TITLE: Difference between cardinal and ordinal
QUESTION [6 upvotes]: I think this question has already been asked, but I do not find any precise trace of it. I do not get the difference between cardinal and ordinal numbers: why aren't they both matching the equivalence classes under the relation "being in bijection with"?

REPLY [6 votes]: This is not easy to understand, because in finite sets there is not much of a difference.
The idea is that for ordinal numbers you also have a relation between the numbers, so you can put them in line.
Let's take as an example the numbers
$$
0,\frac{1}{2},\frac{3}{4},\frac{7}{8},...,\frac{2^n-1}{2^n},...,1.
$$
Now this set has the same cardinality as $\mathbb{N}$ (because there is a bijection) but if you wanted to find a bijection $\Phi$ to the natural numbers preserving the order of numbers, this would not be possible. Preserving order means $ x < y \Longrightarrow \Phi(x) < \Phi(y)$.
It's not possible, as you have the biggest element 1, which would need to be bigger than all natural numbers (but it's not obvious).
In a way the ordinal numbers are a finer tool to describe how big a set is. For example $ 2 \omega + 5$ has the same ordinal number as
$$
0,\frac{1}{2},\frac{3}{4},\frac{7}{8},...,\frac{2^n-1}{2^n},...,1,1+\frac{1}{2},...,1+\frac{2^n-1}{2^n},...,2,3,4,5,6
$$
has the same ordinal number as
$$
\frac{7}{8},...,\frac{2^n-1}{2^n},...,1,1+\frac{1}{2},...,1+\frac{2^n-1}{2^n},...,2,3,4,5,6
$$
because it has to "limit-dents" and afterwards 5 elements.