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I am not a mathematician, but rather a computer engineer with a curious mind.

The continuum hypothesis (CH) has gripped my attention today, and I even asked a question about it earlier today. However, I seem to be unable to grasp the ideas correctly because I've reached a conclusion that I'm sure is wrong but I don't understand why it's wrong.

EDIT: adding the CH for clarification of what I'm trying to prove

Continuum Hypothesis: There does not exist a set $S$ such that $\aleph_0 < |S| < 2^{\aleph_0}$

Something that I learned from my last question: $n^{\aleph_0}$ can be interpreted as the cardinality of the set constructed by enumerating all of the possible sequences of choosing between $n$ choices $\aleph_0$ number of times.

EDIT: It nags me that the continuum theorem is still unproven after over 100 years, so I played with some ideas on paper to hopefully hit a brick wall and understand why the CH is so difficult to prove. I didn't hit a brick wall, but rather I seem to have run off of a cliff and was hoping for some guidance (or maybe a parachute in this metaphor).

I defined a function which describes the cardinality of a set as follows

$J(c+\frac{m}{n}, d \times n) : m < n ; c,d,m,n \in \mathbb{N}$ is the cardinality of the set constructed by choosing between $c$ choices $n-m$ number of times, then between $c+1$ choices $m$ times, repeated $d$ number of times.

Examples:

  • $J(1.5, 4) = J(1 + \frac{1}{2}, 2 \times 2) = (1 \times 2) \times (1 \times 2) = 4$
  • $J(2, 7) = J(2 + \frac{0}{1}, 7 \times 1) = (2) \times (2) \times (2) \times (2) \times (2) \times (2) \times (2) = 128$
  • $J(2.2, 10) = J(2 + \frac{1}{5}, 2 \times 5) = (2 \times 2 \times 2 \times 2 \times 3) \times (2 \times 2 \times 2 \times 2 \times 3) = 2304$
  • $J(3.25, 8) = J(3 + \frac{1}{4}, 2 \times 4) = (3 \times 3 \times 3 \times 4) \times (3 \times 3 \times 3 \times 4) = 11664$

I think some things are clear, but please correct me if one or more of these assertions are wrong.

  • $J(s_0,t) \leq J(s_1,t) \leftrightarrow s_0 \leq s_1$
  • $J(s,t_0) \leq J(s,t_1) \leftrightarrow t_0 \leq t_1$
  • $J(n, m) = n^{m} : n,m \in \mathbb{N}$
  • $J(n, \aleph_0) = n^{\aleph_0}$ by the interpretation of $n^{\aleph_0}$ that I stated earlier in the problem.

I also think that it's clear that $J(1+q, \aleph_0) = 2^{\aleph_0}$ $\forall$ $0 < q < 1, q \in \mathbb{Q}$. I wrote a small proof on my notepad around the idea that finite $q$ will yield a finite $t$ where $(1+q)^t \geq 2$. Correct me if I'm making a mistake here.

Now, for every $\epsilon > 0$, there exists a $q < \epsilon$, $q \in \mathbb{Q}$. Now I plot the function $f(x) = J(x, \aleph_0)$ over the range of say $[1,2]$, $f(x)$ jumps from $1$ to $2^{\aleph_0}$ and right over $\aleph_0$! What's going on here?

More info:

My original try was in a different wording of $J(s,t)$: "... repeated 1 through $d$ number of times. This made $f(x)$ jump from $\aleph_0$ to $2^{\aleph_0}$ between $x=1$ and $x=1 + \epsilon$ for any $\epsilon > 0$. This made me excited until I realized that a different definition of $J(s,t)$ skipped over $\aleph_0$ entirely.

I admit, I was assuming that the CH was true and this basically moved me in the direction of trying to prove that it was true but somewhere I have made an assumption that's false. I was attempting to create a function $J(s,t)$ that resembled the cardinal exponential function $s^t$ in an attempt to get a continuous function over a range to prove that there are no cardinalities between $\aleph_0$ and $2^{\aleph_0}$. This approach seems fundamentally flawed (because one of my definitions "proved" that $\aleph_0$ didn't exist!). Can someone explain the flaw to me?

