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The set of algebraic numbers (let's say $\mathcal{A}$ ) is countable. The set of irrational numbers (let's say $\mathcal{I}$ ) is uncountable. The set of algebraic numbers contains some of irrational numbers and some irrationals are not algebraic.

Now if we choose a subset $\mathcal{I_A}$ of all irrationals from the set of algebraic numbers, will it be countable?

If I consider that $\mathcal{I_A}\subset \mathcal{A}$, then it is countable [ $\because$ subset of a countable set is countable] but again $\mathcal{I_A}\subset \mathcal{I}$ [subsets of uncountable sets can be countable or uncountable].

Hence in this case is $\mathcal{I_A}$ countable? Can we establish a bijection $\mathbb{N}\to\mathcal{I_A}$ ?

Any help or explanation is valuable and highly appreciated.

TopoSet32
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    Any subset of a countable set is countable (or finite). See, e.g., this. – lulu Nov 24 '20 at 09:44
  • Yes. but in this the set of irrationals is the subset of both countable and uncountable set. How do we establish a bijection here? – TopoSet32 Nov 24 '20 at 09:46
  • Your set $\mathcal{I_A}$ is a subset of the algebraic numbers, and those are countable. – lulu Nov 24 '20 at 09:47
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    It is enough that it is a subset of a countable set. That it is also a subset of an uncountable set is then irrelevant. And to show the countability, we do not actually need a bijection. – Peter Nov 24 '20 at 09:47
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    To be clear: you wrote "choose a subset $\mathcal{I_A}$ of all irrationals from the set of algebraic numbers". That's poorly worded. I took it to mean "choose a subset $\mathcal{I_A}$ from the intersection of the irrationals and the algebraic numbers". If you meant something else you should clarify. – lulu Nov 24 '20 at 09:50
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    @lulu Yes, I meant what you have written within the quotation. – TopoSet32 Nov 24 '20 at 09:52
  • To establish a bijection is in general cumbersome. I would only do that if it is explicitely demanded. – Peter Nov 24 '20 at 09:53

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Yes, $\mathcal I_A$ is countable. If you want an injection $f:\mathcal I_A\to\Bbb N$, consider a bijection $g:\Bbb N\to A$ and call $$f(x)=\left\lvert\{n\in\Bbb N\,:\, g(n)\in\mathcal I_A\land 0\le n< g^{-1}(x)\}\right\rvert$$

The image of this function is an interval of $(\Bbb N,\le)$ containinig $0$. Its image is a proper subset of $\Bbb N$ if and only if $\mathcal I_A$ is finite (which, apparently, you are not excluding).

  • Would you like to explain how to interprete the interval $(\mathbb{N}, \leq)$ and what does $\mathrm{g^{-1}(x)}$ imply? – vbm Nov 25 '20 at 19:45