1,1a_1, 1a_1a_2, 1a_1a_2a_3, \ldots\}\in P(\mathbb{N})$. For $\beta\neq \alpha$, the sets $A_{\alpha}$ and $A_{\beta}$ will not agree after some finite number of elements in the sequence because $\alpha$ and $\beta$ stop agreeing after some finite number of digits. Let $S=\{A_{\alpha}\}\subset P(\mathbb{N})$ and we are done.

Dan Rust
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  • You might want $\alpha\in[0.1,1)$, otherwise $\alpha$ and $\alpha/10$ will produce the same set (with zero added if it's not in $A_\alpha$). – Heimdall Dec 18 '14 at 14:33
  • @Heimdall thanks. Easier to fix imo by just prefixing by 1. – Dan Rust Dec 18 '14 at 21:09
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Alternative, from David Mitra's comment:

f is a bijection from rationals to naturals.

For each irrational number $a$ choose one of rational sequences converging to it, $\{x_n\}$ and define the set $A_a=\{f(x_n):n\in N\}$. Then

$$S=\{A_a:a\in R\setminus Q\}$$

As pointed out earlier, any two sequences and therefore any two sets $A_a$ and $A_b$ will only agree in finite elements.

Of course, this solution depends on the Axiom of Choice, whereas Daniel's doesn't.

Heimdall
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