B Question about CH (continuum hypothesis)

jk22

Is it possible to calculate this :

Suppose the iterative root of $2^x$ :

$\phi(\phi(x))=2^x$ (I suppose the Kneser calculation should work, it affirms that there is a real analytic solution)

Then how to compute $\phi(\aleph_0)$ ? (We know that $2^{\aleph_0}=\aleph_1$).

Could this be $\aleph_{1/2}$ ?

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SSequence

I don't know what your wrote, but (assuming standard set-theory) things change a lot depend on whether one assumes AC or not.

For example, under ZFC there would be no such thing as $\aleph_{1/2}$. There would only be $\aleph_{\alpha}$ (where $\alpha \in Ord$). $\aleph_{\alpha}$ basically means (as I understand in layman terms) that:
When you say a set $A$ has cardinality $\aleph_{\alpha}$, then when you well-order it there exists no well-ordering of $A$ which has length (order-type) less than $\omega_{\alpha}$ (the $\alpha$-th uncountable).

Of course, there is also no well-ordering of $A$ which has length greater than $\omega_{\alpha+1}$ (if that were true the cardinality of $A$ would no longer remain $\aleph_{\alpha}$).

What I said before is something you would find in most elementary books (I myself only studied half-way through an elementary book many years ago and haven't been able to return to study further).

=========

Under ZF things are substantially more complicated though (because well-order for a set can't be guaranteed), even more so than under ZFC (as if it wasn't enough!). I don't have any idea about that. Though you might try searching for something like "cardinalities without choice" to get a rough feel for things.

mathman

You are asking about a function $\phi$ which applies to numbers, but is not defined. Your question is whether it is meaningful with cardinal number as an argument. With no further explanation, it is very hard to answer.

stevendaryl

Staff Emeritus
If the continuum hypothesis is true, then there is no cardinal between $\aleph_0$ and $2^{\aleph_0}$. So there would be no way to go "halfway" from one to the other.

SSequence

I think your point of comparison between $\aleph_0$ and $\aleph_1$ is correct. We could probably(?) safely say (in case of CH being true) that there is no cardinality $N$ such that:
$\aleph_0 < N < \aleph_1$

However, in an a priori sense, without assuming a guaranteed well-order for sets, it seems there should be no guarantee that the situation "after this" wouldn't be much more complicated. No?

P.S. [On that note I am not fully sure what the appropriate ordering for cardinals should be in ZF, but that's also probably quite elementary.
For example, in ZFC the cardinals are be well-ordered (so that seems to be the right way to view them in that context). So whenever we write something like "card(A)<card(B)" we know that the "comparison relation" is the one based upon well-order]

mathman

By definition $\aleph_1$ is the second cardinal number. The contuum hypotheses states $\aleph_1=2^{\aleph_0}$.

SSequence

Yes, I understand that:
There is no cardinality $N$ such that $N>\aleph_0$ and $N<2^{\aleph_0}$ (CH)
$\aleph_0<2^{\aleph_0}$ (Cantor theorem)
$card(\omega)={\aleph_0}$
$card(\omega_1)={\aleph_1}$

I was just pointing out that things are complicated in ZF and there are number of peculiarities associated with it (that one should be wary of at least). For example, if I am not mistaken, it is possible in ZF for $\omega_1$ to have countable co-finality. There seem to be a number of things that aren't intuitively obvious. [EDIT:] To back-up my point a bit more, here is a thread I found from a very simple search (https://math.stackexchange.com/questions/404807). Honestly though, it is a bit too difficult/head-spinning for me, but it does illustrate my point. [END]

Personally though, somewhat honestly, learning in detail about the kind of things I wrote in above paragraph might be significantly lower in my list compared to number of other things. A partial reason is simply that even trying to learn (imperfectly) the kind of world(s) ZFC describes is extremely arduous (at least for me) as it is.

