We all know that $|\mathbb{N}| = \aleph_0$. Since $|\{-1\} \cup \mathbb{N}| = \aleph_0$ as well, I guess you could say that $\aleph_0 + 1 = \aleph_0$.

You can go on to derive that $\aleph_0 + \aleph_0 = \aleph_0$, and by induction $n \times \aleph_0 = \aleph_0$ (assuming that $n$ is finite, anyway).

Now, here's a thing: What is $\aleph_0 - \aleph_0$?

Well, $|\mathbb{P}| = \aleph_0$, and $|\mathbb{P} \backslash \mathbb{N}| = \aleph_0$, so perhaps $\aleph_0 - \aleph_0 = \aleph_0$?

No, wait. Consider the set $S = \{n \in \mathbb{N}: n > 5\}$. Now we have $|S| = \aleph_0$, and yet $|\mathbb{N} \backslash S| = 5$. So maybe $\aleph_0 - \aleph_0 = 5$?

But that is absurd, since we can redefine $S$ to make the result any finite number we wish. Or we can define $S$ such that the result is countably infinite.

Does this mean that the notion of $\aleph_0 - \aleph_0$ simply has no definite answer? Or am I just being too simplistic here?

  • 2
    $\begingroup$ That is correct -- subtraction is not defined for infinite cardinals, because there is no way to make it give sensible results. $\endgroup$ – TonyK May 4 '12 at 15:35
  • $\begingroup$ Sometimes it is well-defined and sometimes it isn't: see my answer below. And it does come up in mathematics sometimes: e.g. often one starts with an uncountable set and removes a countable subset: in this case, it is useful to know that what remains is necessarily uncountable (and, perhaps less usefully in practice, is of the same cardinality as the original set). $\endgroup$ – Pete L. Clark May 4 '12 at 16:21
  • $\begingroup$ If $\mathbb{P}$ is the set of primes you have the example backward (it should be $\mathbb{N} \setminus \mathbb{P}$), if it is something else could you define what it is? $\endgroup$ – Paul Plummer Mar 29 '15 at 6:54

Let $\kappa_1$ and $\kappa_2$ be cardinals. It makes sense to say that a cardinal $\kappa_3$ is $\kappa_1 - \kappa_2$ if $\kappa_2 + \kappa_3 = \kappa_1$. But if at least one of $\kappa_2$ and $\kappa_3$ is infinite, then (assuming the Axiom of Choice, as usual) have

$\kappa_2 + \kappa_3 = \max(\kappa_2,\kappa_3)$.

Now $\max(\kappa_2,\kappa_3) = \kappa_1$ iff either ($\kappa_2 < \kappa_1$ and $\kappa_3 = \kappa_1$) or ($\kappa_2 = \kappa_1$ and $\kappa_3 \leq \kappa_1$).

Thus if $\kappa_2 < \kappa_1$, $\kappa_1 - \kappa_2$ is uniquely determined: it is $\kappa_1$.

However, if $\kappa_2 = \kappa_1$, then $\kappa_3$ can be any cardinal less than or equal to $\kappa_1$, i.e., subtraction of a cardinal from itself is not uniquely determined.

(Finally, if $\kappa_2 > \kappa_1$, then there is no cardinal $\kappa_3$ with $\kappa_3 = \kappa_1 - \kappa_2$.)

In particular $\aleph_0 - \aleph_0$ is not uniquely defined: by performing this subtraction in the above sense one can get every finite cardinal and also $\aleph_0$.


Assuming the axiom of choice, or at least that $\kappa$ and $\lambda$ are well ordered cardinals, if $\kappa<\lambda$ then $\lambda-\kappa=\lambda$.

Any other case, as you noted, is not well defined. Without the axiom of choice the structure of cardinals can differ greatly between models and there might be limited examples for when it is well defined.

The same can be said about division. Here we require that $\kappa<\mathrm{cf}(\lambda)$, and it is undefined otherwise.


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