Ratio of whole numbers to even numbers [duplicate]

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)

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.

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.

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.