Mueen Nawaz wrote:
>  In mathematics, you can prove 
> the existence of certain sets, and you can also prove that you can never 
>   explicitly exhibit those sets (i.e. you can *only* show they exist - 
> it is mathematically impossible to demonstrate them, though). So an 
> argument that shows one cannot exhibit/construct a certain set does not 
> imply that they don't exist.
> 
> Some day I have to formally learn how that is.

It may seem a bit like cheating, but there's an easy way to define such 
a set.  Take whatever system you wish to use to construct sets of 
natural numbers, so long as you only have a countable number of axioms 
and rules.  Now, each construction of a set will require a finite set of 
derivation steps to define it.  Since we have only a countable number of 
choices for each derivation step and a finite number of steps in total, 
there are the same number of such derivations as there are natural 
numbers.  Since there are uncountably many subsets of the natural 
numbers, there must be some subsets for which there exists no derivation 
to construct them.  And there you go, a proof of the existence of a set 
which by definition it's impossible to construct!

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