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If $A_i$ is a sequence of sets, define$$\liminf_i A_i = \bigcup_{j = 1}^\infty \bigcap_{i = j}^\infty A_i, \quad \limsup_i A_i = \bigcap_{j = 1}^\infty \bigcup_{i = j}^\infty A_i.$$What is an example where $\liminf_i A_i \neq \limsup_i A_i$?

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    Note: the liminf is "the set of all points in all but finitely many $A_i$", and the limsup is "the set of all points in infinitely many $A_i$" – Ben Grossmann Jun 30 '16 at 19:40
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    You can read the definitions like quantifiers. An element $x$ is in the liminf if and only if there exists $j$ such that for all $i \geq j$, $x \in A_i$. Similarly $x$ is in the limsup if and only if for every $j$ there exists $i \geq j$ such that $x \in A_i$. Thus the liminf is the elements in all but finitely many $A_i$ while the limsup is the elements in infinitely many $A_i$. In particular the liminf is always a subset of the limsup, and it is a proper subset if there is some $x$ which is in infinitely many $A_i$ and also not in infinitely many $A_i$. – Ian Jun 30 '16 at 19:42

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Take $A_{2i} = \emptyset$ and $A_{2i+1} = \{1\}$. You should check that the $\lim \inf$ is then $\emptyset$ while the $\lim \sup$ is $\{1\}$.

Shagnik
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