-6
$\begingroup$

What can 2-adic arithmetic tell us about the Continuum Hypothesis?

More precisely, what happens to the continuum hypothesis if we define an alternative metric on the size of sets saying that two sets are close in size if their sizes differ by a large power of 2 and ask the same question, i.e. "is there some intermediate size between the infinite set and its power set?", based on this metric?

If we think of induction of the superset relation as analogous to counting in integers, then we might consider the infinite set as the limit point of this process in the same way infinity is the limit of the integers.

What set theory appears to lack currently, if we wish to continue the analogy, is some metric by which the difference in size between some set and the infinite set can be measured and thereby some set-theoretic equivalent of Cauchy sequences defined.

If we measure any finite set $X$ and measure the difference between the cardinality of its power set and itself we have $2^{\lvert X\rvert}-\lvert X\rvert$

Obviously $\lim_{n\to\infty}2^n-n$ diverges but what about $\lim_{n\to\infty}\lvert2^n-n\rvert_2$? if this tended to zero then we would have an alternative metric of set size which has many desirable properties, and in which the continuum hypothesis is true.

According to my limited capacity to complete the analogy, should we use this analogous metric for sizes of sets, it would appear to value the size of the power set of the integers as $0$ and therefore in this theory the space between $2^{\aleph_0}$ and ${\aleph_0}$ can accommodate a set the exact size of $\aleph_0$.

$\endgroup$
  • 1
    $\begingroup$ Even if that limit exists why would you think that you had any kind of continuity here? Knowing about $\lim_{n \to \infty} | 2^n -n |$ for any metric $| . |$ tells you nothing about $2^{\aleph_0} - \aleph_0$ (which is easily proved to be equal to $2^{\aleph_0}$) unless you have a certain continuity that allows you to take a limit and still maintain reasonable properties - like cardinality. $\endgroup$ – Stefan Mesken Jul 30 '17 at 14:00
  • $\begingroup$ @StefanMesken I've improved(?) the question perhaps this explains my thinking a bit better. Fundamentally I think I'm probing at the sameness and difference of the two concepts to understand better how they relate to each other. $\endgroup$ – samerivertwice Jul 31 '17 at 21:39
9
$\begingroup$

It tells us absolutely nothing. CH is about cardinal arithmetic. No field arithmetic is compatible with cardinal arithmetic, so there is nothing it can say about it.

Might as well endow the natural numbers with another operation denoted by $2^n$ defined as $2^n=n$ for all $n$, and argue that Cantor's theorem is false because in this structure $2^n=n$.


The point is that the analogy about "metric on sets" is wrong, so there is no point in continuing it any further, which would make it "even more wrong". There is no "metric on sets", and certainly not one which is defined by size.

Set theory, as a theory (from a holistic point of view), might be missing some things, but it most certainly does not miss a metric based on size differences and powers of $2$. That is a myopic view that misses the entire big picture, and most of the small details.

So by hamfisting this analogy in the first place, you get a weird notion that there should be a notion of Cauchy sequences on sets in this metric. Let me postulate, then, the obvious question. Why should this be a metric, and not a valuation which considers even longer sequences? Why not sequences of cofinality $\omega_1$? Why not all type of sequences?

Metric talks about "distance", and so what does it even mean that two sets are close or distant? Are sets of cardinality $500$ and $510$ close? What is the distance between them? Not to mention that since cardinality subtraction is very much undefined, this leads to ultimately odd questions. What is the distance between $\aleph_1$ and $\aleph_0$? What is the distance between $\aleph_{42}$ and $42$? Why don't you just end up with a discrete metric?

The analogy falls apart once you try to ask for details which are not cherry picked for this $2$-adic approach.

$\endgroup$
  • $\begingroup$ Surely the distance $\lvert2^n-n\rvert_2$ being nonzero for every integer makes it not much like that at all? I'm sure some would have said to define $\lvert2^{\infty}\rvert_2=0$ is pointless but Hensel had a valid motivation behind it which turned out to be very powerful. I wondered if there can be some similar motivation for an alternative measure to cardinality. "No field arithmetic is compatible with cardinal arithmetic" makes it sound like you're very much thinking within the bounds of what's been done before, which isn't really in the spirit I intended. $\endgroup$ – samerivertwice Jul 30 '17 at 17:17
  • 2
    $\begingroup$ Surely? Why? This is absolutely unrelated. Even worse, THERE IS NO SUCH THING AS CARDINAL SUBTRACTION. So even writing $2^n-n$ about cardinals is meaningless. $\endgroup$ – Asaf Karagila Jul 30 '17 at 17:19
  • $\begingroup$ If that's unrelated then I misunderstand what you mean by "you might as well endow the natural numbers with another operation defined as $2^n=n$... $\endgroup$ – samerivertwice Jul 30 '17 at 17:25
  • $\begingroup$ Take the natural numbers in the signature $\langle 0,S,+,\cdot,\leq\rangle$ with the usual arithmetic structure. Now add to the language a new operator, denoted by $2^{\bullet}$ and the axiom $\forall n(2^n=n)$. In this extended structure $2^n=n$, despite $2^n$ has little to no relation to the usual exponentiation. Nevertheless, we have a structure where $2^n=n$. Does this refute Cantor's theorem? $\endgroup$ – Asaf Karagila Jul 30 '17 at 17:27
  • 2
    $\begingroup$ Yes, and when I first learned English, I thought about it in terms of Hebrew grammar. This is why it took me a while to get the hang of perfect tenses: I had to stop thinking about English in Hebrew, and start thinking about English in English. Stop thinking about set theory in terms of homotopy and analysis, and think about it in terms of set theory. These are two very different things, with very different ideas underlying to them. $\endgroup$ – Asaf Karagila Aug 1 '17 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.