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Frank Bryce
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    I haven't read it all but the sentence "The CH is still unproven after over 100 years" is simply wrong. It is proven that it is independent from the ZF axioms and if ZF is consistent, then ZF+negation of CH is consistent as well. – Peter Franek Jan 02 '16 at 23:41
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    It's not just that it's "unproven", but that it was proven independent (in first-order logic) of Zermelo-Frankel set theory with or without Axiom of Choice, by Paul Cohen in the 1960s. It had earlier been proven consistent by Kurt Godel in 1938. Further, beginning with Easton's theorem (google-able) it is consistent to assume that $2^{\aleph_0}$ is any of a wide variety of other things than $\aleph_1$. – paul garrett Jan 02 '16 at 23:42
  • hmmm, I didn't understand the distinction. I got the book "The Godelian Puzzle Book" and one of the sentences is "... whether [The Continuum Problem] is true or not remains a mystery. Some, including the author, regard this as the Grand Problem of Mathematics". Page 67, for anyone who has the book. – Frank Bryce Jan 02 '16 at 23:44
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    @JohnCarpenter I haven't read the book but from what you say, it might be that author thought of "mystery" from philosophical/platonical point of view. We don't know which axiomatic system is a "right one". But it is proved that "you cannot prove it from the Peano axioms, neither from ZF axioms, neither from ZFC" which are today the most commonly used deductive systems. – Peter Franek Jan 02 '16 at 23:47
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    In other words: Given the standard axioms for set theory, which suffice to do all the math people typically do (and which would surely suffice for the sort of thing you're doing above, if we could figure out what it all meant), it is impossible to prove CH. And it is impossible to disprove CH. Any proof or disproof is either wrong, or uses some sort of non-standard axioms. – David C. Ullrich Jan 02 '16 at 23:47
  • You need to be a bit careful to make sure your function $J$ is well-defined. Note, for example, that $$J(1.5,4)=J\left(1+\frac24,2\times 2\right)$$ and $$J(2,7)=J\left(2+\frac01,1\times 7\right).$$ – Cameron Buie Jan 02 '16 at 23:48
  • @CameronBuie did you mean in the second line $J(2,7)=J\left(2+\frac07,1\times 7\right).$? – snulty Jan 02 '16 at 23:53
  • @snulty that would work, too. Either way would yield a consistent answer with the definition – Frank Bryce Jan 02 '16 at 23:58
  • @cameronbuie keep in mind that the $n$ is the same $n$ in both cases. I believe that $J(s,t)$ should evaluate to the same thing for any ambiguities in how the fraction can be represented. – Frank Bryce Jan 03 '16 at 00:02
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    Not that I understand exactly how your proof is supposed to go, but that "jump" that surprised you wouldn't surprise me if I did understand. When you're dealing with this sort of thing functions defined by simple formulas simply are discontinuous. Example: Say $f(x)=2^x$ (for cardinals $x$). If $n$ is finite then $f(n)$ is finite, in particular $f(x)<\aleph_0$. But $f(\aleph_0)=2^{\aleph_0}>\aleph_0$. As $n$ tends to infinity the function jumps from values less then $\aleph_0$ to a value strictly larger than $\aleph_0$. What's going on with that jump? The function is simply not continuous. – David C. Ullrich Jan 03 '16 at 00:20
  • @DavidC.Ullrich, I think you have isolated one problem... – paul garrett Jan 03 '16 at 00:23
  • @DavidC.Ullrich Thanks for your answer, this does make sense. Would the same "jump" from 1 to $\aleph_2$ hold true for $J(1+x, \aleph_1)$, assuming that $\aleph_1 = 2^{\aleph_0}$ and $\aleph_2 = 2^{\aleph_1}$ (based on the assumption of the CH). In other words, would this jump over $\aleph_0$ AND $\aleph_1$? This would make sense based on my understanding of your comment. – Frank Bryce Jan 03 '16 at 00:25
  • Also, that book may be a Dover republication of something R. Smullyan wrote prior to P. Cohen's results, since, after all, Smullyan got his PhD in the 1950s. The copyright dates on those Dover reprints are often misleading by half a century or more... – paul garrett Jan 03 '16 at 00:27
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    I have no idea, not having time to figure out what $J$ is, sorry. Was just pointing out that things that sort of look like they should be continuous may simply be discontinuous (so just assuming $J$ is continuous without proof is no good). – David C. Ullrich Jan 03 '16 at 00:34
  • It seems that your definition is $$J\left(c+\frac{m}{n},d\times n\right)=\left[\binom{c}{n-m}\cdot\binom{c+1}{m}\right]^d.$$ Is this correct? If so, then $$J\left(2+\frac07,1\cdot 7\right)=0\neq128=J\left(2+\frac01,7\times1\right).$$ – Cameron Buie Jan 03 '16 at 01:09
  • @snulty: Yes, I did. Thank you! – Cameron Buie Jan 03 '16 at 01:09
  • For another example of a function which might seem like it should be continuous but isn't, consider $f(x)=3^x-2^x$ - or, set-theoretically, $f(x)$ is the least cardinal $\kappa$ such that $2^x+\kappa=3^x$. Then "$\lim_{n\rightarrow\aleph_0}f(x)=\aleph_0$" in the obvious sense, but $f(\aleph_0)=0$ - indeed, $f(\kappa)=0$ for every infinite $\kappa$! – Noah Schweber Jan 03 '16 at 01:44
  • I assume you meant that $$J\left(c+\frac{m}{n},d\times n\right)=\bigl[c\times(n-m)\times(c+1)\times m\bigl]^d.$$ Is that correct? – Cameron Buie Jan 03 '16 at 03:15
  • You might be interested in this: http://math.stackexchange.com/questions/859089/cant-prove-continuum-hypothesis – Asaf Karagila Jan 03 '16 at 11:49
  • If that is, in fact, your definition, then $J(j,k)=0$ for all $j,k\in\Bbb N$ with $k\ne0.$ – Cameron Buie Jan 03 '16 at 15:13
  • Based on the values described, though, you seem to be using the definition $$J\left(c+\frac{m}{n},d\times n\right)=\left[c^{n-m}\times(c+1)^m\right]^d,$$ instead. – Cameron Buie Jan 03 '16 at 15:16
  • @CameronBuie ah, yes, that IS what I'm using. I'm sorry for the invalid comment earlier. – Frank Bryce Jan 04 '16 at 13:39

