# Aleph null !

1. Apr 30, 2008

### Doom of Doom

How might one show that (aleph_null)! = aleph_1?

2. Apr 30, 2008

### HallsofIvy

Staff Emeritus
One might start by defining the "factorial function" for aleph_null!

3. May 1, 2008

### Doom of Doom

Ok, so the person who proposed this problem to me gave me a way to understand (aleph_null)!.

So, consider two sets, A and B. Then |A|*|B|=|A x B|, where AxB is the cartesian product of A and B.

Thus, consider N_m={1,2,3,...m}, and |N_m|=m.

Then (aleph_null)! = |N_1 x N_2 x N_3 x ...... |.

So how can I find a bijection from N_1 x N_2 x N_3 x ...... to, say, P(N), the power set of the naturals?

4. May 1, 2008

### CRGreathouse

Consider {1} x {1, 2} x {1, 2, 3} x ... as the base 1-2-3-... expansion of a number in [0, 1), then biject [0, 1) with the reals by your favorite method. You have at most a countable number of issues with rational numbers. which you can likewise deal with in your preferred method.

This shows that the set has cardinality $\beth_1$, not $\aleph_1$ unless you have the CH.

5. May 1, 2008

### Hubert

Or show that 1x2x3x4.........is greater than or equal to 2x2x2x2x2.... and less than or equal to aleph null^aleph null. Still, this only indicates the factorial is equal to beth_1 without CH, like CRGreathouse said.