Re: Another stab at Cantor

In article <1158699227.198536.286480@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"georgie" <geo_cant@xxxxxxxxx> wrote:

At this stage, it was asserted the resulting (ordered) set contains
all possible strings. That is false. While the set, ordered as given,
is not a list, it can nonetheless be REORDERED

That reordering step is a possible step. It *MUST* be in the set of
all possible steps.

so that the result
->IS<- a list, and we can easily produce strings not on THAT list, and
therefore not in the original (ordered) set.

Except for the fact that it can't be.

Except for the fact that for any listing, as many (in fact "more")
elements are not in that listing than are in it.

Given the set, S, of all binary strings (mappings from the set of
naturals , N, to the set {0,1}, and any mapping, f, from N to S,
then for each member of S there is a (possibly different) member of S
not in the image of f.

The construction is unambiguous but a bit complicated, so I will only
post it if asked.
.

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