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I know that this question has been asked before, but I still can't find a clear answer, I read here this question: Show that the set of all finite subsets of $\mathbb{N}$ is countable. and a guy wrote:
$\emptyset, \{1\}, \{2\}, \{1, 2\}, \{3\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}, \{4\}, \ldots$, this would be the power set of $\mathbb{N}$ listed in a row, why would this create a bijection between $\mathbb{N}$ and this set? (which would prove that the power set of $\mathbb{N}$ isn't equinumerous with $\mathbb{R}$)?

Or for example writing:
$\{1\}, \{2\}, \{3\}, \{4\}, \ldots$
$\{1, 2\}, \{1, 3\}, \{1, 4\}, \ldots$
$\{1, 2, 3\}, \{1, 2, 4\}, \{1, 2, 5\} \ldots$
$\{1, 3, 4\}, \{1, 3, 5\}, \{1, 3, 6\} \ldots$
$\{1, 10, 11\}, \{1, 10, 12\}, \{1, 10, 13\} \ldots$
$\vdots$
$\{2, 3\}, \{2, 4\}, \{2, 5\}, \ldots$
$\{2, 3, 4\}, \{2, 3, 5\}, \{2, 3, 6\} \ldots$
$\vdots$

and then do the diagonalization thing that Cantor used to prove the rational numbers are countable: enter image description here

Why wouldn't this work?

P.s: I know the proof that the power set of a set has a larger cardinality that the first set, and I also know the proof that cantor used to prove that no matter how you list the real numbers you can always find another one that is not inside that list. I know the proof that the power set of $\mathbb{N}$ is equal to $\mathbb{R}$ as well, I'm not saying that my argument is correct and theirs is wrong, I'm just trying to understand why mine is wrong

Arctic Char
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    Which finite index corresponds to an infinite subset of $\mathbb{N}$, such as the subset of even numbers? – Xander Henderson Sep 03 '20 at 11:45
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    If you're allowed to take infinite subsets, then Cantor's diagonal argument creates an infinite subset that is not in the above list. – Chrystomath Sep 03 '20 at 11:46
  • @Chrystomath what do you mean with creates an infinite subset –  Sep 03 '20 at 11:50
  • @XanderHenderson you mean the index corresponding to the sets that start with "$2$"for example? –  Sep 03 '20 at 11:51
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    What the guy wrote is not the power set of $\mathbb{N}$ listed in a row : you just have finite subsets of $\mathbb{N}$, and not every subset of $\mathbb{N}$. – TheSilverDoe Sep 03 '20 at 11:51
  • @TheSilverDoe why not if that list goes to infinity? What elements is missing? –  Sep 03 '20 at 11:54
  • @Vikise Note that, with respect to the argument used to show that the rationals are countable, if you give me a rational number, I can produce the index at which that number appears (though it might take some time to figure it out, I can, in principle, do it). So, pick any infinite subset of $\mathbb{N}$. My favorite infinite subset is the set ${2,4,6,8,10,\dotsc}$. What is the index of that set? – Xander Henderson Sep 03 '20 at 11:55
  • More generally, any set has more sunsets than elements. – J.G. Sep 03 '20 at 11:55
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    @Vikise For exemple, the subsets of $\mathbb{N}$ containing all the even numbers is missing. – TheSilverDoe Sep 03 '20 at 11:56
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    @Vikise "What elements is missing?" As already attempted to have been said, your list does not contain any individual element corresponding to an infinite set. The only elements appearing in your list are finite subsets. Yes, your list will contain sets like ${2,4,6,8,\dots,2n}$ for any desirable but still finite $n$ however it will never contain ${2,4,6,8,\dots}$. Your list may be unbounded in terms of what largest element appears in each set... however each individual element appearing in your list always has some largest finite element. – JMoravitz Sep 03 '20 at 11:59
  • In the argument showing that the rationals are countable, every rational appears in the array at some coordinate: row $m$, column $n$, for finite $m$ and $n$. But in your arrangement of subsets of $\mathbb{N}$ in an array, any set appearing in row $m$ column $n$, for some finite $m$, $n$, is a finite subset of $\mathbb{N}$. So in order to include infinite subsets using this same scheme, you'd need to add even more rows and columns to the array. And you won't be able to do it in a way that keeps only countably many coordinates in the resulting array. – halrankard2 Sep 03 '20 at 12:01
  • @XanderHenderson oh yeah I didn't notice that in the first example (the list of that guy), thank you! –  Sep 03 '20 at 12:04
  • @JMoravitz I think i got it, you mean it's because i put some dots between the elements starting with $1$ and the elements starting with $2$ right? and since the first row after the vertical dots would be the first elements after infinity i can't find an index for those elements. Right? –  Sep 03 '20 at 12:06
  • @halrankard2 Thanks for your answer, i wanted to tag you where i tagget JMoravits but I can't, is that what you meant? –  Sep 03 '20 at 12:07
  • Yes that's right. Or at least you'd need to give a concrete description of what kind of indexing scheme you are using (in analogy to using an $\mathbb{N}\times\mathbb{N}$ grid to index the rationals). Once you do that, however, you'll find that no indexing scheme big enough to capture all subsets of $\mathbb{N}$ can be countable. – halrankard2 Sep 03 '20 at 12:11
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    The thing that's a bit frustrating with these sort of questions is that Cantor's proof is extremely simple and constructive. You just need to throw your function against the construction ${n\in\Bbb N\mid n\notin f(n)}$ to see what goes wrong. And while you're admitting mistake, many people won't. And so after saying it an uncountable number of times, it becomes very exhausting to explain from scratch again. – Asaf Karagila Sep 03 '20 at 12:14

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