# Question about the cardinality of sets and infinity

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 $\displaystyle 0 < x < \lim_{n \rightarrow \infty} n\}$ ?

• I feel as if I have answered this question infinitely many times before. – Asaf Karagila Jul 30 '14 at 5:32
• Do any answers happen to come to mind? – umpub Jul 30 '14 at 5:36
• (Also, many thanks to the editor!) – umpub Jul 30 '14 at 5:36
• Of course. Many answers. I'm just tired of writing them over and over and over and over and over again... Suffice to say, cardinals are not real numbers, and cardinal functions don't have to be continuous, they are usually not continuous. – Asaf Karagila Jul 30 '14 at 5:41

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
$$\lim_{n\rightarrow \infty} card\{N - A_1 - ... - A_n \} = \infty \neq card\{\lim_{n\rightarrow \infty} \{N - A_1 - ... - A_n \} \}$$