Re: Replacement of Cardinality

Am 15.08.2024 um 17:53 schrieb Jim Burns:

On 8/15/2024 8:43 AM, Moebius wrote:

Am 13.08.2024 um 19:02 schrieb Jim Burns:

 

When I first saw WM use 'INVNUF'
he said something that meant
INVNUF(n) = _first_ x with NUF(x) = n

This "definition" is lacking the "domain" of n (i.e. the values it may take).
Hence it's technically not a proper definition i.e. not a definition at all).
In fact, if n e IN = {1, 2, 3, ...} then there IS NO _first_ x e IR with NUF(x) = n (after all in this case there is no x e IR with NUF(x) = n AT ALL).
With the knowledge that img(NUF) = {0, aleph_0} we may consider if there's a _first_ x e IR with NUF(x) = aleph_0. But no, there isn't such an x e IR.
Finally, we may consider if there's a _first_ x e IR with NUF(x) = 0. And yes, there is such an x e IR, namely 0.
So the domain of that "function" is just {0}.
Hence a proper definition of it would have to be stated the following (or a similar or equivalent) way:
       INVNUF(n) = _first_ x with NUF(x) = n     (n e {0}).
So the only value for which INVNUF is defined is 0 (i.e. dom(INVNUF) = {0}).
That's why you CAN'T use "INVNUF(1)" in any of your proofs. Since in this case you would use an undefined "term". And no, you can't just "assume that INVNUF(1) exists" - that's just mumbo-jumbo, but not math.

I've just proved INVNUF(1) not.exists?

Mumbo-Jumbo.*)
Hint: You can't prove that there is no x e IR such that x = 1/0.
Beacuse you can't even USE "1/0" in the statement you want to prove _just because_ the symbol "is undefined".
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*) What you may try to prove is: 1 !e dom(INVNUF). But you can't do that by using "INVNUF(1)" in your proof.
How about: Assume that 1 e dom(INVNUF), then 1 e {0} (since {0} is the domain of INVNUF). But then 1 = 0. Contradiction!