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

On Sep 20, 5:54 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:
On Sep 20, 12:57 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:



On Sep 20, 3:15 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

On Sep 19, 5:42 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 19, 12:26 am, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

On Sep 18, 6:05 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 18, 7:23 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

On Sep 18, 3:33 pm, William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Sep 18, 3:21 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

In ZFC, with standard definitions of the real, rational, and
irrational numbers, let p_i be an irrational number between zero and
one for i from a suitably large well-ordered index set X. With the
well-ordering of the index set, let the i'th element p_{i+1} be an
irrational number between zero and p_i, where i+1 is the least element
of the well-ordering X_i setminus i, that is defined to equal X_{i
+1}. There are uncountably many irrational numbers less than each p_i
and greater than zero, else the irrational numbers are countable (or
the real numbers are not standard). Define P to be comprised of the
p_i's. There exists a rational number q_i between p_i and p_{i+1},
else the rational numbers are not dense in the reals thus that between
any two irrational numbers there is a rational number. For each of
the irrational p_i's, there thus exists at least one unique rational
q_i between p_i and p_{i+1},

This is the fallacy. q_i is not unique. There are an uncountable
number of pairs
p_i and p_{i+1} and only a countable number of q_i available (q_i
must be a rational in [0,1]). Thus, although we can choose a q_i for
each index i, we cannot choose the q_i in such a way that
q_i =/= q_j for i =/= j.

and infinitely many. Let the ordered
pair (p_i, q_i) be an element of a function, as a set, from P to Q.
If there is an uncountable set P of irrational numbers in (0,1), then
there is a 1-to-1 function defined by the set {(p_i, q_i), i E X} from
uncountable P to a subset of Q the rational numbers,

since q_i is not different for each index i, this function is not 1-
to-1.

- William Hughes

No presumptions were made about the countability or lack thereof of
the rationals, only the density (denseness) of the rationals in the
reals. That there is a contradiction is so, I agree, but, to avoid
that contradiction by denying the denseness of the rationals in the
reals leads to another contradiction, in that the rationals are dense
in the reals.

The "uncountably many pairs" of distinct irrational numbers are
generated in a particular way thus that the open intervals (each
containing infinitely many rationals) connecting them are disjoint.

This is not possible. Let's check the construction

let p_i be an irrational number between zero and
one for i from a suitably large well-ordered index set X. With the
well-ordering of the index set, let the i'th element p_{i+1} be an
irrational number between zero and p_i, where i+1 is the least
element
of the well-ordering X_i setminus i, that is defined to equal X_{i
+1}.

You are trying to iterate through an uncountable set. You will not
get very far. In particular, p_j will not be defined if j is not
equal to i+1 for some i in X. Nor will p_k, where k>j. So p_i will
not
be defined for i greater than or equal to the first limit ordinal.

- William Hughes

That's why there is "transfinite induction."

For transfinite induction you need more than a definition of p_j
when j is equal to i+1 and p_i is defined. You need a definition of
p_j when
you have a definition for p_i for all i<j, even when there is no i
such that j=i+1. You do not have this.

- William Hughes

About the application of transfinite induction, consider as above
where even MoeBlee will accept that there can be uncountably many
intervals (0, p_i) in the unit interval.

Yes, there are uncountable many (0,p_i) in the unit interval.
They are not disjoint. This has zilch to do with transfinite
induction.

Those exist, then there is
the consideration of whether the mapping of the irrationals to
elements of an ordinal can be so imposed in manner that happens to
correspond to the reverse of the normal ordering.

Nope. the consideration is whether you can find an uncountable
set of intervals that are disjoint. For this you need
a lot more than just a reversal of the normal ordering.

Due to the denseness of the irrationals in the reals, a well-ordered
sequence of irrationals can be constructed, even if not explicitly
listed, for the successor ordinals of zero, the first limit ordinal.
In this context each subset of P will have a particular greatest
element, in the normal ordering, that is a least element in the
reverse ordering which is an artifact of the direction. (For clarity
it would be simple to use the normal ordering and build the intervals
upwards from a small irrational instead of downwards from an
irrational near one.) Then, that sequence

A sequence is countable, so far we are talking about a countable
subset of P.

can be selected to converge
to but not reach a p_lambda-1 > p_lambda, irrational, which as well
given the uncountability of the irrationals and denseness of the
irrationals in the reals, establishes the first infinite limit
ordinal's case as a consequence of the finite successor ordinals'
cases, because there is a difference between them thus that a rational
exists between them.

So we have now definied a countable number of irrationals,
and defined a value for the first countable limit ordinal.
(note in your original construction, you did not have any limits
at all).

Then, following limit ordinals' cases are established in a similar
manner,

No, you can handle a countable number of limit ordinals in this
fashion, but you do not get to an uncountable limit ordinal.

where again due their denseness and presumed uncountability,
the irrationals are inexhaustible. Is that not sufficient? If not,
please explain why you would think not.

Because you have defined values for a countable number of countable
sequences. This means you have defined values for a countable number
of ordinals. That is all you have done. You still have not defined
values for an uncountable set.

Note this is not trasfinite induction.

- William Hughes

For any of countably many iterations,


Yep iterations again. You will never get further than
countable using interations.

Call gibberish transfinite induction does not make it so.

given that the irrationals are
not countable, there yet exists a p_lambda_1 greater than zero. Then,
for all those successor ordinals of all those (regular, ZFC, von
Neumann) limit ordinals, the case is shown that there is a difference
containing rational numbers. Then, if there are uncountably many
irrationals in the unit interval, and rationals and irrationals are
each dense, there are the uncountably many distinct well-ordered
p_i's, and that many ordered pairs with distinct q_i's. It's shown
that for any successor ordinal of a countable limit ordinal, the case
holds, and that thus the case for the next limit ordinal does as
well. Then, for the same reason that there are enough irrationals for
there to be countably infinitely many, they can always be selected in
a way leaving an interval containing a dense subset of them, for the
countable limit ordinals, that they can be selected in a way not
exhausting the irrationals (leaving an interval of the reals
containing only countably many irrationals), via transfinite induction
the result is shown to hold.

So, in ZFC,with standard definitions of the numbers: the irrationals
are uncountable, irrationals are dense in the reals, or rationals are
dense in the reals. Pick two.

Ross

--
Finlayson Consulting


.

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