Re: More complex numbers than reals?

Am 11.07.2024 um 02:29 schrieb Chris M. Thomasson:

On 7/10/2024 5:28 PM, Chris M. Thomasson wrote:

On 7/10/2024 5:24 PM, Moebius wrote:

If a = b = c, {a, b, c} still has "the same number of elements" as {3, 4, 5 }? :-P

Hint: In this case card({a, b, c}) = 1.
Or with other words: Ex(x e {a, b, c} & Ay(y e {a, b, c} -> x = y)).
Using a special quantifier: E!x(x e {a, b, c}) ("There is exactly one x such that x is in {a, b, c}.")

I see {a, b, c} and {3, 4, 5} and think three elements.

You see (!) there terms in "{a, b, c}" (namely "a", "b" and "c") and 3 terms in "{3, 4, 5}" (namely "3", "4" and "5").

Then I start to examine how the elements are different and their potential similarities, if any.

Right. In this case (i.e. an arithmetic context) we may safely assume that 3 =/= 3, 3 =/= 5 and 4 =/= 5. :-P
On the other hand, since we don't know anything concerning a, b and c, all we can state/say is:
1 <= card({a, b, c}) <= 3
(while card({3, 4, 5}) = 3.)

For some reason, { a, b, c } and { 3, 4, 5 } makes me think of monotonically increasing.

Concerning { 3, 4, 5 } the reason is, that indeed 3 < 4 < 5, though { 3, 4, 5 }  = { 5, 4, 3 } = ... etc.
But concerning { a, b, c } there simply is NO (rational) reason for assuming that.
a may be pi
b may be 0
c may be -i
Then {a, b, c} = {pi, 0, -i}. See?!
Or:
a may be 0
b may be 0
c may be 0
Then {a, b, c} = {0}. etc.
So how can we "compare" sets concerning their "size" (""number of elements"")? :-)
How about using a "measuring stick" (sort of)?