P.S.
Just to clarify a bit it wasn't fully clear to me what the context of post#4 was. If ZFC is assumed then it is obviously trivial that CH implies $\aleph_1=2^{\aleph_0}$.
That's because if we assumed $\aleph_\alpha=2^{\aleph_0}$ (where $\alpha>1$), we could find a cardinality in-between $\aleph_0$ and $\aleph_\alpha$ (for example: $\aleph_1$) contradicting the assumption of CH.

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TeethWhitener

Gold Member
Is it possible to calculate this :

Suppose the iterative root of $2^x$ :

$\phi(\phi(x))=2^x$ (I suppose the Kneser calculation should work, it affirms that there is a real analytic solution)

Then how to compute $\phi(\aleph_0)$ ? (We know that $2^{\aleph_0}=\aleph_1$).

Could this be $\aleph_{1/2}$ ?
There is a lot of notational abuse in this post. The notation $X^Y$ is generally used in set theory as a shorthand for "the set of all functions from $Y$ to $X$." For the specialized case of $2^S$, identifying 2 with the Boolean set $\{0,1\}$, we see that it's a shorthand for "the set of all functions from $S$ to $\{0,1\}$." It is straightforward to show that there is a bijection between this set and the power set of $S$. Thus, the suggestive notation $2^S$, which obeys the law of cardinal exponentiation: $|2^S| = |2|^{|S|}$.

But the iterative root in your post clearly corresponds to a function defined thus: $\phi: S\rightarrow S$, for some $S$, be it natural numbers, reals, etc. Now, functions, themselves being defined as sets (namely, given a function $f:X\rightarrow Y$, $f$ is a set of ordered pairs $(x,y), x\in X, y\in Y$ where each $x\in X$ appears exactly once) do have a cardinality of their own, but that cardinality is equal to the cardinality of the set $X$. So $\phi$, viewed as a set, would have a cardinality equal to whatever set it is operating on; i.e., its domain.

mathman

P.S.
Just to clarify a bit it wasn't fully clear to me what the context of post#4 was. If ZFC is assumed then it is obviously trivial that CH implies $\aleph_1=2^{\aleph_0}$.
That's because if we assumed $\aleph_\alpha=2^{\aleph_0}$ (where $\alpha>1$), we could find a cardinality in-between $\aleph_0$ and $\aleph_\alpha$ (for example: $\aleph_1$) contradicting the assumption of CH.
Statement is confusing. CH statement IS $\aleph_1=2^{\aleph_0}$, not implies that..

TeethWhitener

Gold Member
You are asking about a function ϕϕ\phi which applies to numbers, but is not defined.
Just a small point: the function $\phi$ in the OP is defined. It's the function whose composition with itself yields $2^x$:
\begin{align} \phi \circ \phi : \text{ } \mathbb{R} &\to \mathbb{R} \nonumber \\ x &\mapsto 2^x \nonumber \end{align}
It's sometimes called the "functional square root:"
https://en.wikipedia.org/wiki/Functional_square_root
Whether the function is unique or not is above my pay grade

SSequence

Statement is confusing. CH statement IS $\aleph_1=2^{\aleph_0}$, not implies that..
Yeah the implication is in both directions.

jk22

So maybe it is that there are $\aleph$s that are not cardinalities ?

SSequence

Here is a summary (of the basic scenario) in two or three sentences:
---- In ZFC all cardinalities are of the form $\aleph_{\alpha}$ (where $\alpha \in Ord$). Here $\aleph_{\alpha}$ is cardinality of ordinal denoted as $\omega_{\alpha}$.

---- In ZF one definitely can't assume that all cardinalities are of the form $\aleph_{\alpha}$ (where $\alpha \in Ord$). I think (with high certainty) that, on the very least, it would be consistent that there are no cardinalities of the form as described in previous sentence. One could probably say something stronger than that (but I don't know much, as I mentioned previously).

WWGD

Gold Member
Should help provide context.

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