Why does “the set containing the natural numbers and a sandwich” not have cardinality between the two?
If we're ok with extending that to sets of infinite things (we can still pair elements of each set, we'd just never be able to finish listing all the pairs), then we can say that the natural numbers and "the set containing the natural numbers and a sandwich" are of equal size because we could pair 1 from the first set with the sandwich from the second set, pair 2 from the first set with 1 from the second set, 3 from the first set with 2 from the second set, etc etc.
There's no element of either set without a match in the other set, so they have the same cardinality as the natural numbers, with or without the sandwich.
Saying that two sets have the same cardinality is equivalent to them having a bijection between them.
So the claim is that the natural numbers and the natural numbers plus a sandwich have the same cardinality. This can be proved by the bijection:
0 -> sandwich
1 -> 0
2 -> 1
3 -> 2
.
.
.
n -> n-1
.
.
.
0 -> sandwich_0
1 -> 0
2 -> sandwich_1
3 -> 1
4 -> sandwich_2
5 -> 2
.
.
.
Still enough natural numbers to eat them all, one per.
But still not enough sandwiches to feed all the (so-called) real numbers one sandwich each!
But if sandwiches grew on trees, and we had an infinite branching tree with two branches at each branching point, and every branch has a (pair of) sub-branches, then natural numbers could not eat all the sandwiches, and the sandwiches could feel all the (so-called) real numbers.
As I understand it we can say the cardinality of the reals is 2^aleph_0. Why is it cheating to create a bijection thusly:
0 -> 0
1 -> 1/(2^aleph_0)
2 -> 2/(2^aleph_0)
Problem 2: There's no natural number that maps to (say) 1. Even if you do allow 1/(2^aleph_0), there's no finite number n that would make n/(2^aleph_0) = 1. With any reasonable definitions of the operations involved here, n/(2^aleph_0) would always be infinitesimal, so it would never equal a non-infinitesimal.
Problem 3: You're still skipping over infinitely many numbers. If 1/(2^aleph_0) is a number (and again, this requires going beyond the real numbers) and 1.5 is a number, then 1.5 * 1/(2^aleph_0) = 1.5/(2^aleph_0) is also a number, but no natural number gets mapped to that.
2^{\aleph_0} is a cardinal number, which isn’t really a number in the sense of “an element of a field” or something like that. Dividing by it isn’t a well defined thing.
And, you certainly can’t just multiply any real number (or, any real number between 0 and 1) by 2^{\aleph_0} and get a different integer as a result.
(Now, if you work in the surreal numbers, you can define things like n/(2^{\aleph_0}) (identifying cardinals with the first ordinal of that cardinality), but these would not be real numbers. They would all be infinitesimal , smaller than 1/k for all positive integers k, and yet bigger than 0. Similarly in the surreal numbers, you could multiply real numbers between 0 and 1 by 2^{\aleph_0}, but you would get surreal numbers which are larger than every integer (in fact, larger than any countable ordinal))
Summary: What you wrote doesn’t define a mapping from the integers to the real numbers . (It can be interpreted as defining a map from integers to something else though.)
It's even possible to add an infinite number of elements to an infinite set and retain the same cardinality (the Integers set has the same cardinality as the Naturals set).
0 <-> sandwich
1 <-> 0
...
n+1 <-> n
By definition, if you can biject two sets, they have the same cardinality.