Let's say we have $\mathbb{N}$, the set of natural numbers:
$\{1, 2, 3, 4, 5...\}$
...which has a cardinality of infinity, and the set $A_x$ which consists of the variable "$x$" (so $\{x\}$).
If I did this:
$N - A_1 - A_2 - A_3 - \cdots - A_x$
...for the limit as $x \rightarrow \infty$, then does the resulting set have a cardinality of $0$? Or does the fact that $x$ is a limit mean that it is not the same type of infinity and that the remaining set will still have a cardinality of infinity?
On a similar note:
Is the cardinality of the set of natural numbers the same as the cardinality of the set: $\{x: x$ is an integer such that $\displaystyle 0 < x < \lim_{n \rightarrow \infty} n\}$ ?
The final set will be empty set. Let's see how. The only elements in $\mathbb{N}$ are, well, positive integers. But for every such $n$, we are subtracting $A_n$, thus $n$ cannot be in the limiting set. Thus the cardinality is $0$. Observe here that
$$ \lim_{n\rightarrow \infty} card\{N - A_1 - ... - A_n \} = \infty \neq card\{\lim_{n\rightarrow \infty} \{N - A_1 - ... - A_n \} \} $$