Re: Another stab at Cantor

In sci.math, msadkins04@xxxxxxxxx
<msadkins04@xxxxxxxxx>
wrote
on 18 Sep 2006 09:50:13 -0700
<1158598213.849000.127730@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
Let R be the set of all infinite binary strings which eventually settle
into a repeating digit or pattern of digits. Let L1 be a well-defined
ordering of R. Let D1 be the diagonal number obtained from L1 by the
application of Cantor's diagonal process. D1 is not a member of R.

Let L2 be the list obtained by putting D1 at the head of L1 (that is,
by adding one to the index number of each member of L1, and placing D1
in the first position of the newly indexed list). Let D2 be the
diagonal number obtained from L2 by the application of Cantor's
diagonal process. D2 is not a member of R, and is not D1. Let L3 be
the list obtained by putting D2 at the head of L2.

Let L_Omega be the list defined by the totality of all possible steps
of this procedure. There is no non-repeating infinite binary string
excluded by the procedure. L_Omega therefore contains all
non-repeating infinite binary strings, which is to say all irrationals.

Mark Adkins
msadkins04@xxxxxxxxx


This cannot be done.

Here's a thought in ordering your prospective (if doomed) list.

If L1, L2, ..., Ln, ....

is your original list (and there *is* such a list; Q is
denumerable), then one can diagonalize this list in the
usual way, getting D1. Simply put D1 second:

L1, D1, L2, ..., Ln, ....

Now one can diagonalize *this* list and put that fourth:

L1, D1, L2, D2, ..., Ln, ....

and so on. Eventually, one gets a constructive procedure L'(L) such
that L'(x,y) is the yth digit of the x'th element in the list; this
procedure is computable if L is computable.

However, one can easily diagonalize *this* list, resulting in a D'1.
Where does he go? Nowhere obvious, that's for sure.

One could try triangles, quadrangles, quintangles, etc.,
but we've failed already by just assuming a list of all
reals (this is not your R, but a union of L_omega and
your R). For *any* such L, if L(x,y) is the yth digit
of the x'th element therein, in a base b, then D(x) =
notD(L(x,x)), where notD(u) is not b-1 and not u (for b=2,
take base 4 or 8 instead by grouping in 2 or 3 bits),
cannot possibly be in the list and therefore the list is
incomplete -- *no matter what's in the list to begin with*.
Adding D(x) somewhere in the list simply constructs a new
list L', which takes one back to the beginning of the proof.

So no, L_Omega can't possibly be the list of all reals, or
the list of all irrationals. It just won't work.

--
#191, ewill3@xxxxxxxxxxxxx
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