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Re: Distinguishability of paths of the Infinite Binary tree???
Posted:
Dec 29, 2012 2:03 AM
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On Dec 29, 4:06 am, Virgil <vir...@ligriv.com> wrote:
> In article
> <51eb8729-b50a-4136-8af6-2c527c93b...@x20g2000vbf.googlegroups.com>,
>
> Zuhair <zaljo...@gmail.com> wrote:
> > By the way you said it is not obvious to you what I meant by parameter
> > free definable
>
> An example of a mathematical definition ( or even a non mathematical
> one) which you regard as being "parameter free" and of one which you
> regard as not being "parameter free" might clear the air.
>
> I am not totally clear in my owm mind what you mean by the phrase.
>
> Do you, for instance, regard the standard definition of countability of
> a set (set S is countably if and only if there exists a surjection from
> |N to S) as being parameer free or not, and why?
> --
This is a well known subject. There is no problem if some people don't
know about it. The problem is if some people professing big claims
like refuting Cantor or saying that THOUSANDS of mathematician for a
CENTURY long time are acting fools, and then it turns that those
people themselves don't know basic definitions? really strange!
Now we come to your example:
Is countability of a set parameter free definable or not?
The answer is YES.
Why?
Because there is a parameter free formula "phi(S)" such that
For all S. Countable(S) iff phi(S)
And what is meant by phi(S) being parameter free is that phi(S) is a
formula in which only the symbol S occurs free, i.e. all other
variable symbols in phi(S) are quantified within the formula, and of
course S is free.
Now lets explicitly examine this
take phi(S) to be the following formula
Exist f. Exist N. (for all y. y in N iff y is a finite ordinal) & f: S
--> N & f is injective.
The open expansion of this formula show that only the variable symbol
S occurs free.
QED
So for example in a theory like NF where we do have the set of all
countable sets. So this set is parameter free definable set, because
membership of this set follows satisfaction of a parameter free
formula.
Now let me give you an example of an object that cannot be definable
in a parameter free manner.
Lets take some theory that provides sufficient material to define
'definable real' as:
x is a definable real <-> iff x is a real & Exist phi. for all y. y in
x <-> phi(y)
where phi(y) is a parameter free formula (i.e. only y occurs free in
phi(y)).
Now one can prove using Cantor's argument that ANY bijection F between
the set R* of ALL definable reals and the set N of all naturals IS non
parameter free definable
The reason is that if we suppose the contrary i.e. the existence of
such a bijection that is parameter free definable, then the diagonal
defined after it would be a parameter free definable real that is not
in the set of ALL definable reals, which is a clear contradiction.
So although there is a bijection between R* and N, yet it is provable
that any such bijection is non parameter free definable!
In other words there is no parameter free formula phi(y) such that the
above bijection have all its membership determined after satisfaction
of phi(y).
Hope that is helpful and clear
Zuhair
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