2 Answers2

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The biggest issue here is the following misconception: $$\huge\textbf{Cardinals are not real numbers!}$$

The fact that for finite cardinals the basic arithmetic coincides with that of the natural numbers and thus with the reals does not grant you the same toolkit with cardinals in general. Even less so when limits are considered.

So either you define a function on the rational numbers which gives back cardinals, which is fine, but then $\aleph_0$ is not a rational number and you cannot put it into the domain of the function, or you defined a function from cardinals to cardinals but then you have no business feeding it rational numbers. There is no set whose cardinality is a fraction.

One of the factors to this confusion is an ill equipped toolkit. The infinity dealt with as far as real analysis goes is absolutely unrelated to cardinal numbers and requires a different kind of machinery to work. Things become even more troublesome if you throw ordinal arithmetic into the pot, which is also very different.

In short, as was mentioned in the comments, there is no reason to expect that cardinal exponentiation is continuous at limits. But the problems begin on a deeper level, as written in large, boldface letters in the beginning of my answer.

Daniel Fischer
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Asaf Karagila
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There are many problems with this kind of reasoning. Honestly, I don't understand the definition of $J$. Even if I understood, it is not clear what is its domain. Even if it is well-defined in some sense, it is not clear why it should be continuous. (Note that already $2^x$, considered as a function from cardinals to cardinals, is discontinuous at $\aleph_0$, as $\lim_{x\to\aleph_0-} 2^x=\aleph_0\neq 2^{\aleph_0}$.)

The lack of formalism is not necessarily a drawback if your "proof" had a clear idea (that either could, or could not, be formalized). But if it has one, you failed to describe that idea and/or its intuition and indstead of it, used a very technical language from the start. And this technical language, unfortunately, doesn't look like the technical language of mathematics.

Peter Franek
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