## Cardinality, The aleph() Function

### The aleph() Function

If S is any set,
the aleph function, written ℵ(S),
is the least ordinal that will not fit into S.
We know from an
earlier theorem
that this ordinal exists, and that it is a limit ordinal;
in fact it is a cardinal.
Think of ω as ω0,
and let ω1 = ℵ(ω0).
In other words, ω1 is the least cardinal that won't fit into ω0.
Establish the recursive definition ωn+1 = ℵ(ωn),
the "next" cardinal number.
Here n is any ordinal, and n+1 is the successor of n.

If b is a limit ordinal, let ωb be the union of ωc for all c ∈ b.
This is an ordinal, but is it a cardinal?
Suppose ωb maps onto an earlier ordinal w.
Remember that w is a member of a cardinal class,
and ωb maps onto any set of the same size.
This includes the cardinal that represents w,
which we will call ωc.
Let d be the successor of c,
and note that the same map,
from ωb onto ωc,
embeds ωd into ωc.
Yet ωd is, by definition, the cardinal that won't map 1-1 into ωc.
Therefore ωb doesn't map onto a lower ordinal, and is a cardinal.
In fact it is the least cardinal that is larger than ωc for all c ∈ b.

apply the
recursion theorem
and we have a function from the ordinals into the infinite cardinals.
This should be called the ω() function, but it's not.
Just to confuse you,
this function is called ℵx, with x as subscript,
which is distinctly different from
the earlier function
ℵ(x), where x is in parentheses.
To clarify, ℵ(x) is the least cardinal that won't map into the set x;
ℵx is the cardinal associated with the ordinal x,
i.e. the xth infinite cardinal.

Note that ℵx is an increasing function.
For x < y, ℵx < ℵy.
Therefore the map is 1-1.
If it is not onto,
let w be the least infinite cardinal with no preimage.
Either there is a largest cardinal below w or there isn't.
If z is the largest cardinal below w, let ℵx = z,
whence ℵx+1 = w.
If w is a limit cardinal,
take the unionn of the preimages of the lower cardinals to get an ordinal x,
whence ℵx = w.
The map is 1-1 and onto,
hence ordinals and infinite cardinals correspond.

This doesn't mean they are the same size sets;
they aren't sets at all.
We already showed the ordinals don't form a set,
hence the infinite cardinals don't form a set either.