Cardinal number subtraction

Solution 1:

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 $\aleph_0$.

Solution 2:

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.