What is larger — the set of all positive even numbers, or the set of all positive integers?

We will call the set of all positive even numbers E and the set of all positive integers N.

At first glance, it seems obvious that E is smaller than N, because for E is basically N with half of its terms taken out. The size of E is the size of N divided by two.

You could see this as, for every item in E, two items in N could be matched (the item x and x-1). This implies that N is twice as large as E

On second glance though, it seems less obvious. Each item in N could be mapped with one item in E (the item x*2).

Which is larger, then? Or are they both equal in size? Why?

(My background in Set theory is quite extremely scant)

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For every item in N, two items in E could be matched (4n and 4n-2), so by your logic, E is also twice as large as N. – BlueRaja - Danny Pflughoeft Aug 2 '10 at 23:18
It might be a good exercise to try and define what exactly one means by one infinite set being larger than another set. – Aryabhata Oct 1 '10 at 17:14
Galileo raised more or less the same question. – bof May 6 '14 at 3:33

Mathematics is the art of clever forgetting. The first mathematical breakthrough, numbers, came about when people realized that if you just forgot about whether it was 5 cows + 3 cows, or 5 rocks + 3 rocks or whatever you always got 8. Numbers are what you get when you look at collections of objects and forget what kind of object they are.

When you say "as sets" you mean you're forgetting a lot of information, in particular you don't care about what the names of the elements in that set are or what properties those elements have. As sets the positive numbers and the positive even numbers are "the same" (that is are in bijection) because you can take 1,2,3,... and just rename 1 to 2, and 2 to 4, and n to 2n, and you've just renamed all the elements and got the even numbers!

However, if you want to remember more about these sets, for example that they're not arbitrary sets they're both subsets of the natural numbers, then they become distinguishable. Depending on how you want to measure "size of a subset of the natural numbers" they might be different sizes. For example, a common way to measure "size of a subset of the natural numbers" is by its "density." That is you look at the first N numbers and calculate what fraction of them are in your set, and then take the limit as N goes to infinity (warning for sufficiently complicated sets this limit might not exist). So for your two examples, one has density 1 and the other has density 1/2, which is one way to make precise the intuition that the former is bigger as a subset of the natural numbers (though not as a set) than the latter.

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Thanks for addressing the distinction between the two sets :) I think this answer is the most complete. – Justin L. Jul 22 '10 at 0:23

The word 'size' doesn't have a intuitive meaning for set of infinite items.

Mathematicians defined cardinality by one-to-one correspondence(bijection), and it's generally what it means by 'size'.

If there exist a bijection between A and B, then the two sets have the same cardinality. You have shown the existence of a bijection, therefore E and N have the same cardinality.

You might mean the 'density' of N is twice as large as E. density of A(a subset of natural number) is limit of |{a<=n|a\in A}|/n as n goes to infinity.

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They are both the same size, the size being 'countable infinity' or 'aleph-null'. The reasoning behind it is exactly that which you have already identified - you can assign each item in E to a single value in N. This is true for the Natural numbers, the Integers, the Rationals but not the Reals (see the Diagonal Slash argument for details on this result).

-- Added explanation from comment --

The first reasoning is invalid because the cardinality of infinite sets doesn't follow 'normal' multiplication rules. If you multiply a set with cardinality of aleph-0 by 2, you still have aleph-0. The same is true if you divide it, add to it, subtract from it by any finite amount.

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But what about the first reasoning? Is it invalid? Why? – Justin L. Jul 21 '10 at 22:21
because infinity is not a number; 2*"infinity" = "infinity" = "infinity + 1" = "infinity squared" ... they all have the same cardinality if the infinities are all of cardinality aleph-0. – Jason S Jul 21 '10 at 22:23
Thanks @Jason, I've updated my answer to include that point :) – workmad3 Jul 21 '10 at 22:36
@Jason: But $2^{\aleph_0} > \aleph_0$ so not all operations behave this way with infinite cardinals. – François G. Dorais Jul 21 '10 at 23:10
@Francois: I've heard that one... what does that correspond to? the mapping between an infinite set S and its subsets? – Jason S Jul 22 '10 at 1:28

Each item in N could be mapped with one item in E (the item x*2).

Yes. Both sets have cardinality aleph-0.

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Does this not imply that N is twice as large as E? – Justin L. Jul 21 '10 at 22:22
Jason S is correct. – user126 Jul 21 '10 at 22:24
No, this does not imply that N is twice as large as E. The sets are in a one-to-one correspondence. Each 2*x in your set E has a unique "pullback" to N. – Tom Stephens Jul 21 '10 at 22:25
I misquoted the wrong part of the original post first, sorry about that. – Jason S Jul 22 '10 at 1:29

Your example is an example of a countable sets: http://en.wikipedia.org/wiki/Countable_set

As others have explained, if you find a bijection between these two sets then they are of the same cardinality or have the same number of elements.(I hope that statement is true :)) But in this case you have found one with N which is what makes it a countable set. f(x)=2x would be an example of a bijection between N and 2N. Since for every x in the set of natural numbers there is a corresponding number in the set of 2N (only one) then you can say that they have the same cardinality.

A cooler example would be the interval of the real numbers (0, 1) and R itself, those are not countable but their cardinality is the same.

Other notes, if you are wondering what aleph is: http://en.wikipedia.org/wiki/Aleph_number

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Almost all of this misses the idea that you can't do traditional arithmetic with infinity that way that most of us think of it. 10/2 = 5, five is half of 10, no problem. If Judy has five apples and Joe has 10 apples, then it is proper to say "Judy has half as many apples as Joe." But to say "there are half as many even positive integers as there are positive intergers" is to infer something about infinity/2, or half of infinity, which is not entirely kosher.

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(I've reverted a superfluous comment that only accomplished to remove the answer from its personality.) – Andrés Caicedo May 6 '14 at 3:28

The density argument above is not quite to the point, since it involves not only a set but an ordering.

In fact, if you play with orderings, you can get even stranger results, such as the integers and rationals can both be ordered as a continuum (according to the ordering, between any two members in the set there is another member) and a non-continuum (there are two members with no other member in between). Or worse, between any two members is at most a finite set of other members (non-continuum) or an infinite set of other members (continuum). That's kind of an extreme form of density.

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You can also get it the other way around -- N is (kind of) a subset of E.

Map each integer n into 4n. Then there's a bijection (1-1, onto function) between N and (let's call it) N4, the set of all integers divisible by 4, so N and N4 are, in your intuition, of the same "size". But N4 is a proper subset of E ! Of course, you could just as well have mapped N directly to E with bijection f(n) = 2n.

Anyway, that shows you how slippery it is to apply concepts of finite size to infinite sets, and why bijection is the way to go.

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I think your understanding of bijection is not correct here, Bijection means one-to-one and a surjection (i.e., "onto"). – iMath Sep 9 at 7:07