Is infinity all black and white?

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 Joined: Sun Aug 31, 2014 7:39 am
Is infinity all black and white?
Are all levels of infinity natural numbers plus the null set or is there something in between?
This is one of the most vexing questions in math that haunted Cantor and other mathematicians. All I can say for now is that math is as consistent with the idea as well as not being with the idea. Here is a Wiki article with more details:
http://en.m.wikipedia.org/wiki/Continuum_hypothesis
PhilX
This is one of the most vexing questions in math that haunted Cantor and other mathematicians. All I can say for now is that math is as consistent with the idea as well as not being with the idea. Here is a Wiki article with more details:
http://en.m.wikipedia.org/wiki/Continuum_hypothesis
PhilX
Re: Is infinity all black and white?
Here is a naive question  If the continuum hypothesis is independent of set theory (can't be proven or disproven), then how could it ever be proven or disproven? Isn't number theory based upon set theory?
It is like the parallel postulate. Euclid's parallel postulate is independent of the other axioms. So, neither it nor the noneuclidean versions can be proven or disproven.
It is like the parallel postulate. Euclid's parallel postulate is independent of the other axioms. So, neither it nor the noneuclidean versions can be proven or disproven.

 Posts: 5621
 Joined: Sun Aug 31, 2014 7:39 am
Re: Is infinity all black and white?
I guess Godel is having the last laugh at us.Wyman wrote:Here is a naive question  If the continuum hypothesis is independent of set theory (can't be proven or disproven), then how could it ever be proven or disproven? Isn't number theory based upon set theory?
It is like the parallel postulate. Euclid's parallel postulate is independent of the other axioms. So, neither it nor the noneuclidean versions can be proven or disproven.
PhilX

 Posts: 5621
 Joined: Sun Aug 31, 2014 7:39 am
Re: Is infinity all black and white?
Technically speaking, no one called me out for indirectly referring to Godel's incompleteness theorems in connection with the continuum hypothesis (which isn't a theorem).
Still, however, they're statements about math theorems and a hypothesis that math can be as consistent without as well as with those statements. What irks me about Godel's proof as that it doesn't specify which theorems it applies to (at one time, it was speculated that Fermat's Last Theorem or FLT was such a theorem). Based on Godel, if one asks whether everything can be known, with respect to math the answer is NO which is my philosophy.
Here's an article from Wiki:
http://en.m.wikipedia.org/wiki/G%C3%B6d ... s_theorems
PhilX
Still, however, they're statements about math theorems and a hypothesis that math can be as consistent without as well as with those statements. What irks me about Godel's proof as that it doesn't specify which theorems it applies to (at one time, it was speculated that Fermat's Last Theorem or FLT was such a theorem). Based on Godel, if one asks whether everything can be known, with respect to math the answer is NO which is my philosophy.
Here's an article from Wiki:
http://en.m.wikipedia.org/wiki/G%C3%B6d ... s_theorems
PhilX
Re: Is infinity all black and white?
I thought you were referring to Godel proving that the continuum hypothesis couldn't be proven in set theory  half of the proof of independence, as Cohen proved that it couldn't be disproven (I may have those two switched).Philosophy Explorer wrote:Technically speaking, no one called me out for indirectly referring to Godel's incompleteness theorems in connection with the continuum hypothesis (which isn't a theorem).
Still, however, they're statements about math theorems and a hypothesis that math can be as consistent without as well as with those statements. What irks me about Godel's proof as that it doesn't specify which theorems it applies to (at one time, it was speculated that Fermat's Last Theorem or FLT was such a theorem). Based on Godel, if one asks whether everything can be known, with respect to math the answer is NO which is my philosophy.
Here's an article from Wiki:
http://en.m.wikipedia.org/wiki/G%C3%B6d ... s_theorems
PhilX
I always thought the type of statement that is undecidable according to the incompleteness theorems was something like 'this theorem cannot be proven in this axiom system'  kind of like the liar paradox. But I am definitely not an expert on the matter.
Re: Is infinity all black and white?
@PhilX, I resurrected this monthsold thread because you have a lot of similar questions about transfinite cardinals in this and other threads and I wanted to unpack some of these issues if you're still interested.
The question of whether every infinite set is cardinally equivalent to an Aleph is actually the Axiom of Choice (AC), not the Continuum Hypothesis (CH).
First, two quibbles.
(Quibble 1) "The null set" is sometimes used as another name for the empty set. And in measure theory, a null set is any set of measure zero. I believe by "the null set" you intend to mean Aleph_0. Your usage is misleading. In no way is Aleph_0 "the" null set nor is it "a" null set.
(Quibble 2) It's also not true that all the Alephs are indexed by natural numbers. In fact they're indexed by ordinals. So after Aleph_0, Aleph_1, Aleph_2, ... we have Aleph_w, where w (lowercase omega in the literature) is the ordinal corresponding to the cardinal Aleph_0. The Alephs keep going forever, and the entire collecton of Alephs is a proper class.
Back to your question.
Every Aleph is an ordinal, by definition. An Aleph is the least ordinal that's cardinally equivalent to a given set. In other words if we have a set X, then if X is cardinally equivalent to any ordinal at all, there must be some least ordinal with that property (since any nonempty set of ordinals is wellordered). We name that least ordinal the cardinal of that set. Since the ordinals are wellordered, so are the cardinals: Aleph_0, Aleph_1, etc. There is no cardinal strictly between two consecutive Alephs, by the very definition of the Alephs.
Repeating this point, which is the crux of the matter: Every cardinal is an ordinal, and the ordinals are wellordered. So the cardinals are wellordered. There's a first one, a second one, and so forth. Nothing in between.
If you accept AC, then any set whatsoever may be wellordered. Therefore every infinite set is cardinally equivalent to some Aleph.
If you don't accept AC, then there is some set that can't be wellordered, hence it is not cardinally equivalent to any ordinal, hence it's not an Aleph.
In other words if you deny AC, then there is some infinite set that can't possibly have any cardinality assigned to it at all! It's not less than some cardinal, it's not greater than some cardinal. It's just out there on the side somewhere with no sensible way of assigning it a size. People are always complaining that AC leads to counterintuitive results (BanachTarski, the Hat paradox, etc); but so does the negation of AC!
That's the short answer. To provide a longer explanation I'd need to write up a tutorial on ordinals, cardinals, and Alephs, I'd better ask first if this is still of interest to you or if you've sorted it all out since you wrote your post a few months ago.
Here are some Wiki links of interest. I'd suggest reading them in this order.
https://en.wikipedia.org/wiki/Wellorder
https://en.wikipedia.org/wiki/Ordinal_number
https://en.wikipedia.org/wiki/Aleph_number
https://en.wikipedia.org/wiki/Cardinali ... _continuum
I also wanted to mention that you are misunderstanding CH. CH says that between Aleph_0 and 2^Aleph_0 there are no other Alephs. In other words CH says that 2^Aleph_0 = Aleph_1.
It's easy to show that the cardinality of the real numbers is 2^Aleph_0. [Ask me for the proof if you're curious.] But the question is, which Aleph is this? It could be Aleph_1 or it could be Aleph_47 or for all we know it could be Aleph_(Aleph_47).
Note that your belief (expressed in some of your other posts) that the reals have cardinality Aleph_1 is in fact a restatement of CH. It's an open question, if one is a Platonist. It's a meaningless question, if one is a formalist. Paul Cohen, who proved that AC and CH are independent of ZF, was on record as thinking that the cardinality of the reals must be far larger than Aleph_1, because taking power sets is such a powerful operation.
https://en.wikipedia.org/wiki/Paul_Cohe ... ematician)
The question of whether every infinite set is cardinally equivalent to an Aleph is actually the Axiom of Choice (AC), not the Continuum Hypothesis (CH).
First, two quibbles.
(Quibble 1) "The null set" is sometimes used as another name for the empty set. And in measure theory, a null set is any set of measure zero. I believe by "the null set" you intend to mean Aleph_0. Your usage is misleading. In no way is Aleph_0 "the" null set nor is it "a" null set.
(Quibble 2) It's also not true that all the Alephs are indexed by natural numbers. In fact they're indexed by ordinals. So after Aleph_0, Aleph_1, Aleph_2, ... we have Aleph_w, where w (lowercase omega in the literature) is the ordinal corresponding to the cardinal Aleph_0. The Alephs keep going forever, and the entire collecton of Alephs is a proper class.
Back to your question.
Every Aleph is an ordinal, by definition. An Aleph is the least ordinal that's cardinally equivalent to a given set. In other words if we have a set X, then if X is cardinally equivalent to any ordinal at all, there must be some least ordinal with that property (since any nonempty set of ordinals is wellordered). We name that least ordinal the cardinal of that set. Since the ordinals are wellordered, so are the cardinals: Aleph_0, Aleph_1, etc. There is no cardinal strictly between two consecutive Alephs, by the very definition of the Alephs.
Repeating this point, which is the crux of the matter: Every cardinal is an ordinal, and the ordinals are wellordered. So the cardinals are wellordered. There's a first one, a second one, and so forth. Nothing in between.
If you accept AC, then any set whatsoever may be wellordered. Therefore every infinite set is cardinally equivalent to some Aleph.
If you don't accept AC, then there is some set that can't be wellordered, hence it is not cardinally equivalent to any ordinal, hence it's not an Aleph.
In other words if you deny AC, then there is some infinite set that can't possibly have any cardinality assigned to it at all! It's not less than some cardinal, it's not greater than some cardinal. It's just out there on the side somewhere with no sensible way of assigning it a size. People are always complaining that AC leads to counterintuitive results (BanachTarski, the Hat paradox, etc); but so does the negation of AC!
That's the short answer. To provide a longer explanation I'd need to write up a tutorial on ordinals, cardinals, and Alephs, I'd better ask first if this is still of interest to you or if you've sorted it all out since you wrote your post a few months ago.
Here are some Wiki links of interest. I'd suggest reading them in this order.
https://en.wikipedia.org/wiki/Wellorder
https://en.wikipedia.org/wiki/Ordinal_number
https://en.wikipedia.org/wiki/Aleph_number
https://en.wikipedia.org/wiki/Cardinali ... _continuum
I also wanted to mention that you are misunderstanding CH. CH says that between Aleph_0 and 2^Aleph_0 there are no other Alephs. In other words CH says that 2^Aleph_0 = Aleph_1.
It's easy to show that the cardinality of the real numbers is 2^Aleph_0. [Ask me for the proof if you're curious.] But the question is, which Aleph is this? It could be Aleph_1 or it could be Aleph_47 or for all we know it could be Aleph_(Aleph_47).
Note that your belief (expressed in some of your other posts) that the reals have cardinality Aleph_1 is in fact a restatement of CH. It's an open question, if one is a Platonist. It's a meaningless question, if one is a formalist. Paul Cohen, who proved that AC and CH are independent of ZF, was on record as thinking that the cardinality of the reals must be far larger than Aleph_1, because taking power sets is such a powerful operation.
https://en.wikipedia.org/wiki/Paul_Cohe ... ematician)
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