I'm little confused about the whole concept of limits of sets. I'm working with the definition:
$$\liminf_{n\to\infty}(A_n) := \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}(A_k).$$
As an example consider the sequence of sets $A_n=\{2k | k\ge n\}$, for all $n \in \mathbb{N}$. I'm assuming all the sets $A_n$ include $+\infty$ (I'm not sure my notation is correct for that).
Then, for all $m \in \mathbb{N}$, the $\sup_{n\ge m}(A_n) = +\infty$, and so: $$\limsup_{n\to\infty}(A_n) = \{+\infty\}.$$
However $\inf_{n\ge m}(A_n) \ne {+\infty}$, because for every $n$, $$\inf (A_n) = 2n < +\infty,$$
which would lead to a contradiction. Given that, it seems $$\liminf_{n\to\infty}(A_n)=\varnothing.$$
But this is wrong because $\{-\infty\} \notin \varnothing$. Could someone please help me understand where I went wrong in my argument?
Best Answer
It may be useful to point out that the definitions "supremum" and "infimum" make sense in any partially ordered set $P$: If $S \subset P$ is any subset of the poset $P$, then $\mathrm{sup}(S)$ (if it exists) is the unique element $t\in P$ such that
(i) $s \leq t$ for all $s \in S$ (i.e. it is an upper bound for S) and
(ii) if $t'$ is any element of $P$ satisfying $s\leq t'$ for all $s\in S$ then $t\leq t'$ (i.e. it is the least upper bound).
You can define the infimum in the same way. If, as the poset, you take the power set $\mathcal P(X)$ of a set $X$, i.e. the set of all subsets of $X$, where the partial order is given by containment, then if $\{S_i:i \in I\}$ is a subset of $\mathcal P(X)$, clearly $\mathrm{sup}\{S_i: i\in I\} = \bigcup_{i \in I} S_i$, and similarly $\mathrm{inf}\{S_i: i \in I\} = \bigcap_{i \in I} S_i$.
If we have a sequence $(A_n)$ of subsets of $X$, then $$ \mathrm{limsup}_n A_n = \mathrm{inf}_n(\mathrm{sup}\{A_k: k \geq n\}) = \bigcap_{n \geq 0}\bigcup_{k \geq n}A_k $$ and similarly $$ \mathrm{liminf}_n (A_n) = \mathrm{sup}_n(\mathrm{inf}\{A_k: k \geq n\}) = \bigcup_{n \geq 0}\bigcap_{k \geq n}A_k $$
The $\mathrm{limsup}_n(S_n)$ are elements of $X$ which occur in infinitely many of the $S_n$s, while $\mathrm{liminf}_n(A_n)$ are the elements of $X$ which occur in all but finitely many of the $A_n$s.
If you take $X = \mathbb N$ and $A_n = \{2k: k \geq n\}$ then $\mathrm{limsup}_n(A_n) = \mathrm{liminf}_n(A_n) = \emptyset$, since any even integer $2k$ lies in only finitely many of the subsets $\{A_n\}$.
The $\mathrm{limsup}$ and $\mathrm{liminf}$ of sequences of real numbers is a completely different notion, using the total order on the real numbers rather than partial order of containment.