Re: Replacement of Cardinality (infinite middle)

On 08/29/2024 10:24 PM, Jim Burns wrote:

On 8/29/2024 8:28 PM, Ross Finlayson wrote:

On 08/29/2024 05:12 PM, Ross Finlayson wrote:

On 08/29/2024 04:32 PM, Jim Burns wrote:

On 8/29/2024 6:46 PM, Ross Finlayson wrote:

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..., REQUIRED, ....

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Things missing from my neat little hedgerow are
missing because I intend for them to be missing.
My neat little hedgerow has no weeds.
It has not had and will not have weeds.
And weeds would not be an improvement.
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My neat little hedgerow is well.ordered;
each non.empty subset holds a minimum.
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In my neat little hedgerow,
each Little Bunny Foo Foo has a successor,
scooping up the field mice and bopping them on the head,
and is a successor, except the first, named 0.
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Successors are non.0 non.doppelgänger non.final.
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You are welcome to talk about something else, Ross.
Note, though, that,
if you are talking about something else,
then you are talking about something else.
Non.triangles are not counter.examples to triangles.
Non.Bunny.Foo.Foos are not counter.examples to Bunny.Foo.Foos.
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Have a nice day.

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Sort of like you don't apply the inductive cases
that each stay "nope" and instead only affirm
that each one "goes", where, it goes.
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Where it _goes_.

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No inductive case _goes_
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if   P(0) and ∀j ∈ ℕ: P(j)⇒P(j+1)
then P(0) and ¬∃j ∈ ℕ: P(j)∧¬P(j+1)
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Each non.0 of ℕ has a predecessor.
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if   P(0) and ¬∃j ∈ ℕ: P(j)∧¬P(j+1)
then the set {i ∈ ℕ: ¬P(i)} holds no first.
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ℕ is well.ordered.
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if   the set {i ∈ ℕ: ¬P(i)} holds no first
then {i ∈ ℕ: ¬P(i)} = {}
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if   {i ∈ ℕ: ¬P(i)} = {}
then ∀k ∈ ℕ: P(k)
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if   P(0) and ∀j ∈ ℕ: P(j)⇒P(j+1)
then ∀k ∈ ℕ: P(k)
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No inductive case _goes_
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Then a claim like "I don't pick wrong"

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Note:
My claim is not like "I don't pick wrong".
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My claim is like "I am talking about Little Bunny Foo Foos",
with their famous well.order, successors, predecessors,
and field.mice.bopping.
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The support for this claim that I'm talking about that
is my continued talking about it.
It's possible that this argument is hard to see
because it's microscopic.
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Nevertheless, like the moons of Jupiter,
the argument persists, with or without our looking.
Eppur si muove.
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So, well-order the reals.

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If an inaccessible cardinal exists,
then the reals can be well.ordered.
So?
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Regularity of _difference_, and, regularity of _dispersion_,
both _increment_, and _modularity_, are examples of two
various kinds of regularity, one from the _increment_ the
other from _partition_, indeed some have that "inductive set"
by itself is _not_ sufficient and even "successor" the usual
case is _not_ sufficient and Peano's and Presburger's increment,
repeated increment, and repeated repeated increment, are
_not_ sufficient, to result a model of integers, modular,
and regular in difference, and regular in dispersion.
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Where they go, ....
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So, for sufficiently large m, for sufficiently larger n,
that (m+1 - m) = (n+1 - n), is not a given without both
regularity of difference and regularity of dispersion,
REQUIRED.
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Or, that is why, ....
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The reals actually give a well-ordering, though,
it's their normal ordering as via a model of line-reals.
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Of course any other one you'd give would have taking a
subset of ordinals, which of course are _always_ well-ordered,
with those being an uncountable subset's, of the reals,
_also in their normal ordering_.
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So, ....
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Anyways the "only-diagonal and/or anti-diagonal: both or none",
here has also this quite strong point: "arithmetic is increment
and division". I'm saying usual algebra's is wrongly one-sided.
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