# Is $n <\aleph_0, n \in \mathbb{N}$ well defined?

Is $n <\aleph_0, n \in \mathbb{N}$ well defined?

That is, can I "compare" a natural number to a transfinite number in a strict sense?

My intuition says yes since if $S$ is a finite set then we have $|S| < \aleph_0$ but I want to be really sure.

## 4 Answers

Yes. Of course. Recall that if $a,b$ are cardinals, $a\leq b$ if there are sets $A$ and $B$ of cardinalities $a$ and $b$ respectively such that $A\subseteq B$. And $a<b$ if $a\leq b$ and $a\neq b$.

Clearly $n<\aleph_0$ because taking the first $n$ natural numbers gives us a subset of $\Bbb N$ which has cardinality $n$; and that set is finite, whereas $\Bbb N$ is not.

Yes you can since all the natural numbers are cardinal numbers as well, corresponding to the cardinalities of the finite sets.

Yes. Notice that $K_n=\{0,1,2,\dots ,n-1\}\subseteq \mathbb{N}$, therefore $n\leq \aleph_0$. Now, there is no onto function $f:K_n\to\mathbb{N}$, and the reason for that is: if $f:K_n\to\mathbb{N}$, we know that $f(K_n)$ is finite set with less than $n$ elements. Therefore $f(K_n)\neq \mathbb{N}$, hence $f$ is not onto, thus $n\neq \aleph_0$. And we have $n< \aleph_0$.

To offer a counterpoint, I'll agree that yes you can, but you have to be careful to understand (and sometimes to make sure that whomever you're communicating with understands) exactly what you mean. We use the less-than relation $\lt$ so casually that it's easy to forget there are multiple different definitions of it, all of which happen to agree with our 'common-sense' notion of less-than on the most natural domain $\mathbb{N}$; there's the cardinal notion of 'less-than' (there's an injection but no bijection), there's the ordinal notion of 'less-than' (one well-ordering is an initial segment of the other), and then of course there are various order relations on domains like $\mathbb{Q}$, etc. that formally translate to relations on equivalence classes.

While in this case, the use of $\aleph_0$ makes it fairly clear that you're intending the cardinal ordering (and the answer is the same regardless of which meaning is used in this case), there are enough small subtleties that it's worth taking care.

• I don't think that anyone, in the past century or so, have used $\aleph_0$ for denoting an ordinal. – Asaf Karagila Dec 9 '13 at 23:34