Re: The non-existence of "dark numbers"

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

On 12.03.2025 22:31, Alan Mackenzie wrote:

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

On 12.03.2025 18:42, Alan Mackenzie wrote:

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

[ .... ]

Learn what potential infinity is.

I know what it is.  It's an outmoded notion of infinity, popular in the
1880s, but which is entirely unneeded in modern mathematics.

That makes "modern mathematics" worthless.

What do you know about modern mathematics?

I know that it is self-contradictory because it cannot distinguish
potential and actual infinity.

It can, but doesn't need to.  Potential and actual infinity are needless
concepts which only serve to confuse and obfuscate.  If you disagree,
feel free to cite a standard result in standard mathematics which depends
on these notions.

When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, ....

Do you ever bother to check what you write?  The difference operator \
applies to sets, not to cardinal numbers.  I can guess what you mean, but
your readers shouldn't have to guess that.

.... then |ℕ| \ |{1, 2, 3, ..., n+1}| = ℵo. This holds for all elements
of the inductive set, i.e., all FISONs F(n) or numbers n which have
more successors than predecessors.

I.e. all natural numbers.

Only those contribute to the inductive set!

The inductive set is all natural numbers.  Why must you make such a song
and dance about it?

Modern mathematics must claim that contrary to the definition ℵo
vanishes to 0 because ℕ \ {1, 2, 3, ...} = { }.  That is blatantly
wrong and shows that modern mathematicians believe in miracles.
Matheology.

Modern mathematics need not and does not claim such a ridiculous thing.
Your understanding of it is what's lacking.

You may recall me challenging others in another recent thread to cite
some mathematical result where the notion of potential/actual infinity
made a difference.  There came no coherent reply (just one from Ross
Finlayson I couldn't make head nor tail of).  Potential infinity isn't
helpful and isn't needed anymore.

3. The least element of the set of dark numbers, by its very
      definition, has been "named", "addressed", "defined", and
      "instantiated".

It is named but has no FISON. That is the crucial condition.

What the heck does it mean for a number to "have" a FISON?  Assuming you
can define that, you need to prove that the least "dark number" "has" no
FISON.  And assuming you can do that (which I very much doubt), you then
have to clarify what that condition is crucial to and how.

So you counter my proof by silently snipping elements 4, 5 and 6 of it?
That's not a nice thing to do.

They were based on the mistaken 3 and therefore useless.

You didn't point out any mistake in 3.  I doubt you can.

I told you that potential infinity has no last element, therefore there
is no first dark number.

The second part of your sentence does not follow clearly from the first,
therefore the sentence is false.  And even if it were not false, it has
no bearing on my item 3.

But I can agree with you that there is no first "dark number".  That is
what I have proven.  There is a theorem that every non-empty subset of
the natural numbers has a least member.  On the assumption (yours) that
"dark numbers" are a subset of the natural numbers, that proves that
there are no "dark numbers" at all.

Try to remove all numbers individually from the harmonic series such
that none remains. If you can't, find the first one which resists.

Why should I want to do that?

In order to experience that dark numbers exist and can't be manipulated.

Dark numbers don't exist, as Jim and I have proven.

When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, then |ℕ| \ |{1, 2, 3, ..., n+1}| =
ℵo. How do the ℵo dark numbers get visible?

There is no such thing as a "dark number".  It's a figment of your
imagination and faulty intuition.

Induction cannot cover all natural numbers but only less than remain
uncovered.

The second part of that sentence is gibberish.  Nobody has been talking
about "uncovering" numbers, whatever that might mean.  Induction
encompasses all natural numbers.  Anything it doesn't cover is not a
natural number, by definition.

Every defined number leaves ℵo undefined numbers. Try to find a
counterexample. Fail.

What the heck are you talking about?  What does it even mean for a number
to "leave" a set of numbers?  Quite aside from the fact that there is no
mathematical definition of a "defined" number.  The "definition" you gave
a few posts back was sociological (talking about how people interacted
with eachother) not mathematical.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).