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Recently I learned about the infinite cardinal $\aleph_0$, and stumbled upon a seeming contradiction. Here are my assumptions based on what I learned:

  1. $\aleph_0$ is the cardinality of the natural numbers
  2. $\aleph_0$ is larger than all finite numbers, and thus cannot be reached simply by counting up from 1.

But then I started wondering: the cardinality of the set $\{1\}$ is $1$, the cardinality of the set $\{1, 2\}$ is $2$, the cardinality of the set $\{1, 2, 3\}$ is 3, and so on. So I drew the conclusion that the cardinality of the set $\{1, 2, \ldots n\}$ is $n$.

Based on this conclusion, if the cardinality of the natural numbers is $\aleph_0$, then the set of natural numbers could be denoted as $\{1, 2, \ldots \aleph_0\}$. But such a set implies that $\aleph_0$ can be reached by counting up from $1$, which contradicts my assumption #2 above.

This question has been bugging me for a while now... I'm not sure where I've made a mistake in my reasoning or if I have even used the correct mathematical terms/question title/tags to describe it, but I'd sure appreciate your help.

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    $\begingroup$ Can you count to $\aleph_0$?. I am not even going to start to see if I can., $\endgroup$
    – user328032
    Apr 28, 2016 at 1:19
  • $\begingroup$ It seems to me that you want this to be an ordered set, but it does not really make sense to tack on $\aleph_0$ to the end in the way that you want. $\endgroup$
    – Cameron Williams
    Apr 28, 2016 at 1:20
  • $\begingroup$ @CameronWilliams Yes, but then what would be the last element of the set? $\endgroup$
    – Timtech
    Apr 28, 2016 at 1:21
  • $\begingroup$ I can count up to $\aleph_0$. Just give me $\aleph_0$ seconds added to my life and I hope I will be able to be patient enough to do this... Countable doesn't mean you can count to it, it just means it contains the whole numbers excluding all the rational decimals between them. $\endgroup$
    – KKZiomek
    Apr 28, 2016 at 1:22
  • $\begingroup$ @Timtech That's the thing. There isn't a "last" element here. There is a maximal element, but not a last. Last implies that you can reach that element in finitely many steps. "Last" is somewhat of a colloquialism. $\endgroup$
    – Cameron Williams
    Apr 28, 2016 at 1:22

2 Answers 2

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This is a good example where intuition about a pattern breaks down; what is true of finite sets is not true of infinite sets in general. The natural numbers $\textit{cannot}$ be denoted by the set $A=\{1,2,...,\aleph_0\}$ as the set $\aleph_0$ is not a natural number. It is true that the cardinality of $A$ is $\aleph_0$ (a good exercise), but it contains more than just natural numbers.

If $\aleph_0$ were a natural number then, as you point out, we would have a contradiction. However $\aleph_0$ is the $\textit{cardinality}$ of the natural numbers, and not a natural number itself. By definition, $\aleph_0$ is the least ordinal number with which the set $\omega$ of natural numbers may be put into bijection.

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  • $\begingroup$ So is ℵ0 a set or a cardinal number, or both? "the set ℵ0" or "ℵ0 is the cardinality of the natural numbers" $\endgroup$
    – Timtech
    Apr 28, 2016 at 17:26
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    $\begingroup$ Both... In $ZFC$ $\textit{everything}$ is a set, but more explicitly, the definition of cardinal numbers I know is this: Let $A$ be a set. Then the cardinal number of $A$ is the least ordinal $\kappa$ such that there exists a bijection $f: \kappa \to A$. Now by definition, ordinals are transitive sets that are well-ordered by $"\in"$, and since cardinals are in particular ordinals, they are sets. Since $\aleph_0$ is a cardinal number, it is also a set. $\endgroup$
    – Exit path
    Apr 28, 2016 at 22:01
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Good question! You are free to include aleph null in your set, but it still contains an infinite set of natural numbers. So you can't count up to it. What you have is

$$\{1,2,\ldots,\text{ an infinite list of numbers },\ldots , \aleph_0\}$$

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  • $\begingroup$ Ah, makes sense, thanks. $\endgroup$
    – Timtech
    Apr 28, 2016 at 22:05

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