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The set in the title is presented in this answer as an example of a similar set to the $P(\mathbb{N})$ (in the context of explaining the necessity of the axiom of choice in the existence of a well ordering on reals), but the definition of it in terms of enumerating the $\mathbb{N}$ gives me the impression it's actually a countable set.

(This should actually have been a comment to the original answer but I don't have enough reputation. Feel free to remove if I'm violating any guidelines).

lucas
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 30 '24 at 17:07
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    And once again the community bot is way off base. The problem that is asked about here is crystal clear. The question post has other issues, such as a lack of individual work and thoughts on the problem, but a vague problem statement is not one of them. – Arthur Mar 30 '24 at 17:08
  • Thanks for clarifying on the issues. Working on it. – lucas Mar 30 '24 at 17:13
  • Not every countable set is well ordered. The rational numbers, for example, are quite far from being well ordered. But every countable set can be well ordered. This set, too. But if we look at it as a partial order under the subset relation, it is not a well ordered set. That's the point of the answer you've linked. – Asaf Karagila Mar 30 '24 at 17:25
  • @AsafKaragila Thanks!! This clarifies everything. Thank you for taking the time to read the original answer. Sorry to have decoupled the threads with a new question. – lucas Mar 30 '24 at 17:29
  • Rereading the original question I got confused because it asked "So, why isn't this a well ordering on the uncountable of reals?" and the answer was using a countable set to illustrate it, but I guess the point of that particular order not working is valid on both countable/uncountable sets. – lucas Mar 30 '24 at 17:36
  • @AsafKaragila The linked answer talks about lexicographic ordering, not subset ordering. – Arthur Mar 30 '24 at 18:08

1 Answers1

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By definition, a set is countable if there exists a bijection from itself to the naturals. In other words, if there is a one-to-one function from $A$ to $\mathbb{N}$ that does not lefts out any natural number. In this case, it is easy to find such a function: $$ f: \mathbb{N} \rightarrow A, \ f(n) = A_n $$ It is easy to show that $f$ is one-to-one and for every $a \in A$, there exists $n_a \in \mathbb{N}$ such that $f(n_a) = a$

In conclusion, the set $A_n$ is countable

A. Person
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