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

On Sep 20, 10:17 pm, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

A well ordering is an ordering relation on elements of a set such that
each subset of the set has a least element by the ordering.

Close.

R is a well ordering on S <-> (R is a strict total ordering on S &
every nonempty subset S has an R-least member).

(At least I consider each element to be in the universe,

Obviously any object mentioned in a theory is a member of any universe
of a model of the theory.

and when
collections are defined by their elements

The axiom of extensionality stipulates that a set is determined only
by its elements, if that's what you mean.

and have the element-of and
subset defined that they're sets.)

I don't know what that is supposed to mean.

What's your point?

Your 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}."

Take it step by step:

Let R be a well ordering of X non-empty.

Let p be a function from X into {r | r is irrational & r e (0 1)}.

After that, your formulation is incoherent (but it's still not the
fatal flaw, since we could still fix your incorehent formulation and
find a deeper mistake): You say "i+1 is the least element of the well-
ordering X_i setminus i". In other words, you're defining a 1-place
operation on X. But the notation 'X_i' indicates that X itself is also
a function. But you've not said what function X is. Moreover, you say
"setminus i" when you must mean "setminus (i.e. complement) {y+ | y R-
less than i}.

So we might address that mess this way (and from now on instead, of 'R-
less' and things like that, I'll just say '<' etc. instead of 'R-less'
as R is understood by context:

What you want is to say what "i+" is for any element i of X.

So, since R is a well ordering of X, there is a successor relation in
X (whether or not it's the ordinary successor relation of the ordinals
- i u {i} - is not crucial; what's important is that we can define 'R-
successor' and thus have our successor relation on X on that basis.
(By context, we can just say 'successor' instead 'R-successor)

So i+ = the successor of i.

Then, let c be the least member of X. So p(c) is some irrational in (0
1).

Then, for any i in X, p(i+) is some irrational in (0 1)\{p(y) | y <
i}.

And we'll add a stipulation about p that you mentioned previously: For
every i+, we have p(i+) < p(i).

Hint here: What is missing at this point?

Then you say, "There are uncountably many irrational numbers less than
each p_i and greater than zero".

Yes, for any r>0, there are uncountably many irrational numbers in any
interval (0 r). But, back to the hint, something is not accounted for
in your construction so far.

Then you define P as the range of p.

Okay, but think about the hint again now.

Then you say, "There exists a rational number q_i between p_i and p_{i
+1}"

Yes, for any i in X, there is a rational number between p(i) and p(i
+1).

You continue, "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}, and infinitely
many."

Uncountably many indeed.

Then you say, "Let the ordered pair (p_i, q_i) be an element of a
function, as a set, from P to Q."

I'd make that:

Let q be a choice function on P such that q(p(i)) = an irrational
between p(i) and p(i+).

So p(i+) < q < p(i+).

If you want this to be on X, then:

Let q' be a choice function on X such that q(p(i)) = an irrational
between p(i) and p(i+).

Then you say, "If there is an uncountable set P of irrational numbers
in (0,1)"

IF. Is P uncountable? P is the range of p. Go back to the hint now!

You finish, "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"

I don't even have to check whether the function is 1-1. You just need
to go back to the hint, which will lead you to reexamine your claim
that P is uncountable.

MoeBlee

.

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