On 5/30/2024 3:44 AM, WM wrote:
Le 29/05/2024 à 20:29, Jim Burns a écrit :
On 5/29/2024 1:11 PM, WM wrote:
Le 28/05/2024 à 21:42, Jim Burns a écrit :
On 5/27/2024 4:10 PM, WM wrote:
If a set of real numbers has no elements below 0,
then it ends
>
Support your claim.
>
The set has ended where no elements follow.
>
Consider a set B of real numbers such that
real number x exists such that
x ≤ each element of B
>
Are you (WM) claiming that
⎛ B.element b′ exists such that
⎝ b′ ≤ each element of B
?
>
The answer is no/yes.
b is impossible to find.
b exists.
There are nonempty.sets B bounded.below x≤ᴬB
about which we know enough that
we can follow the claim that
B holds a least.B.element b′
with a finite not.first.false sequence of claims
which lead to a contradiction.
If b′ exists,
then the contradiction is true.
The contradiction isn't true.
b′ not.exists.
Information about finding or not.finding b′
doesn't change any of that.
It is superfluous information.
If, yes, that is your claim,
then please support that claim.
>
I did so already using unit fractions.
NUF(x) = 1 between x = 0 and x = 1/10^10^10^100000.
No, NUF(x) ≠ 1
| Assume you are correct.
| Assume NUF(x) = 1 for
| 0 < x < 1/10^10^10^100000.
|
| unit fraction ⅟n exists: 0 < ⅟n < x
| no unit fraction exists between 0 and ⅟n
|
| However,
| unit fraction ⅟(n+1) exists: 0 < ⅟(n+1) < ⅟n
| Contradiction.
Therefore, NUF(x) ≠ 1
This is proven by the mathematics of unit fractions
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
Never two appear simultaneously.
⅟n and ⅟(n+1) don't appear simultaneously.
And yet, NUF(x) ≠ 1
The set is not where no elements are.
If the set was before,
then it has ended meanwhile.
>
Are you claiming that
the set holds an element ≤ each element?
>
If, yes, you claim that,
then please support that claim.
>
Bob!
King Bob's time is fully engaged reigning over England,
but he sends all of sci.math his best wishes.
https://www.youtube.com/watch?v=TjAg-8qqR3gisn't logic.
Finite sequences of not.first.false claims
are logic.
>
Your sequences of not.first.false claims leads to
the result that elements can disappear by exchange.
Thank you.
A finite sequence S which holds a false claim
holds a first.false claim.
We consider a set B and
a sequence S of claims about set B
which, in various ways, we have verified
are only not.first.false claims.
One of the claims in S is
0. B contains an element but not a first.element.
Here are the options:
1.
1.1 B holds an element and a first.element.
1.2 S holds a false.claim and a first.false.claim.
Scratch option 1.
We have verified that 1.2 can't be true.
2.
2.1 B holds an element and a first.element.
2.2 S holds a false.claim but not a first.false.claim.
I, for one, welcome the scratching of option 2,
since I do not make infinite sequences of claims.
However,
you (WM) should note that
your "invincible logic" which denies the claim in S
_should_ force you (invincibly) to deny claim 2.2
and option 2.
A false.claim but not a first.false.claim?
Matheology!
3.
3.1 B holds an element but not a first.element.
3.2 S holds no first.false claim and no false claim.
Pick any option from {3}
Conclusion:
Some sets are infinite.
You should recognize that this is impossible.
Therefore,
if your chain contains only correct conclusions, then your start must be wrong.
Here is the most timid start I know of
which provides all the matheology which
you rail against:
_Empty, Adjunct, eXtensionality_
E. The empty set ∅ exists.
A. For sets x and y, adjunct x⨭y = x∪{y} exists.
X. Two equi.membered sets are equal sets.
From EAX follows the usual natural.number arithmetic,
arithmetic which leaves your darkᵂᴹ numbers
out of the picture entirely.
0 = ∅
k⁺¹ = k⨭k = k∪{k}
j < k ⇔ j ∈ k
j⁻¹ ∈ k ⇔ ∃i ∈ k: i⁺¹ = j
ℕ(k) ⇔ 0 ≤ k ∧ ∀j: 0 < j ≤ k ⇒ 0 ≤ j⁻¹ < k
ℕ(k) does not assert that an infinite set exists.
ℕ(k) asserts that k is a countable.to natural --
of which infinitely.many exist, whether
I assert they exist or I assert they not.exist.
After more definitions and more not.first.false claims,
we get to disappearances of Bob and so forth.
if your chain contains only correct conclusions, then your start must be wrong.
Either one of EAX is wrong
⎛ E. The empty set ∅ exists.
⎜ A. For sets x and y, adjunct x⨭y = x∪{y} exists.
⎝ X. Two equi.membered sets are equal sets.
or you are wrong.
Hmmm. It is a puzzlement.
V
Re: V