Re: The infintely small number b

On Nov 23, 10:07 am, Venkat Reddy <vred...@xxxxxxxxx> wrote:
On Nov 23, 2:16 am, Marshall <marshall.spi...@xxxxxxxxx> wrote:



On Nov 22, 9:16 am, Venkat Reddy <vred...@xxxxxxxxx> wrote:

Yes I understand that. But we are trying to find the inclusive bounds
for that open set D, and they are not in R according to you. I would
like to correct you here. They are still in R but we just can't know
them. As we knew, it boils down to the fact of not being able to
navigate from a point to next one.

Let P(x) mean that x is a real number greater than 0:

1) for all x, P(x) if and only if (x > 0 and
x is a real number)

We call some l a lower bound of some P
if for every x that satisfies P(x), l <= x.
We call some lower bound l of P an
inclusive lower bound if P(l).

Assumption A) the inclusive lower bound l of P exists:

A1) exists l, P(l).
A2) for all x, P(x) means l <= x.

Let m = l/2

Since P(l), l > 0. Division of nonzero numbers is closed
over the reals, so m is real. For all x, x > 0 means
(x/2) > 0, so m > 0. m is real and m > 0, so P(m)
from 1).

However, m < l, which contradicts 2). Since we have
reached a contradiction starting from our assumption,
our assumption must be false. Therefore l does not
exist.

This proves that there exists a quality of the real
numbers, namely P, for which there is no smallest
number exhibiting that quality.

So, P is a quality where P(x) means x is a real number greater than 0.
And you have proved that there is no smallest number exhibiting that
quality.

If we remove the redundant words from the above, you mean to say there
is no smallest number greater than 0. This could have been much
simpler to proove. So you mean this proof contradicts my saying that
smallest number exists in R. Fine. Thats not my core argument, anyway.

My argument was - every finite interval in continuum (not modeled with
single kind of points or numbers yet) always has its bounds just like
any other finite interval.


1. This is "ex cathedra". You give no reason that an interval
should "have its bounds" (whatever that means) (You had a confused
argument having to do with cutting and hence "duplicating" points,
but you backed off that when it was pointed out that saying
y was a bound for more than one segment, did not mean y had
to be duplicated)

2 You do not say what you mean by "have its bounds". You do
not mean "have bounds" because you agree that the open set
(a,b) has bounds even though it does not contain them.
You don't mean that the bounds are elements of the interval because
you have been explicit in saying that an interval is not
made up of bounds.
- William Hughes


.

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