Cardinal number subtraction 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?

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 $\aleph_0$.  
