I've been learning about transfinite cardinals and examining the expression $\aleph_0 – \aleph_0$. I've read online that subtraction in this case is not defined because it can result in any finite cardinal as well as $\aleph_0$.
However, is it reasonable to simply say that $\aleph_0 – \aleph_0 \le \aleph_0$?
If not, why not? Thanks!
EDIT: I have seen texts write $\aleph_0 – n = \aleph_0$ where $n \in \mathbb{Z}_{\ge 0}$, which makes sense. So then it appears subtraction is defined between a transfinite cardinal and finite cardinal but not on two transfinite cardinals?
Best Answer
Subtraction of cardinals is defined only when the result is unique: If $\kappa,\lambda$ are cardinals, and there is a unique cardinal $\mu$ such that $\lambda+\mu=\kappa$, then $\kappa-\lambda$ is defined, and its value is $\mu$. The notation is not very common, since the situations where the expression is well-defined are either uninteresting, or rare, depending on whether we assume the axiom of choice.
An example is the case you suggest: $\aleph_0-n=\aleph_0$, for any natural number $n$. Also, for transfinite cardinals (alephs) we have that $\kappa-\lambda$ is defined precisely when $\lambda<\kappa$, in which case $\kappa-\lambda=\kappa$.
A less trivial example is that $2^{\aleph_0}-\aleph_0=2^{\aleph_0}$. This is interesting in that the axiom of choice is not needed to verify it. See here for other examples and references.
As for your question regarding $\aleph_0-\aleph_0\le\aleph_0$, I suppose you could define this expression in a way that is meaningful, even though $\aleph_0-\aleph_0$ by itself is not. I would take it to mean that whenever $A,B$ are countably infinite sets, with $B\subseteq A$, then $A\smallsetminus B$ is countable.