Cardinality of Infinite Sets

First the ...

Rules

1. Set A and set B have the same cardinality IFF the elements of set A can be put in a 1-to-1 correspondence with the elements of set B. for example {s, 7, h} can be put in a 1-to-1 correspondence with the set {5, z, %} and so the two have the same cardinality.

2. Set A has a smaller cardinality than set B IFF set A can be put in a 1-to-1 correspondence with a proper subset of set B. For example, the set {2, g, &} can be put in 1-to-1 correspondence with the proper subset, the set {\, k, f} of the set {p, f, \, k} and so the cardinality of {2, g, &} is less than the cardinality of {p, f, \, k}.

Then the ...

Notes:
a) elements of a set A possess a 1-to-1 correspondence with the elements of a set B IFF each element of set A is matched with exactly one element of set B.

b) Set A is a proper subset of set B if set B contains all the elements of set A and has at least one element that is not in set A. For example, {1,$} is a proper subset of {d, $, 1}

c) The cardinality of a set A, n(A), is the number of elements in set A. If A = {e, ¥, 3} then n(A) = 3.

Lastly, the argument ...

N = The set of natural numbers = {1, 2, 3, ...}
E = The set of even numbers = {2, 4, 6, ...}
O = The set of odd numbers = {1, 3, 5, ...}

1. Rule 1 implies E has the same cardinality as N: (1, 2), (2, 4), ..., (n, 2n).

2. Rule 2 implies E's cardinality is less than N's cardinality as E can be put in a 1-to-1 correspondence with O [like so (1, 2), (3, 4), ..., (n, n + 1)] and O is a proper subset of N.

What gives?