B A question about cardinalities

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The discussion explores the concept of infinity, questioning whether there are infinite infinities due to the existence of infinite whole numbers and decimals. It explains that the cardinality of whole numbers is aleph null, while the cardinality of non-terminating decimals is the cardinality of the continuum. The conversation also touches on the power set of an infinite set, which has a strictly greater cardinality than the set itself. Additionally, it raises concerns about the coherence of the set of all cardinalities, referencing Russell's paradox and the distinction between bounded and unbounded comprehension. Ultimately, the discussion concludes that unbounded comprehension leads to contradictions and is therefore rejected.
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TL;DR Summary
a question about infinity
If there are an infinite number of whole numbers, and an infinite number of decimals between any two whole numbers, and an infinite number of decimals in between any two decimals, does that mean that there are infinite infinities? And an infinite number of those infinities? And an infinite number of those infinities? And an infinite number of those infinities? And an infinite number of those infinities? And… (Infinitely times. And that infinitely times. and that infinitely times. and that infinitely times. And..)...

 
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phil335 said:
TL;DR Summary: a question about infinity

If there are an infinite number of whole numbers, and an infinite number of decimals between any two whole numbers, and an infinite number of decimals in between any two decimals, does that mean that there are infinite infinities? And an infinite number of those infinities? And an infinite number of those infinities? And an infinite number of those infinities? And an infinite number of those infinities? And… (Infinitely times. And that infinitely times. and that infinitely times. and that infinitely times. And..)...

Yes.

If we have a set ##X## with infinitely many elements ##|X|## then its power set ##P(X)##, i.e. the set of all subsets of ##X## has ##|P(x)|=2^{|X|}## many elements which is strictly bigger than ##|X|## because ##P(X)## contains all subsets ##\{x\}\in P(X)## for ##x\in X## and many more sets. There is no one-on-one map between ##X## and ##P(X).##

It is not clear whether there is another proper infinity class ##c(X)## between them, i.e. whether
$$
|X| \leq c(X) \leq |P(X)|=2^{|X|}
$$
implies ##|X|=c(X)## or ##c(X)=2^{|X|}## or whether this is not the case.
 
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Look up "aleph null" and go from there.
 
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phil335 said:
TL;DR Summary: a question about infinity

If there are an infinite number of whole numbers,
The cardinality of the whole numbers is normally taken as aleph null.
phil335 said:
and an infinite number of decimals between any two whole numbers
The cardinality of the set of terminating decimals between any two whole numbers is also aleph null.
The cardinality of the set of not necessarily terminating decimals between any two whole numbers is the cardinality of the continuum -- the cardinality of the power set of the naturals.

phil335 said:
and an infinite number of decimals in between any two decimals
Again, the cardinality of the set of terminating decimals between any two distinct decimals is aleph null while the cardinality of the set of non-terminating decimals between them is the cardinality of the continuum.
phil335 said:
does that mean that there are infinite infinities?
There are at least aleph null infinities. Because for each infinite cardinality there is an obvious successor.

However, I am not entirely sure that the cardinality of the set of all cardinalities is a coherent notion.
 
How many elements are there in the set of all elements that are not part of a set?
 
DaveC426913 said:
How many elements are there in the set of all elements that are not part of a set?
Have you been exposed to Russell's paradox and the distinction between bounded and unbounded comprehension?

Bounded comprehension or the Axiom Schema of specification says that if you have a set and a predicate (a yes/no function that operates on set members) then a set that contains exactly those set members that satisfy the predicate exists.

Unbounded comprehension is pretty much the same thing. But it does not require a set as a starting point. Just the predicate. It would assert that if you have a predicate then the set of all elements that satisfy the predicate exists. Naively, this seems sensible enough. Generations of mathematicians were perfectly willing to accept this idea.

The problem with unbounded comprehension is Russell's paradox. Let the predicate be "does not contain itself as a member" and you have the set of all sets that do not contain themselves. To avoid the paradox, one approach is to accept only bounded comprehension.

Your question above uses unbounded comprehension. So we reject it.
 
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