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I am going through Bartle and Sherbert's Introduction to Real Analysis and doing all the exercises. I have just tried the last of section 1.3, which requires you to prove the aforesaid statement. I think I've come up with something satisfactory, but it looks nothing like the hint the book offers at the end.

I start by building a grid that includes all the finite subsets of N as follows. Each finite subset has a natural number n of elements (except the empty set, which has none), so each row includes all the subsets with the same number of elements (except the first row, which also includes the empty set. First row would start like this: ∅, {1}, {2}, {3} ... Second row would start like this: {1,2}, {1,3}, {1,4} ... Position within the row is determined by two aspects: we add up the elements of the sets and start with those who sum is the smallest possible natural number m that results from the sum of distinct elements of the sets in the row. For the first row, this is the empty set; for the second row, this is 3 (1+2) and so on. If two sets in the same row add up to the same m, their ordering is determined by checking the minimum element of those sets. The one with the smallest minimum goes first. If the sets share the same minimum, we proceed to check the next minimum of both sets once we've removed the first minimum, and so on recursively. For example, in row 2, both {1,4} and {2,3} add to 5, but as 1<2, {1,4} goes before the other.

I claim this grid includes all finite subsets of N and uniquely locates them through row and column position. Using this fact, a bijection be established between this set and the set NxN or ordered pairs of natural numbers. ∅ maps to {1,1},{1} maps to {1,2}, {1,2} maps to {2,1} and so forth.

Now the textbook chapter showed there's a bijection from N to NxN (call it f). We have just produced a bijection from P(N) to NxN (call it g), and as the inverse of a bijection is also a bijection, by composing f and the inverse of g we have created a bijection from N to P(N), which proves that P(N) is countable.

Is this correct? Have I missed something?

Anne Bauval
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Manuel Del Río Rodríguez
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1 Answers1

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Your argument looks fine to me. You can try to formalize it, but at this point the idea is more important than writing every rigorous detail down.

One suggestion if you want to formalize your argument a bit more: prove that the order you define on the rows is a well-order, meaning that any subset (of the row) has a minimum element. In other words, this ensures that there is always a well-defined "next" element in the sequence.

You can also prove that within a single row, you reach any set belonging to it in a finite number of steps. This is to ensure your map from $\mathbb{N} \times \mathbb{N}$ is surjective.

Anakhand
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  • Thanks. The textbook hasn't actually explained order or well order (it'll do it in the next chapter I'm about to start, number 2, with axioms of the Real number field). I didn't feel the need to prove the well-order of the subsets, but I guess it shouldn't be too difficult to derive through induction. As for surjectivity, isn't it enough to just state that given any ordered pair (m,n) of NxN and our description of our set P(N), we can guarantee that there's a path from (m,n) to a set A in N such that A has m elements and occupies the nth position in row m? – Manuel Del Río Rodríguez Feb 29 '24 at 17:01
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    Yes, what you say is enough, but what I meant to say is that as it's written now it's a relatively informal argument: you are essentially just restating what it means for that map to be surjective. If you want to write a (more) formal proof you can try to formalize this step; but again, I don't think you need to. – Anakhand Feb 29 '24 at 17:09
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    The reason I mention the well-order is that you are implicitly using this property when you implicitly define the map from $\mathbb{N}$ to the row by "counting" each entry in the row successively in the order you have established: you can only make this definition if the order is a well-order. That is, if I want to define a map $f: \mathbb{N} \to S$, where $S$ is an ordered set, by defining $f(1) = \text{first element}$, $f(2) = \text{second element}$, etc., I need the notions of "first element", second element", etc. to make sense, which is precisely the well-ordered property. – Anakhand Feb 29 '24 at 17:14