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Im a beginner in (infinite) set theory, So keep that in mind. I know cantor's diagonal, cantors pairing function , the ZFC axioms and basic stuff like that.

Im very intrested in infinite sets and in particular cardinality and the continuüm hypothesis.

Now I read an article ( see link ) where They mention that - If I understand correctly - that two types of cardinality are equal : card(p) = card(t).

Im not sure If $p$ and $t$ are the standard notation for these sets / Cardinals / ordinals.

But the article says who worked on it , So I hope you guys can explain $p$ and $t$ to a beginner like me.

It is Said - I think - that $p$ and $t$ are candidates to break CH.

In case the article goes away here is a quote :


INFINITY

Mathematicians Measure Infinities and Find They’re Equal By KEVIN HARTNETT September 12, 2017

Two mathematicians have proved that two different infinities are equal in size, settling a long-standing question. Their proof rests on a surprising link between the sizes of infinities and the complexity of mathematical theories. 70

Colors Collective for Quanta Magazine In a breakthrough that disproves decades of conventional wisdom, two mathematicians have shown that two different variants of infinity are actually the same size. The advance touches on one of the most famous and intractable problems in mathematics: whether there exist infinities between the infinite size of the natural numbers and the larger infinite size of the real numbers.

The problem was first identified over a century ago. At the time, mathematicians knew that “the real numbers are bigger than the natural numbers, but not how much bigger. Is it the next biggest size, or is there a size in between?” said Maryanthe Malliaris of the University of Chicago, co-author of the new work along with Saharon Shelah of the Hebrew University of Jerusalem and Rutgers University.

In their new work, Malliaris and Shelah resolve a related 70-year-old question about whether one infinity (call it p) is smaller than another infinity (call it t). They proved the two are in fact equal, much to the surprise of mathematicians.

“It was certainly my opinion, and the general opinion, that p should be less than t,” Shelah said.


Here is the article

https://www.quantamagazine.org/mathematicians-measure-infinities-find-theyre-equal-20170912/

So What is this all about ?

I looked on arxiv but I was not sure If there was a connection.

Also quantamagazine says it is a big result , but I have not heard anyone mention it ??

Now that $p = t$ I wonder If They are both countable or both uncountable ... or still candidates to break CH ??

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marked as duplicate by Mauro ALLEGRANZA, Mark S., Parcly Taxel, Matthew Towers, Noah Schweber set-theory Sep 20 '17 at 18:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I have no idea what the specific cardinalities are, but I am fairly certain you should disregard most of what is written in that article. Especially the connection to CH seems like it is unlikely to be relevant, seeing as CH is known to be independent of ZFC, so certainly these cardinalities will not be able to "break" CH. $\endgroup$ – Tobias Kildetoft Sep 18 '17 at 8:59
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    $\begingroup$ This is answered in the answers to Is there a bijection.... $\endgroup$ – Mark S. Sep 18 '17 at 9:04
  • $\begingroup$ The connection with CH is this - $\mathfrak{p}$ and $\mathfrak{t}$ are both examples of cardinal characteristics of the continuum (see my answer to the linked question above), which in particular are (i) necessarily uncountable and of size at most continuum, but (ii) consistently strictly less than the continuum. Saying that they are "candidates to break CH" is missing the point ... what's true is that they are amongst the "naturally definable" infinite cardinalities which could be strictly between $\aleph_0$ and $2^{\aleph_0}$. But this really isn't rare, e.g. $\aleph_1$ is such a cardinal! $\endgroup$ – Noah Schweber Sep 20 '17 at 18:21
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    $\begingroup$ @RobertFrost CCCs can be counterexamples to CH - it's just that they aren't provably strictly less than the continuum, only consistently less than the continuum. Similarly, it is consistent that $\omega_1$ is a counterexample to CH, it's just not provable. $\endgroup$ – Noah Schweber Nov 12 '17 at 20:38
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    $\begingroup$ Note that it's even possible for there to be a definable set of reals of size in between $\aleph_0$ and continuum - for instance, it's consistent that the constructible reals form such a set. We just can't prove that they do, so there's no problem. $\endgroup$ – Noah Schweber Nov 12 '17 at 20:42