Limit of a Sequence of Sets

In sequences of sets, it seems to me that ‘size’ is thought about as the number of elements in a set, not some idea of the smallest element in the set. Consider a sequence of sets (A_n)_{n=1}^\infty where A_n \in \Omega for all n. Note:
* \cap_{n=k}^\infty A_n consists of elements that are in every A_n with n \geq k. (note: if an element is only in finitely many A_n, then it is outside \cap_{n=k}^\infty A_nfor all k sufficiently large.
* \cup_{n=k}^\infty A_n consists of elements that are in some A_n with n \geq k. (note: if an element is only in finitely many A_n, then it is outside \cup_{n=k}^\infty A_nfor all k sufficiently large.
* If x \in \cap_{n=k}^\infty A_n then all A_nhave x in it, not just an infinite subsequence. Hence, \cap_{n=k}^\infty A_n is a lower bound (in the sense of number of elements) for all A_n for all n \geq k.
* If x \in \cup_{n=k}^\infty A_n if x is in infinitely many of the A_nit is in the union \forall k. Hence, \cup_{n=k}^\infty A_n is an upper bound (in the sense of number of elements) for all A_n for all n \geq k.
There are two notions for the limit of a sequence of sets: the limit infimum, and the limit supremum. 
# Limit Infimum: \lim \inf A_n := \bigcup_{k=0}^\infty \Bigl( \bigcap_{n=k}^\infty A_n \Bigr) this is the collection of x‘s that are (1) in infinitely many of the A_n
, and (2) in all A_n for n sufficiently large.
# Limit Supremum: \lim \sup A_n := \bigcap_{k=0}^\infty \Bigl( \bigcup_{n=k}^\infty A_n \Bigr) this is the collection of all x‘s that are in infinitely many of the A_n.
As can be seen from the definitions, the limit infimum is a subset of the limit supremum,\lim \inf A_n \subseteq \lim \sup A_n. [ADD PROOF]