Re: Replacement of Cardinality (infinite middle)

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:

..., REQUIRED, ....

>
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.
>
My neat little hedgerow is well.ordered;
each non.empty subset holds a minimum.
>
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.
>
Successors are non.0 non.doppelgänger non.final.
>
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.
>
Have a nice day.

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.
>
Where it _goes_.

No inductive case _goes_
if   P(0) and ∀j ∈ ℕ: P(j)⇒P(j+1)
then P(0) and ¬∃j ∈ ℕ: P(j)∧¬P(j+1)
Each non.0 of ℕ has a predecessor.
if   P(0) and ¬∃j ∈ ℕ: P(j)∧¬P(j+1)
then the set {i ∈ ℕ: ¬P(i)} holds no first.
ℕ is well.ordered.
if   the set {i ∈ ℕ: ¬P(i)} holds no first
then {i ∈ ℕ: ¬P(i)} = {}
if   {i ∈ ℕ: ¬P(i)} = {}
then ∀k ∈ ℕ: P(k)
if   P(0) and ∀j ∈ ℕ: P(j)⇒P(j+1)
then ∀k ∈ ℕ: P(k)
No inductive case _goes_

Then a claim like "I don't pick wrong"

Note:
My claim is not like "I don't pick wrong".
My claim is like "I am talking about Little Bunny Foo Foos",
with their famous well.order, successors, predecessors,
and field.mice.bopping.
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.
Nevertheless, like the moons of Jupiter,
the argument persists, with or without our looking.
Eppur si muove.

So, well-order the reals.

If an inaccessible cardinal exists,
then the reals can be well.ordered.
So?