Re: An uncountable countable set

In article <45210227@xxxxxxxxxxxxxxxxxxx>,
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:

Virgil wrote:
In article <45205fa9@xxxxxxxxxxxxxxxxxxx>,
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:

Virgil wrote:
In article <45203919@xxxxxxxxxxxxxxxxxxx>,
Tony Orlow <tony@xxxxxxxxxxxxx> wrote:



Since ordinals are, by definition, well ordered, they cannot contain any
endlessly decreasing sequences, which TO's models require.
Neither can the reals.
How about the set of negative integers?
How is that not an endlessly decreasing sequence of reals?
The origin is at a finite location. Order starts from the bottom, if
"decreasing" has any meaning.

The set of negative integers has no "bottom".

TO seems to be changing his tune when it is used against him.

According to TO every set of numbers has a natural order, and it is
within that natural order that we must view it, but now he wants to
reject the natural order because it runs counter to another of his
claims.

TO blows hot and cold with the same breath.

I'm saying the if you iterate the negative integers starting at 0, in
that order, there is no infinite descending sequence.


But that is not their "natural" order, and TO elsewhere insists that we
follow natural orderings.

On the other hand
I don't know why I said "neither can the reals". In any case, the only
way the ordinals manage to be "well ordered" is because they're defined
with predecessor discontinuities at the limit ordinals, including 0.
That doesn't seem "real"

In what sense of "real". There are subsets of the reals which are order
isomorphic to every countable ordinal, including those with limit
ordinals, so until one posits uncountable ordinals there are no problems.




and the axiom of choice aside, I don't see
there being any well ordering of the reals.

The point is that no one can see it even if, given the AC, it is there.

The closest one can come is
the H-riffic numbers. :)

Which have long since been shown not to be anything like well ordered.
.

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