# Is $\aleph_0 = \mathbb{N}$?

Some very wise people here have just told me that $\aleph_0 = \mathbb{N}$, i.e. that the cardinality of the set of natural numbers is just the set of natural numbers itself. Is this now the general consensus in mathematics, real analysis in particular? Or have I got completely the wrong end of the stick as usual?

• I would write $\aleph_0= | \mathbb N |$. – wythagoras Aug 30 '15 at 16:07
• @Renato, It may be if $a, b, c, d$ happen to equal $0$, $1$, $2$ and $3$. – Henning Makholm Aug 30 '15 at 16:11
• Why would you care what people in real analysis have to say about $\aleph_0$? It would be as relevant as asking a set theorist what is $\pi$. – Asaf Karagila Aug 30 '15 at 16:12
• @GitGud: There's Scott's Trick, which is useful when working in ZF without choice but with foundation. In this representation $\aleph_0$ is the set of all countably infinite sets of rank $\omega$, so instead we have $\omega\in\aleph_0$. – Henning Makholm Aug 30 '15 at 16:17
• I think it is probably true that if you asked a bunch of real analysis "Is it true that $\{0,1\}=2$?" they would probably look at you like you are crazy, but a set theorist would say "Of course." If you take numbers as primitive elements in a theory, there is no need to worry about how they are defined and what they 'really' are. And if you are comfortable doing that with numbers, why not with infinite cardinals? – mweiss Aug 30 '15 at 16:34

There's nothing deep going on here. Its just that:

• Its often convenient to identify $\mathbb{N}$ with the least infinite ordinal $\omega$.
• Its often convenient to identify each well-orderable cardinal number $\kappa$ with the least ordinal $\alpha$ such that $|\alpha| = \kappa$. This is called the von Neumann cardinal assigment.

Under these identifications, we find that $\mathbb{N} = \omega = \aleph_0 = |\mathbb{N}|.$ I wouldn't read too much into it though.

• Interesting. Another instance were it seems the conventions of ZFC theory are not widely adopted outside the narrow specialty of set theory itself. – Dan Christensen Aug 30 '15 at 16:15
• @DanChristensen The issue is much more serious than field-dependent conventions, for in asserting that $\mathbb N\neq \aleph_0$, one must be able to tell what $\aleph_0$ is and, of course, when one defines it, the equality is trivial. – Git Gud Aug 30 '15 at 16:22
• Let $\mathbb{N}'$ be $\bigcap\{x:\emptyset\in x\wedge\{\{z\}:z\in x\}\subseteq x\}$. One can recursively define all the usual arithmetic, and its not identical with $\aleph_0$. But that's not the set that's conventionally used for the naturals. – Malice Vidrine Aug 30 '15 at 16:58
• Peano's axioms by themselves? That wouldn't be very helpful if analysis is your concern. – Malice Vidrine Aug 30 '15 at 22:04
• Not to mention I think you missed the point. Curiosity was expressed about implementations of $\aleph_0$ that differ from one for $\mathbb{N}$; I was offering an example of the converse, which is easier to find, not recommending the Zermelo naturals. – Malice Vidrine Aug 30 '15 at 22:15

The objects in set theory are sets. Only sets in $\sf ZFC$ and its related theories. This means that if you want to interpret a mathematical object in set theory you need to assign it a set.

Of course you are free to assign to it any set that you wish, as long as you have the axiom of replacement set theory is more or less interpretation agnostic (in the sense that it proves that two interpretations are generally exchangeable). But still you need to pick some interpretation, to at least show one way of doing this is possible.

So just like the standard method for interpreting ordered pairs is via the definition by Kuratowski, the standard way of assigning cardinals to well-orderable sets is by picking the least ordinal of that cardinality as a representative. So we define the cardinals for infinite [well-orderable] sets by transfinite induction,

• $\aleph_0=\omega$ (the least infinite ordinal which exists from the axiom of infinity, power set and separation),
• $\aleph_{\alpha+1}$ is the least ordinal whose cardinality is not smaller than $\aleph_\alpha$ (and such ordinal exists by Hartogs theorem),
• If $\delta$ is a limit ordinal, then $\aleph_\delta=\sup\{\aleph_\alpha\mid\alpha<\delta\}$ (which exists from replacement and union).

There is nothing more to it, and it's just one possible way of interpreting these cardinals. You can also represent cardinals in other ways (e.g. Scott's trick give you Scott cardinals, but it will fail you if choice and foundation fail, e.g. if you allow urelements and choice failed), or choose not to represent the cardinals internally and work, awkwardly I might add, with equivalence classes of sets.

It's up to you. But the consensus of what object $\aleph_0$ is exists only in set theory, and not in analysis. So asking analysts for their opinion is quite irrelevant for this question.

And so, if you identify in set theory $\Bbb N$ with $\omega$ you get that $|\Bbb N|=\Bbb N$. Whether you choose to identify the natural numbers with the finite ordinals is up to you, but most set theorists [working in $\sf ZF$ and the like] do.

At the end of the day you can argue whether or not functions are sets of ordered pairs, and whether or not $0$ is a natural number or not. At the end of the day this is just missing the point of a foundational theory. To give a foundation to mathematics. We have yet to find a theory that does it as well as $\sf ZFC$ (in my opinion anyway, some might argue differently).

If you're unhappy with the current foundations of mathematics, there are only two options: (1) study it and learn to accept it's supposed flaws; or (2) find a different foundation.

• It seems that ZFC cannot adequately model (if that is the right word) what most mathematicians mean by $\aleph_0$. No great loss, I suppose. – Dan Christensen Aug 30 '15 at 20:34
• It seems that you have no idea what you're talking about. But I suppose you're entitled to your opinion. No great loss, I suppose. – Asaf Karagila Aug 30 '15 at 20:45
• ZFC doesn't try to model what mathematicians imagine numbers, cardinals, etc "mean." It does try to model a cumulative hierarchy of pure sets in a first-order finite language with relatively simple axioms-- and the reason it is worth studying is that (virtually) all mathematical notions can be formally interpreted in ZFC. No one imagines that an ordered pair (a,b) "is really" {{a,b},{b}}, but it shows that we don't need an "outside of pure sets" primitive for ordered pairs even though the intuitive notion lies outside of pure sets (IMO anyway) . Same for cardinality. That's all. – Ned Aug 30 '15 at 21:01

You have N≠ℵ0, Infact one is the set (N), while ℵ0 is a transfinite number. Asking ℵ0=N, is somewhat.. you are asking.. Is 2={1,2} ? which is obviously not.

• Obviously not, since a set cannot contain itself. In axiomatic set theory we have $2=\{0,1\}$ by definition, but $2\ne\{1,2\}$. – Henning Makholm Aug 30 '15 at 16:12
• This answer isn't so bad that it should be accruing downvotes in my opinion. Under "usual" or "naive" mathematical conventions, we have $\mathbb{N} \neq \aleph_0$ (again in my own opinion). And I'm betting that Martin Lof Constructive Type Theory (which I know almost nothing about) doesn't even allow $\mathbb{N} = \aleph_0$ as a well-formed term. – goblin Aug 30 '15 at 16:22
• @goblin: I would argue that in naive mathematics the question simply isn't asked, not that is answered in the negative. This is certainly the case in type theories (not being able to phase it in good syntax is not a negative answer). – Malice Vidrine Aug 30 '15 at 16:33
• @HenningMakholm "since a set cannot contain itself" is not always true. – quid Aug 30 '15 at 17:30