How to prove aleph^aleph=2^aleph in set theory
water
Dave L. Renfro
> How to prove aleph^aleph=2^aleph in set theory
It's not true that we always have b^b = 2^b, even for
infinite cardinals b. (It's false for every finite
cardinal except for 2.)
More specifically, it's false for b = aleph_0, although
it's true for b = 2^(aleph_0). I believe somewhere in
Sierpinski's book "Cardinal and Ordinal Numbers" there
is a paragraph where he shows there are arbitrarily large
cardinals b such that b^b = 2^b and arbitrarily large
cardinals b such that b^b > 2^b, but I don't have his
book with me right now to give a more specific reference.
Dave L. Renfro
Tonico
> as title
**************************************************
Use Cantor'Bernstein's Theorem and some cardinal inequalities:
Denote A = Aleph_zero, so:
A^A <= (2^A)^A = 2^(A*A) = 2*A <= A^A
Since clearly 2 < A and also A < 2^A accoridng to Cantor's Theorem.
Regards
Tonio
Dave L. Renfro
> Use Cantor'Bernstein's Theorem and some cardinal inequalities:
>
> Denote A = Aleph_zero, so:
>
> A^A <= (2^A)^A = 2^(A*A) = 2*A <= A^A
>
> Since clearly 2 < A and also A < 2^A accoridng to Cantor's Theorem.
Hummm...I don't know what the result is that I cited Sierpinski's
book for, but it's obviously not this. Indeed, what you wrote
is correct for any infinite cardinal A, not just A = alpeh_0.
In my previous post I was incorrect when I said 2^b < b^b
for b = aleph_0.
Dave L. Renfro
Denis Feldmann
> Tonico wrote:
>
>> Use Cantor'Bernstein's Theorem and some cardinal inequalities:
>>
>> Denote A = Aleph_zero, so:
>>
>> A^A <= (2^A)^A = 2^(A*A) = 2*A <= A^A
>>
>> Since clearly 2 < A and also A < 2^A accoridng to Cantor's Theorem.
>
> Hummm...I don't know what the result is that I cited Sierpinski's
> book for,
Probably that A*A=A is not always true (without choice). Or (but that
would be a stranger confusion) that aleph_w<>2^A=c, as (aleph_w)^aleph_0
>aleph w, while c^aleph_0=c
Dave L. Renfro
>> Hummm...I don't know what the result is that I cited
>> Sierpinski's book for, [...]
Denis Feldmann wrote:
> Probably that A*A=A is not always true (without choice).
> Or (but that would be a stranger confusion) that
> aleph_w<>2^A=c, as (aleph_w)^aleph_0 >aleph w, while c^aleph_0=c
No, I definitely was not thinking of A*A not always
being A in the absence of the axiom of choice. What I
was thinking of is definitely a ZFC result and it is
definitely the kind of thing you almost never see
mentioned in texts on set theory.
Something you wrote reminds me better about what it was.
I'm still not sure, but I think it was something like
there being arbitrarily large cardinals b such that
b^(aleph_0) and (aleph_0)^b are equal and arbitrarily
large cardinals b such that b^(aleph_0) and (aleph_0)^b
are not equal.
Dave L. Renfro
Denis Feldmann
Yes, this looks better :-)
>
> Dave L. Renfro
Adrian Duma
for all infinite cardinal (*including* aleph_0, the smallest infinite cardinal).
http://en.wikipedia.org/wiki/Cardinal_number
Now use Schroeder-Bernstein in order to conclude that 2^b = b^b for any infinite b.
http://en.wikipedia.org/wiki/Cantor-Bernstein-Schroeder_theorem
Thus, b^b = 2^b iff b = 0 (by convention), b = 2, or b is infinite.
David R Tribble
>> Something you wrote reminds me better about what it was.
>> I'm still not sure, but I think it was something like
>> there being arbitrarily large cardinals b such that
>> b^(aleph_0) and (aleph_0)^b are equal and arbitrarily
>> large cardinals b such that b^(aleph_0) and (aleph_0)^b
>> are not equal.
>
Denis Feldmann wrote:
> Yes, this looks better :-)
I'm not disagreeing, but if a^b (for cardinals a and b) is the
set of all functions from b to a (the usual definition of cardinal
exponentiation), when would this not be equal to b^a?
(I assume it's not a simple answer.)
quasi
<da...@tribble.com> wrote:
>... but if a^b (for cardinals a and b) is the
>set of all functions from b to a (the usual definition of cardinal
>exponentiation), when would this not be equal to b^a?
Of course, if a = b, then a^b = b^a.
If a,b are infinite cardinals with a < b, then
a^b >= 2^b = b^b >= b^a
so a^b >= b^a.
