Re: The non-existence of "dark numbers"

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:

 

If the numbers are definable.

 

Meaningless.  Or are you admitting that your "dark numbers" aren't
natural numbers after all?

 

They

 They?
 

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.
When |ℕ| \ |{1, 2, 3, ..., n}| = ℵo, 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. Only those contribute to the inductive set! 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.
   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.

 

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.

 

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?

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.
Regards, WM