Re: Logarithm of transfinite numbers

In article <MPG.1e8b557d7e89a7b898ab62@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:

Jonathan Hoyle said:
Hi, Tony. Just wanted to jump in here. :-)

Be my guest! It's an open party... :)


The point is that there is no finite n in N for which the set
achieves infinity.

If by "achieving infinity", you mean "is of infinite size", then you
are correct. Infinite sizes cannot be achieved by finite increments of
finite sizes. You must have either an infinite number of increments or
an increment of infinite size.

Yes, that sounds right as stated, both for the set size and for the values of
the naturals infinitely incremented. But, do you consider it to actually be
an
infinite sequence of increments if none of those increments has an infinite
number of predecessors?

If each "increment" has an infinite number of others following it, as is
the case, it does no matter how many or how few it has preceding it.


This is true for every finite natural, that it has this relation
to the set size.

You are correct, this is true for every finite natural.

So, this inductively proven property of the element in
relation to the set size has implications for the set size
as a whole.

No, it inductively proves it for all set sizes which are finite natural
numbers. It is not proven (nor is it even true) for sets of infinite
size.

No, it proves for every natural number in the set that the set up to and
including that natural number cannot be infinite. So, the set never is
anything
but finite, as long as you are adding elements which are never anything but
finite.
It is not the size of those elements but hw many of them ther are. To
seems hypnotized by his notion that a set of finite sized objects cannot
be other than finite itself, a notion that is obviously false.




As long as the elements are all finite, so is the set
size.

Untrue, as explained above. Although this proposition is does hold for
all sets of finite naturals which has a greatest element. This is the
operative difference between sets like {1,2,3,...,n} and sets like
{1,2,3,...}.

Your explanation above does not pertain to the property proven for the entire
set of elements regarding the set size at each of those elemenets.

Every set of naturals having a lagest element is Dedeking finite.
Every set of naturals, except the empty set, NOt having a largest
element is provably Dedekind infinite.

As Dedekind infiniteness for sets is the only mathematically valid
definition of infiniteness for sets, TO is full of crap.



If at every point in the set the value of the element at that point is
equal to the count of elements up to that point, and if at every
point in the set the value of the element at that point is finite, then
at every point in the set the number of elements up to that point is
finite, so there is NO point in the set where the number of
elements up to that point is infinite.

You need to define what you mean by "point". If you mean stopping at
some finite natural n and looking at the subset {1,2,...,n}, then you
are correct, there is no "point" (subset up to some finite natural) at
which the set becomes infinite. Likewise, at no such "point" soes the
complement {n+1,n+2,...} ever becomes finite. To get to the infinite,
you must transcend finite "points".

But, if all of the elements in the set are finite, and in positions within
the
set equal to their finite value, then there exists no element in the set
which
marks anything but a finite set

Does TO claim that a finite number must to mark the end of an
unboundedly increasing sequence? Stupidity squared!



You
do not have an infinite set of naturals in the quantitiative sense until you
have infinite quantities in the set.

In that case, TO has no infinite sets at all.




Sure, with the Dedekind set-theoretic definition that's true, as I
said above, but


But nothing. Dedekind infiniteness is the only infiniteness we recognise
as having any set theoretic meaning. TO's attempts to confuse the issue
are mere trolldom.
.

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