Re: The non-existence of "dark numbers"

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

On 17.03.2025 12:56, Alan Mackenzie wrote:

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

On 16.03.2025 21:08, Alan Mackenzie wrote:

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

[ .... ]

N is defined as the smallest inductive set.

But that definition is impossible to satisfy. Sets are fixed, inductive
"sets" are variable collections.

Wrong.  An inductive set exist by the axiom of infinity.

The set of FISONs is an inductive set. But it is not ℕ because
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo..

More garbage.  That "because" doesn't hold, and merely repeating it ad
infinitum won't make it become true.  As Jim is painstakingly pointing
out, you don't understand the infinite.

The inductive set which the axiom of infinity causes to exist is N.

The subtraction of the set of all FISONs all of which cannot empty ℕ
cannot empty ℕ.

Gibberish.

But just how do you think inductive sets vary?  Do they vary by the day
of the week, the phases of the moon, or what?  Can you give two
"variations" of an inductive set, and specify an element which is in one
of these variations, but not the other?

[ .... ]

So, an irrelevant cite in German, and no answer to the question.  Please
try in your next reply to give a substantive answer, or admit you were
mistaken.

A proof exists that they are there. ℕ_def contains all numbers the
subtraction of which from ℕ does not result in the empty set.

That's not a mathematical statement.

The numbers 1, 2, 3 are such numbers. They are elements of that set.

You clearly know the meaning of these words.

And their meaninglessness is clear.

Your following statements prove that you understand the meaning.

You're doing a quantifier shift

Of course. Here it is justified since the subtraction of all FISONs
which cannot empty ℕ cannot empty ℕ.

That's gibberish.  For a start, there's the illegitimate use of "empty"
as a verb, which I've already admonished you over.  For another thing you
are (possibly deliberately) being ambiguous in the meaning of "all".  You
might mean "the subtraction of EACH FISON", you might mean "the
subtraction of EVERY FISON together".

Are you deliberately writing this ambiguity?  If so, you have my
contempt.  If not, you would be advised to take advice from a
mathematician or other clear thinking person about how to write clear,
meaningful text.

Also, you are describing a set of FISONs, without proving it exists
(which is not difficult), and without proving it is unique.  It is
clearly not unique, there being an uncountably infinite number of sets
which would satisfy the condition you refuse to write clearly.

 And your N_def, as you have "defined" it, is satisfied by any
proper subset of N.

No, it is the union of FISONs.

Non-responsive answer.  Any proper subset of N will satisfy your
"definition" of N_def.

Or in a different interpretation, N_def = N,

No. ℕ_def is a proper subset.

Wrong.  The topic of this subthread is the non-existence of dark numbers
(or at least it was until you derailled it).

Where we're at at the moment is:
1. You're asserting there exists an inductive N_def which is a proper
  subset of N.
2. N \ N_def is thus non-empty.
3. In that case there must exist a least member of N \ N_def.

Please state what this least member is, or at least describe how you'd go
about identifying it.

since An e N, N\{n} is non-empty.

Here we use
∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo

Either you're incapable of writing mathematically what you mean,

I did it frequently:

You're incapable of doing it all the time.  Or unwilling.

∀n ∈ UF: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0

Obviously the subtraction of all numbers which cannot empty ℕ cannot
empty ℕ.

"Empty" in this sense is meaningless.

You are not unable to understand the meaning. But you are dishonest.

I refuse to discuss things expressed in sloppy meaningless language, as
"empty" used as a verb here is.  A number cannot "empty" a set, because
the number is not an agent; it is not an operator;

Subtraction is an operator.

Non-responsive answer.  A number cannot "empty" a set, because the number
is not an agent of any kind.

it is not a function.  Such sloppy language allows you to reason
sloppily, and possibly to derive falsehoods as if they were facts.

You are unable to read or to understand. You criticise your reading or
your incoherent thinking, not my writing.

My reading and understanding are fine.  Nobody else in this newsgroup
criticises them.  Your writing is full of falsehoods, ambiguity, and
meaninglessness.  Many people on this newsgroup criticise it for those
reasons.

I think you are capable of expressing your thoughts in a mathematical
fashion.  I wish I could be sure, though.

Simply read what I write. It is ridiculous.

I can agree with that.

[ .... ]

Subtraction is a function on two numbers mapping to a number of the same
type.  What you appear to be talking about is "removal" of an element or
subset of a set from that set.  Do you really use the word "subtrahieren"
in German for this?

Set subtraction is also used in English.

Please answer the question: Is the verb "subtrahieren" used in German to
denote the removal of an element or a subset of a set from that set?

Regards, WM

--
Alan Mackenzie (Nuremberg, Germany).