Re: The non-existence of "dark numbers"

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?  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".

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.

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.

Jim has supplied at least one other proof.

He claims that lossless exchange can produce losses. He is in
contradiction with logic.

Irrelevant to the current discussion.  He has supplied at least one other
proof of the non-existence of "dark numbers".

As invalid as yours.

No.  You have failed to identify any invalidity.

If you were able to learn, then you would have the chance here:
ℕ \ {1} = ℵo

Where do you get that from?  You're trying to say a subset of N is
identical to the first transfinite cardinal.  That cannot be true.

and if ℕ \ {1, 2, 3, ..., n} = ℵo
then ℕ \ {1, 2, 3, ..., n+1} = ℵo.

Where do you get that from?  It is clearly false - the first of these
sets contains an element, n+1, that the second one doesn't.  Therefore
they are distinct sets.

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.

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).