jsfisher said:
Revisit your identity mapping for S = {a, b, c, ...}. The mapping is f(x) : x -> {x}. You correctly observed that {} is outside the range of this mapping. So is {a,b}, and so is {a,c}, and so is....
Exactly one is nowhere to be found.
In
http://www.internationalskeptics.com/forums/showpost.php?p=10950983&postcount=794 I wrote:
doronshadmi said:
S is an infinite set.
H is a set that its members are not mapped with all |S| members of set S, and is has exactly one member for any mapping between S and P(S).
Here is an example, without loss of generality:
If S={a,b,c,...} then the distinct result of the distinct mapping
a --> {a}
b --> {b}
c --> {c}
...
is {}.
In that case H={{}}, where a member of P(S) like {a,b} is not a member of H because a --> {a} or b --> {b}, so what you wrote above is irrelevant to my "weaker" argument.
It has to be clarify that {} as a single member of H, is the result of a
group of infinitely many possible mappings from S to P(S), where in the case of jsfisher's example, {} is some possible single member of H, by (for example) the three following mappings of such group,
without a loss of generality (about {} or any other P(S) member, as a possible single member of set H):
If S={a,b,c,...} then the distinct result of the distinct mappings
a --> {a}
b --> {b}
c --> {c}
...
or
a --> {a,b}
b --> {b}
c --> {c}
...
or
a --> {a,c}
b --> {b}
c --> {c}
...
and so on ... ad infinitum
is {}.
So we are not talking about a single particular mapping, but about a group of mappings that their result is some single and distinct member of P(S), as a member of H.
Since any possible member of P(S) can be a member of H, one can't formally\logically define the set of all members of P(S) that are not mapped with all S members, simply because any particular member of H (which is some member of P(S)) can be mapped with some member of S in another group of mappings from S to P(S) that does not determine it as a member of set H.
In order to understand it better, there is a group of infinitely many arrangements of the members of the diagonal set w.r.t to the members of N={1,2,3,...), which provide a
single diagonal member that is not mapped with any member of N.
But this provided
single diagonal member can't be a member of the set of all diagonal members that are not mapped with any member of N, simply because this
single diagonal member can be mapped with some member of N, in another group of infinitely many arrangements of the members of the diagonal set w.r.t to the members of N={1,2,3,...}.
----------------
So each
single member of set H is the result of a group of infinitely many things, where each group provides another
single H member, such that these
single members can't be gathered into a one set, which has more than a
single member.
What is written in
http://www.internationalskeptics.com/forums/showpost.php?p=10950983&postcount=794 has to be taken in terms
group for any
single member of set H, so my example there (a --> {a} or b --> {b}) is insufficient, and has to be replace in terms of
groups, as explained here.
