What's the diference between $A<\infty$ and $A<\aleph_0$? In my topology class the teacher gave some examples of topologies, and I'm trying to prove that they really are topologies. If $X$ is a set then:


*

*$\mathcal C=\{A:\# (X-A)<\infty\}$ is a topology in $X$.

*$\mathcal C'=\{A:\# (X-A)<\aleph_0\}$ is a topology in $X$.


I've already proved that $\mathcal C$ is a topology, but I don't know what to do for $\mathcal C'$, what's the diference?
 A: There must have been some kind of misunderstanding (either on your part or your instructor's).  $\aleph_0$ is the cardinality of the natural numbers which [assuming the Axiom of Choice...] is the smallest infinite cardinal.  Thus the second expression is just a more formal way of writing the first one.
Maybe what was intended was $\{ A : \# (X \setminus A) \leq \aleph_0\}$, namely the subsets which are "cocountable": their complement is finite or countably infinite.
Note that there is another mistake: when $X$ is infinite, the empty set does not belong to $\mathcal{C}$, but it should in order to give a topology.  Similarly, assuming the above correction, when $X$ is uncountably infinite the empty set does not belong to $\mathcal{C}'$, but it should.  
One can fix and even generalize the question as follows: let $\kappa \geq \aleph_0$ be any infinite cardinal.  Then for any set $X$, the family $\mathcal{C}_{\kappa}$ consisting of the empty set together with all subsets $A$ with $\# (X \setminus A) \leq \kappa$ gives a topology on $X$.  
A: In fact, the two collections are one and the same. To see this, note first that if $S$ is any set, $\#S<\infty$ means that $S$ has finitely many elements, while $\
\#S<\aleph_0$ means that there is an injection from $S$ to the set of natural numbers, but there is no surjection. This is equivalent to $S$ being finite, because if $S$ were infinite, then there would exist a surjective function from $S$ to $\mathbb N$ (that is, you could assign numeric labels, one on each element of $S$, such that every natural number is labeled onto some element; this is possible if you assume the axiom of choice), which is a contradiction.

As pointed out by @PeteL.Clark, my previous answer was incorrect, in which I erroneously interpreted $\aleph_0$ as $\#\mathbb R$, whereas, in reality, $\aleph_0=\#\mathbb N$. I apologize.
