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I have been reading about cardinal arithmetic in an introduction to set theory and have the following questions that are unclear to me after working through some problems:

  • How is it that $\aleph_1=2^{\aleph_0}$?
  • Does $\aleph_1=\aleph_0^2$?

The motivation behind the question is that I have seen conflicting sources on whether or not $2^{\aleph_0}$ is the same as $\aleph_1$ and so I would like to have a definitive answer on the difference (and if there is one, what is this distinction).

  • Would the limit of $2^x$ as $x$ approaches $+\infty$ be equal to $\aleph_1$?

Here, I am unsure about the relation between the infinite cardinal numbers and the infinite limits dealt with in a course on real analysis. Does a limit to infinity in a real analysis context mean a limit to $\aleph_0$ or some other cardinal number? And if not, can we define limits in the context of cardinal arithmetic?

In other words, my questions ask: would the limit of $2^x$ as $x$ approaches infinity ($\aleph_0$) be equal to $\aleph_1$? Also would it be correct to say $\aleph_1=\aleph_0^2$? Or is $\aleph_0^2=\aleph_0$?

FD_bfa
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Adithya
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 30 '23 at 12:31
  • $2^{\aleph_0}=\beth_1$; $\aleph_0^2=\aleph_0$ and $\sup_{x<\aleph_0} 2^x=\aleph_0$. $\aleph_1$ can be described in a few ways, one being the cardinality of the set of isomorphism classes of well-orderings on subsets of $\Bbb N$. – Sassatelli Giulio May 30 '23 at 12:32
  • so basically the power set of N0. i just dont get how 2^N0 is N1. shouldnt the sum of 2 countably infinite times equal aleph 0? – Adithya May 30 '23 at 12:51
  • @Adithya a) Basically a completely different object. b) Neither of us had mentioned a sum. – Sassatelli Giulio May 30 '23 at 12:55
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    Does this answer your question? Is $2^{\aleph_0} = \aleph_1$? – MJD May 30 '23 at 12:56
  • By $x$ do you mean a natural number or a real number? If it's a real number then "the limit of $2^x$" doesn't make sense; in calculus it's defined to be $\infty$, which is really just an abbreviation of "as $x$ becomes large $2^x$ gets arbitrarily large". If $x$ means natural number and $2^x$ means the (cardinality of ) power set of a $x$-element set, and we are taking limit of cardinal numbers, then the result is $\aleph_0$. $2^{\aleph_0}$ is yet another thing, the cardinality of the power set of $\mathbb{N}$ – Lxm May 30 '23 at 12:56
  • my bad i made a mistake – Adithya May 30 '23 at 12:56
  • I've expanded upon this in my answer. Let me know if that helps to clarify the difference between infinity in an analysis context versus in a set theory context @Adithya – FD_bfa May 30 '23 at 16:26

2 Answers2

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Claim 1: "How is $\aleph _1 = 2^{\aleph_0}?$"

This claim is not true in general. We only say this is true if you accept the Continuum Hypothesis. This hypothesis states that there is no set with cardinality strictly between the integers and the real numbers.

Proving or disproving the Continuum Hypothesis has been shown to be impossible using the axioms of $ZF$ or indeed $ZFC$. Whether or not you accept the hypothesis affects whether or not this claim is true.

Therefore, whether or not the claim is true really just comes down to whether or not you want it to be true (of course it is a bit more nuanced than that, but in practice there is nothing wrong with accepting or rejecting the Continuum Hypothesis).

Claim 2: "Does $\aleph _0^2 = \aleph _1$?"

Using Hessenberg's Theorem (for every infinite cardinal $m$ the product $\color{red}{m \cdot m}$ is equal to $\color{blue}m$) we choose $m = \aleph_0$ to get the following result:

$$ \color{red}{(\aleph _0 \cdot \aleph _0)} \space = \aleph^2_0 \space = \color{blue}{\aleph_0}$$

Therefore, $\aleph_0^2 = \aleph_0$. In fact, we can adapt the above argument slightly to show that $(\aleph_0)^x = \aleph_0$ for any positive integer $x$ (where we consider $x=2$ to address the case in the question).

Claim 3: "So would the limit of $2^x$ as $x$ approaches infinity be equal to $\aleph_1$?"

This seems to be a misunderstanding regarding cardinal numbers. Cardinal numbers tell us the size of sets. When we take a limit as $x$ approaches infinity, we are using the term "infinity" here in a fundamentally different way. We are no longer referring to the size of a set and so infinity in this context has no meaning in the language of cardinal arithmetic. We simply say $2^x \rightarrow + \infty$ as $x \rightarrow + \infty$. Cardinal numbers do not come into play when dealing with limits.

FD_bfa
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The Hessenberg theorem states that $$\kappa+\nu=\operatorname{max}\{\kappa,\nu\}=\kappa\cdot\nu,$$ so $\aleph_0\cdot \aleph_0=\aleph_0,$ if they are both infinite cardinals.

Guillermo García Sáez
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