Wikipedia:Reference desk/Archives/Mathematics/2017 July 7

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July 7

Implications of the Goldbach conjecture

Are there any useful implications of the Goldbach conjecture being (a) true or (b) false? Loraof (talk) 22:33, 7 July 2017 (UTC)

It implies Goldbach's weak conjecture and probably others. Bubba73 ^{You talkin' to me?} 00:21, 11 July 2017 (UTC)

Considering that Goldbach's weak conjecture has already been proven, though, it may not count as a useful implication. Double sharp (talk) 02:15, 11 July 2017 (UTC)

The weak version has been proven to hold for numbers > $e^{3100}\approx 2\times 10^{1346}$.Bubba73 ^{You talkin' to me?} 02:33, 11 July 2017 (UTC)

No, read the article carefully: further down it says "In 2012 and 2013, Peruvian mathematician Harald Helfgott released a pair of papers improving major and minor arc estimates sufficiently to unconditionally prove the weak Goldbach conjecture", citing those papers mentioned. The lede is a bit poorly written: the bound you state was the best known prior to his work. I'll edit it for clarity. Double sharp (talk) 02:49, 11 July 2017 (UTC)

Thanks, I didn't know that (or read far enough). Bubba73 ^{You talkin' to me?} 03:47, 11 July 2017 (UTC)

You're welcome! Double sharp (talk) 04:22, 11 July 2017 (UTC)

Implications of the twin prime conjecture

Are there any useful implications of the twin prime conjecture being (a) true or (b) false (along the lines of "if it's true/false, then this other conjecture is true/false")? Loraof (talk) 22:37, 7 July 2017 (UTC)

ℵ₁

In the course I took in which it was covered, $\aleph _{1}$ was explained as the next larger infinite cardinal after $\aleph _{0}$. The very notion of "the next larger infinite cardinal" presupposes that there's a smallest one among all the infinite cardinals larger than the given one.

1. Is it generally true that for any infinite cardinal, there's always a well-defined next larger cardinal? If so, is there a simple proof for that?
2. If the above is not true, how do you prove that there's a next larger cardinal in the case of $\aleph _{0}$?

--134.242.92.97 (talk) 23:01, 7 July 2017 (UTC)

Given the axiom of choice, every set can be wellordered, which gives a bijection between that set and some (von Neumann) ordinal number. Since there is some ordinal number in bijection with the set, there is a least such ordinal. That is called the "initial ordinal" of that cardinality, and is usually identified with the cardinal number. The cardinality of the initial ordinal is the initial ordinal itself.

For any cardinal, there is a larger cardinal (for example, the cardinality of the powerset of the first cardinal). That cardinal is identified with (equal to) some ordinal. Therefore there is a least ordinal that has larger cardinality than the given cardinal. --Trovatore (talk) 23:17, 7 July 2017 (UTC)

 

By the way, if for some reason you don't want to assume the axiom of choice, you can still define ℵ₁. In that case, you define it as the cardinality of the set of all countable ordinals (which equals the smallest uncountable ordinal, as an initial ordinal). In that case, it may no longer be the smallest cardinal larger than ℵ₀, but it's still a minimal cardinal larger than ℵ₀. --Trovatore (talk) 23:29, 7 July 2017 (UTC)

Thanks. I noticed that the article on Aleph number has the statement that "$\aleph _{1}$ is the cardinality of the set of all countable ordinal numbers", just like in the alternative treatment you suggested. How do we know this is just an alternative characterization of the same thing as "the next larger cardinal after $\aleph _{0}$", and not something really different? --134.242.92.97 (talk) 00:09, 8 July 2017 (UTC)

The set of countable ordinals cannot be countable, because if it were, you could take the union of all countable ordinals, and that would still be countable (the union of countably many countable sets is countable). But then it would itself be an ordinal, and countable. Then that ordinal would be a countable ordinal, and therefore an element of itself, contradicting the axiom of foundation (or the definition of von Neumann ordinal if you prefer not to use foundation).

So it's not hard to see that if you take the union of all countable ordinals, you get the first uncountable ordinal, which is an initial ordinal and thus a cardinal. Any ordinal less than that one is countable, and therefore there is no smaller uncountable cardinal. --Trovatore (talk) 00:41, 8 July 2017 (UTC)

My apologies, I made that way more complicated than it has to be, and also I made it look like it depends on the union of countably many countable sets being countable (which in turn depends on a weak fragment of the axiom of choice).

Let's start over. The set of all countable ordinals is wellordered by set membership, because that's true of all sets of ordinals. It's also a transitive set, because any ordinal that's an element of a countable ordinal α is smaller than α, therefore also a countable ordinal, and therefore in the set of all countable ordinals.

So the set of all countable ordinals is an ordinal. It can't be a countable ordinal, because then it would be an element of itself, violating foundation (or the fact that it's wellordered by set membership). So it's an uncountable ordinal. It's the smallest uncountable ordinal, because any smaller ordinal is an element of it, and therefore a countable ordinal.

Therefore it's an initial ordinal, so it's a cardinal. Any smaller cardinal would be an initial ordinal in the set of all countable ordinals, thus countable. So it's the smallest uncountable cardinal, QED. --Trovatore (talk) 08:50, 8 July 2017 (UTC)

Question 1 is asking about is the continuum hypothesis, which has been proved undecidable within Zermelo-Frankel set theory, which is to say, the answer to 1 is no. Either the cardinality of the real numbers is $\aleph _{1}$, the next larger cardinal above $\aleph _{0}$, or it isn't, and that question is undecidable. I would think this means that the answer to question 2 is "you can't", but this is beyond my knowledge. --76.71.5.114 (talk) —Preceding undated comment added 00:12, 8 July 2017 (UTC)

No, neither question has anything directly to do with the continuum hypothesis. The answer to question 1 is yes (assuming the axiom of choice). --Trovatore (talk) 00:26, 8 July 2017 (UTC)

For question 2, Cantor's diagonal argument proves that there is one larger than $\aleph _{0}$. Bubba73 ^{You talkin' to me?} 21:54, 9 July 2017 (UTC)

whoops, but that doesn't necessarily mean that there is a well-defined next one. Bubba73 ^{You talkin' to me?} 21:57, 9 July 2017 (UTC)