Is $\aleph_0 - n$ defined for finite $n$?

Question

The reason I am asking this is since $\aleph_0 - n$ = $- n+\aleph_0$, which would be undefined in cardinal arithmetic (since all cardinal numbers are positive). However, if we take two infinite sets $a$ and $b$, where $a$ has $n$ less elements than $b$, there is still a bijection between the two sets, so $|a| = |b| = \aleph_0$. Any help on this would be greatly appreciated.

Answer 1

If $\kappa,\,\lambda$ are cardinals, $\kappa-\lambda$ is defined iff exactly one cardinal $\mu$ satisfies $\mu+\lambda=\kappa$, in which case that $\mu$ is our definition of $\kappa-\lambda$. (Equivalently, the condition is that size-$\lambda$ subsets of a given size-$\kappa$ set exist, and all have equally large complements with respect to it.) Therefore, $\aleph_0-n=\aleph_0$ for all finite $n$.

What you cannot do, however - for nonzero cardinals $n$, finite or otherwise - is write $-n+\aleph_0$, or even $-n$. That's an illegal operation for the aforementioned $n$. It would be equivalent to $0-n$, and that's undefined because $\emptyset$ has no size-$n$ subsets. In fact, this same logic implies that, whereas cardinal arithmetic lets you write $4-3=1$, it defines neither $-3+4$ nor $-3$.