Re: How many different unit fractions are lessorequal than all unit fractions? (infinitary)

WM <wolfgang.mueckenheim@tha.de> wrote:

On 09.10.2024 17:11, Alan Mackenzie wrote:

WM <wolfgang.mueckenheim@tha.de> wrote:

No. When we *in actual infinity* multiply all |ℕ|natural numbers by 2,
then we keep |ℕ| numbers but only half of them are smaller than ω, i.e.,
are natural numbers. The other half is larger than ω.

Ha ha ha ha!  This is garbage.  If you think doubling some numbers gives
results which are "larger than ω" you'd better be prepared to give an
example of such a number.  But you're surely going to tell me that these
are "dark numbers

{1, 2, 3, ..., ω}*2 = {2, 4, 6, ..., ω*2} .

You've misunderstood the nature of N.  The set is not
{1, 2, 3, ..., ω}, it is {1, 2, 3, ...}.  There is no last element ω in
it.

Should all places ω+2, ω+4, ω+6, ... remain empty?

It's not clear what you mean by this.  There are no such "places".

Should the even numbers in spite of doubling remain below ω?

Yes, of course.

Then they must occupy places not existing before.

No.  Remember the set is infinite, so you cannot use finite intuition to
reason about it.

That means the original set had not contained all natural numbers. That
mans no actual or complete infinity.

Nonsense.

Theorem: If every endsegment has infinitely many numbers, then
infinitely many numbers are in all endsegments.

That is simply false.  You cannot specify a single number which is in
all endsegments.

True. This proves dark numbers.

Dark numbers don't exist, or at least they're not natural numbers.  There
is no number in each and every end segment of N.

True. But those endsegments which have lost only finitely many numbers
and yet contain infinitely many, have an infinite intersection.

End segments don't "lose" anything.  They are what they are, namely well
defined sets.  Note that your "True" in your last paragraph, agrees that
the intersection of all end segments is empty, which you immediately
contradict by asserting it is not empty.

[ .... ]

Note: The shrinking endsegments cannot acquire new numbers.

An end segment is what it is.  It doesn't change.

But the terms of the sequence do. Here is a simple finite example:

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
{2, 3, 4, 5, 6, 7, 8, 9, 10}
{3, 4, 5, 6, 7, 8, 9, 10}
{4, 5, 6, 7, 8, 9, 10}
{5, 6, 7, 8, 9, 10}
{6, 7, 8, 9, 10}
{7, 8, 9, 10}
{8, 9, 10}
{9, 10}
{10}
{ } .

Example of what?  The reasoning you might do on finite sets mostly isn't
applicable to infinite sets.

Why not? The essence is that only finitely many numbers have been lost
and the rest is remaining.

Your massive misunderstanding of infinite sets is surely a good reason.
Using your vernacular "lost", it is clear that in the infinite
intersection of the end segments of N, an infinite number of numbers has
been "lost".  There are none left over.

Theorem: Every set that contains at least 3 numbers (call it TN-set)
holds these numbers in common with all TN-sets.

Not even an ignorant schoolboy would maintain this.  The two TN-sets {0,
1, 2} and {3, 4, 5} have no numbers in common.

These sets do not belong to the above example. They are not TN-sets.

You should be more careful with your definitions.

Quantifier shift: There is a subset of three elements common to all
TN-sets. Understood?

Yes, I understand completely.

No.

Now complete all sets by the natural numbers > 10 and complete the
sequence.

You can't "complete" a set.

You can add elements.

Then you get different sets, which weren't the ones you were trying to
reason about.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).