0
$\begingroup$

In Kolmogorov and Fomin's Introduction to Real Analysis, there are a pair of problems which seem to be asking the reader to prove the Continuum Hypothesis. These are in Section 3, problems 12 and 13, which I shall reproduce below.

Problem 12: Prove that the set $M$ of all ordinals corresponding to a countable set is itself uncountable.

Problem 13: Let $\aleph_1$ be the power (cardinality) of the set $M$ in the preceding problem. Prove that there is no power (cardinality) $m$ such that $\aleph_0 < m < \aleph_1$.

To me, this seems to offer two equally absurd interpretations: the first, where you assume $\mathbb{c} <\aleph_1$, and then you have to "prove" that the power of the continuum doesn't exist, the second is assuming $\mathbb{c}=\aleph_1$, and then you have to prove something that was shown to be undecidable in ZFC!

Does anyone have any insight on the nature of what this problem is asking? I've been wracking my brain trying to figure it out with some of my classmates (this problem was not assigned as homework), and none of us can crack it.

$\endgroup$
8
  • $\begingroup$ I don't see CH here. Uncountable does not mean of cardinality $\ge c$. $\endgroup$
    – André Nicolas
    Oct 22, 2014 at 5:17
  • $\begingroup$ I'm still learning all this stuff, so I'm shaky on what you mean. The way I understand the notion of uncountability is that you cannot have an uncountable set with a cardinality less than $c$. Can you explain more, or have a link to where I could read more? $\endgroup$
    – epsilonics
    Oct 22, 2014 at 5:24
  • 1
    $\begingroup$ If we assume Axiom of Choice, which we usually do, then for any two cardinals $\lambda$ and $\kappa$, we have $\lambda\lt \kappa$ or $\lambda=\kappa$ or $\kappa\lt \lambda$. It can be shown fairly easily that $\aleph_1\le c$. But the axioms of ZFC do not settle the question of whether it is $\lt c$ or equal to $c$. $\endgroup$
    – André Nicolas
    Oct 22, 2014 at 5:49
  • 1
    $\begingroup$ @epsilonics: Your fundamental misconception is this: ‘you cannot have an uncountable set with a cardinality less than $c$’. That statement is precisely CH. The correct statement is that you cannot have an uncountable set with a cardinality less than $\aleph_1$. Your statement is true if and only if $\mathfrak{c}=\aleph_1$, and that assertion is CH. If CH is false, $\aleph_1$ is an uncountable cardinality less than $\mathfrak{c}$. $\endgroup$
    – Brian M. Scott
    Oct 22, 2014 at 5:49
  • 1
    $\begingroup$ @epsilonics: You’re welcome. No, CH is completely irrelevant to the question. In the first problem you simply have to prove that the set $M$ is neither finite nor countably infinite. In the second you have to show that if $X$ is a set such that $|X|<|M|$, then $X$ is either finite or countably infinite. I would definitely concentrate on the first problem first. (HINT: $M$ is an ordinal; if it were countable, it would be a member of itself ...) $\endgroup$
    – Brian M. Scott
    Oct 22, 2014 at 6:51

1 Answer 1

Reset to default
1
$\begingroup$

Note that $\mathbf c$ does not appear at all in the problem statements. Problem 12 asks you to show that $\aleph_1:=|M|>\aleph_0$, Problem 13 asks you to show that there is no $m$ with $\aleph_0<m<\aleph_1$. The Continuum Hypothesis is concerned with the question if there exist cardinalities $m$ with $\aleph_0<m<\mathbb c$ where $\mathbb c:=2^{\aleph_0}$ (so with a negative answer to it, we would have $\mathbb c=\aleph_1$)

$\endgroup$
2
  • $\begingroup$ If I understand you, you are saying that it ignores the Continuum Hypothesis and this question in the book is sort of a thought experiment? My understanding of cardinals is still very limited, and I am under the impression that the hierarchy of set sizes is finite $\to$ $\aleph_0$ $\to$ $c$ $\to$ ... $\endgroup$
    – epsilonics
    Oct 22, 2014 at 5:28
  • $\begingroup$ @epsilonics: The $\aleph$ numbers are not defined by using the power set, but rather a more limit process of "minimal increment". The continuum hypothesis is exactly the statement $\aleph_1=\frak c$. But both symbols have an independent definition. $\endgroup$
    – Asaf Karagila ♦
    Oct 29, 2014 at 13:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .