# Is it true that $\aleph_1=2^{\aleph_0}$, and if so, would the limit of $2^x$ as $x$ approaches $+\infty$ be equal to $\aleph_1$? [duplicate]

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 $$\aleph_0^2=\aleph_0$$?

• 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 Bot
May 30, 2023 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 $\Bbb N$. May 30, 2023 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? May 30, 2023 at 12:51
• @Adithya a) Basically a completely different object. b) Neither of us had mentioned a sum. May 30, 2023 at 12:55
• Does this answer your question? Is $2^{\aleph_0} = \aleph_1$?
– MJD
May 30, 2023 at 12:56

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.

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.

• so then how is 2^ℵ0 equal to a higher infinity? isnt it just equal to the sum of 2 infinite times? should that approach ℵ0? May 30, 2023 at 12:49
• First of all $aleph_1=2^{\aleph_0}$ is only true if You're accept the continuum Hypoteshis. What is always true un ZFC is that $\kappa<2^{\kappa}$ for every cardinal $\kappa$, which is Cantor's theorem May 30, 2023 at 12:54
• could you elaborate on how hessenbergs theory implies ℵ0⋅ℵ0? I don’t quite understand how it works May 30, 2023 at 12:59
• Take $\kappa=\aleph_0=\nu$. $\operatorname{max}\{\aleph_0, \aleph_0\}=\aleph_0$ May 30, 2023 at 13:02
• i didnt quite understand but ill look more into it. thanks! i dont get what k and v represent or why k+v = kv. or what the proof of your last statement is. May 30, 2023 at 13:11