A Tale of Two Sets

Given 2 sets, A and B, if A contains all of the members of B AND has members B does not, A's cardinality has to be greater than B's.
And the predicted unsupported and cowardly response of "Joe doesn't understand infinity", is duly noted.
Let the flailing begin...
Given 2 sets, A and B, if A contains all of the members of B AND has members B does not, A's cardinality has to be greater than B's.
And the predicted unsupported and cowardly response of "Joe doesn't understand infinity", is duly noted.
Let the flailing begin...
19 Comments:
At 12:18 PM, socle said…
And the predicted unsupported and cowardly response of "Joe doesn't understand infinity", is duly noted.
Another possible response: You actually do understand infinity to some extent, but are mostly just trolling.
I'm genuinely interested to know the answer to this question: In JoeMath, what is the largest cardinality of any set?
At 1:22 PM, Joe G said…
Another possible response: You actually do understand infinity to some extent, but are mostly just trolling.
I would say I understand infinity better than most and it is telling that you avoided the point of the post.
In JoeMath, what is the largest cardinality of any set?
Still working on a name for that. I like "aleph zeta". What do you think?
What I do know is that given 2 sets, A and B, if A contains all of the members of B AND has members B does not, A's cardinality has to be greater than B's.
At 1:44 PM, Joe G said…
And the only Aleph Zeta set is the set of everything. All other sets are proper subsets of that.
At 2:07 PM, socle said…
And the only Aleph Zeta set is the set of everything. All other sets are proper subsets of that.
Is this Aleph Zeta set a member of the set of everything then?
At 2:31 PM, Joe G said…
The Aleph Zeta set is the set of everything.
However, I have been told, albeit without suppport, that infinity makes things different somehow. So perhaps somewhere along the infinite journey the Alpha Zeta set will become a member of itself.
As soon as I have some subordinates, I will have them look into that.
At 3:05 PM, socle said…
The Aleph Zeta set is the set of everything.
But as you know, if A is a set and P(A) is its power set, then P(A) > A. This holds for all finite subsets of N, therefore it must hold for all sets, as your inductive principle states.
So P(Aleph Zeta) > Aleph Zeta, which contradicts the assumption that Aleph Zeta is the set of maximal cardinality.
At 3:17 PM, Joe G said…
The Aleph Zeta set is the set of everything.
But as you know, if A is a set and P(A) is its power set, then P(A) > A.
A is a proper subset of the Aleph Zeta set.
This holds for all finite subsets of N, therefore it must hold for all sets, as your inductive principle states.
N is also a proper subset of the Aleph Zeta set.
Very good socle.
So P(Aleph Zeta) > Aleph Zeta, which contradicts the assumption that Aleph Zeta is the set of maximal cardinality.
The Aleph Zeta set contains its power set, as the power set would also be part of everything.
At 5:08 PM, socle said…
The Aleph Zeta set contains its power set, as the power set would also be part of everything.
Is Aleph Zeta an infinite set?
At 9:55 PM, Joe G said…
Everything infinite is in it.
At 10:12 AM, socle said…
Everything infinite is in it.
But is it infinite itself?
Answer: Yes. AZ is equivalent to a proper subset of P(AZ). You are saying that P(AZ) is a proper subset of AZ. Therefore AZ is equivalent to a proper subset of itself, which means AZ is Dedekindinfinite. It follows that AZ is infinite in the normal sense.
At 1:16 PM, Joe G said…
Infinite in the normal sense as in it will cease to exist when agencies capable of thinking about it cease to exist.
The sorta, kinda finite kind of infinite.
Got it.
At 2:45 PM, socle said…
Infinite in the normal sense as in it will cease to exist when agencies capable of thinking about it cease to exist.
Is that a problem? As I mentioned before, if this is true, then CSI, the Explanatory Filter, and minus log base 2 will also cease to exist at the same time.
That doesn't stop Dembski, Corny Hunter, and the other leading ID thinkers from busting their asses publishing papers in peerreviewed journals. Maybe you should sack up and follow their example.
At 4:47 PM, Joe G said…
As I mentioned before, if this is true, then CSI, the Explanatory Filter, and minus log base 2 will also cease to exist at the same time.
And as I mentioned before, so what?
Do you have a point? Or just an infinite amount of pointless points?
At 5:40 PM, socle said…
So if you have a problem with "infinity" for this reason, you also have a problem with CSI. You should tell KF that FSCIO/I will cease to exist when there are no longer agencies capable of thinking about it, therefore it's a bogus concept.
At 5:56 PM, Joe G said…
LoL! The design inference actually matters as it is all about the real world.
Infinity may be a useful concept, but it still only exists in our minds.
At 7:03 PM, socle said…
LoL! The design inference actually matters as it is all about the real world.
And the role CSI plays in design inferences is roughly zero. Except for caek and baseballs.
At 8:20 PM, Joe G said…
And the role CSI plays in design inferences is roughly zero.
Sez the ignorant sock puppet...
At 1:59 AM, Unknown said…
" 'And the role CSI plays in design inferences is roughly zero. '
Sez the ignorant sock puppetâ€¦"
Perhaps you'd like to show us how it is used then. A calculation of Dr Dembski's metric would be good.
At 9:12 AM, Joe G said…
Been there, done that. Again your ignorance means nothing to me.
OTOH you cannot demonstrate that natural selection can do anything.
Post a Comment
<< Home