Re: Aleph_Aleph_1

On Jun 25, 5:44 am, zuhair <zaljo...@xxxxxxxxx> wrote:
On Jun 25, 6:33 am, Aatu Koskensilta <aatu.koskensi...@xxxxxx> wrote:

zuhair <zaljo...@xxxxxxxxx> writes:
Is aleph_aleph_1 provable in ZFC?

No -- only formulas are provable in ZFC.

Ok then, I'll rephrase my question:

 is the following a theorem of ZFC?

Ex: x=aleph_(aleph_1).

For any term T of the (extended) language, we have the theorem of
identity theory:

Ex x=T.

And 'aleph_(aleph_1)' is a term of the (extended) language.

So this is not the exact question you need to ask.

Or even more generally:

 is the following a theorem in ZFC?

forall x ( x is an ordinal -> there exist y ( y= aleph_x ))

Essentially the same answer as I gave above.

I mentioned already, the aleph operation is defined by transfinite
recursion. I'll use 'H' instead of 'aleph' and parentheses instead of
the underscore character:

'H' is a 1-place function symbol added to the language by definition
by transfinite recursion.

If 'T' is a term, then 'H(T)' is a term.

Using the Fregean method,

If ZF |- T is an ordinal

then

ZF |- H(T) ~= 0

and if

ZF |- ~ T is an ordinal

then

ZF |- H(T) = 0.

Moreover, from ZF we prove that for all ordinals j and k, if j in k,
then H(j) is a cardinal less than H(k).

Also, in ZFC we prove that every infinite cardinal is H(k) for some
ordinal k.

But back to your specific question. H(1) is a cardinal and thus an
ordinal, so H(H(1)) is a cardinal.

However, as David Ullruch and I were discussing, using alephs
themselves as arguments for the aleph operation is not ordinary or
such good notation, even though it it is technically allowed.

MoeBlee

.

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