Re: RAF: Rational numbers, irrational numbers: each dense in real numbers

On Oct 21, 9:05 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

With regards to the proof (largely in sketch, but assembled piece by
piece) of the existence via choice of a transfinite "sequence" of
descending irrationals, I would like to hear more arguments as to why
that is not possible, so they can be addressed.



Let X be an ordinal, and for every x in X, let r_x
be a real number. 0 < r_x < 1

Let C={r_x| x in X}

Note that if x is in X, then x has
a successor, call it x+1.

Assume that 0<r_(x+1)<r_x.

C is a strictly descending "sequence".

Let N be the set of natural numbers.
Let n be an element of N.

Let U_n = {r_x| x in X, 1/n >= (r_x-r_(x+1)) > 1/(n+1)}

U_n has fewer than n+1, elements, so U_n is finite.

C = union n in N U_n

C is the union of a countable number of finite sets, so
C is countable.

Thus any strictly descending "sequence" is countable.

Note that all we need for this proof is that every element
of the "sequence" is strictly greater than its successor.
Since every element has a successor, we do not have
to distinguish betweeen limit and non-limit ordinals.

- William Hughes

.

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