The inequality is at least sometimes strict, I think.
Suppose a,b are infinite cardinals with a < b. Must 2^a < 2^b? I think
so, but I'm not sure. To avoid worrying about that, I'll only consider
infinite cardinals a,b such that 2^a < 2^b.
Let a,b be infinite cardinals with 2^a < 2^b. If _either_ of the
following conditions holds, then a^b > b^a.
(1) b <= 2^a
(2) b = 2^s
proof of (1):
Assume a,b are infinite cardinals with b <= 2^a < 2^b.
a^b = 2^b
b^a <= (2^a)^a = 2^a
hence a^b > b^a
proof of (2):
Assume a,b are infinite cardinals with 2^a < 2^b and b = 2^s.
Then
a^b = 2^b
b^a = (2^s)^a = (2^s or 2^a) = (b or 2^a)
Thus, either b^a = b or b^a = 2^a, but both are less than 2^b,
hence a^b > b^a.
As claimed, we have shown that the inequality is sometimes strict.
However I don't know if there are infinite cardinals a,b with
a < b and a^b = b^a
It's conceivable that the existence of such a pair of cardinals is
independent of the standard axioms of set theory.
quasi
Dave L. Renfro
Dave L. Renfro wrote (in part):
> Something you wrote reminds me better about what it was.
> I'm still not sure, but I think it was something like
> there being arbitrarily large cardinals b such that
> b^(aleph_0) and (aleph_0)^b are equal and arbitrarily
> large cardinals b such that b^(aleph_0) and (aleph_0)^b
> are not equal.
This morning, before I left home, I looked for this in a
handwritten manuscript of mine (written around 1990-91).
The correct statement, and a proof, are below. My notes
don't say where I got it from, but I couldn't find it
in the obvious place -- Sierpinski's "Cardinal and Ordinal
Numbers", which I said yesterday is where it was. I'm
not certain Sierpinski's book doesn't have it, but I
did look pretty carefully all through it. I finally
managed to find the result in Chelsea's 1978 reprint of
the English translation of Hausdorff's 1935 3'rd edition
of his book "Set Theory", at the end of Section 7
(top of p. 41). [Incidentally, the front matter of
the Chelsea edition says Hausdorff's 3'rd edition came out
in 1937, but this is incorrect. A well known Russian
translation of the 1927 2'nd edition (sometimes incorrectly
stated in bibliographies as a translation of the 1935
3'rd edition) came out in 1937, for whatever that's worth.]
THEOREM: Let b' be any infinite cardinal number.
(A) There exists a cardinal number b > b' such that
b < b^(aleph_0).
(B) There exists a cardinal number b > b' such that
b = b^(aleph_0).
NOTE: It is easy to see that every cardinal number b satisfies
b < b^(aleph_0) or b = b^(aleph_0).
PROOF:
(A) Let b_1 = b', b_2 = 2^(b_1), b_3 = 2^(b_2), . . .
and let b = sum{b_n: n = 1, 2, 3, ...}. Then we have
b < product{b_n: n = 1, 2, 3, ...} [Konig's inequality]
<= product{b: n = 1, 2, 3, ...} [i.e. (b)(b)(b)...]
= b^(aleph_0).
(B) Let b = [2^(b')] ^ (aleph_0). Then we have
b^(aleph_0) = { [2^(b')] ^ (aleph_0) } ^ (aleph_0)
= [2^(b')] ^ (aleph_0) = b.
Dave L. Renfro
quasi
Here is an improved sufficient condition for strict inequality ...
Proposition:
If a,b are infinite cardinals such that
2^a <= 2^s < 2^b <= 2^(2^s)
for some cardinal s, then
a^b > b^a.
proof:
Assume the hypothesis.
2^b <= 2^(2^s) implies b <= 2^s.
As proved in my previous reply,
2^s < 2^b and b <= 2^s
implies b^s > s^b.
2^a <= 2^s < 2^b implies a <= s < b, hence
a^b >= b^s > s^b >= a^b
so a^b > b^a,
as was to be shown.
quasi
quasi
If we assume a model of set theory such that, for cardinals x,y,
x < y implies 2^x < 2^y,
then the statement of the above proposition can be simplified to:
Proposition:
If a,b are infinite cardinals such that
a <= s < b <= 2^s
for some cardinal s, then
a^b > b^a.
quasi
Butch Malahide
> [. . .]
> However I don't know if there are infinite cardinals a,b with
>
> a < b and a^b = b^a
Let b_0 = a = aleph_{omega}, let b_{n+1} = 2^{b_n} for n < omega, and
let b = b_0 + b_1 +...+ b_n +....
I.e., b = beth_{omega}.
Then a < b and a^b = b^a = 2^b.