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This is a question on what is written here.

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 $\liminf_i A_i \neq \limsup_i A_i$?

Take $A_{2i} = \emptyset$ and $A_{2i+1} = \{1\}$. Why is it that the $\liminf$ is $\emptyset$ while the $\limsup$ is $\{1\}$? How do I see these two things?

Community
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  • Can you compute, for example, $\cap_{i=6}^{8} A_i$? – Michael Sep 14 '16 at 03:25
  • Some useful explanations of how to think about limsup and liminf of sets can be found here: http://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Hans Lundmark Sep 14 '16 at 10:28
  • Just remember that $\liminf_iA_i$ is the set of points that are in all of the $A_i$ from some point on, while $\limsup_iA_i$ is the set of points that are in infinitely many of the $A_i$. It’s easy to arrange matters so that some point is in infinitely many of the $A_i$ but not in any ‘tail’ of them. – Brian M. Scott Sep 14 '16 at 21:35

4 Answers4

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For any $n$, $A_{2n} \cap A_{2n+1} = \emptyset$ and $A_{2n}\cup A_{2n+1}=\{1\}$.

So $\cap_{i\geq j}A_i\subset A_{i}\cap A_{i+1}=\emptyset$ hence $\cap_{i\geq j}A_i=\emptyset$ so $\cup_{j\geq1} \cap_{i\geq j}A_i=\cup_{j\geq1}\emptyset= \emptyset$.

Again, $\cup_{i\geq j}A_i\supset A_i\cup A_{i+1}=\{1\}$ and since each $A_i$ is either $\emptyset$ or $\{1\}$ we have that $\cup_{i\geq j}A_i=\{1\}$ Thus $\cap_{j\geq1}\cup_{i\geq j}A_i=\cap_{j\geq1}\{1\}=\{1\}$.

Landon Carter
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Intuitive answer.

The sequence $A_n$ is as follows:

$$\emptyset, \{1\},\emptyset, \{1\},\emptyset, \{1\},\emptyset, \{1\},\emptyset, \{1\},...$$

The limit inferior of the sequence is the elements of $\Omega$ s.t. after some index $m$ the said elements will be in $A_m, A_{m+1}, A_{m+2},...$

Let's try $m=1$. What elements of $\Omega$ are in $A_1$, $A_2$,...? None! Precisely, the elements of $\emptyset$ are in $A_1$, $A_2$,...

The same is true for $m=2$, $m=3$, etc.

The limit superior of the sequence is the elements of $\Omega$ s.t. for every index $m$, the said elements are in some future set (set with index greater than $m$).

For example, let's try $m=1$, what are the elements of $\Omega$ that are in some future set? $1$ seems to work.

Let's try $m=1$, what are the elements of $\Omega$ that are in some future set? $1$ seems to work. It is in $A_2$. It is also in $A_4$, $A_{100}$ and $A_{645245444}$

Let's try $m=2$, what are the elements of $\Omega$ that are in some future set? $1$ seems to work. It is in $A_{6554}$.

BCLC
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$\lim \inf A_i$ can be interpreted as "the set of elements that are eventually contained in all sets beyond a certain index" while $\lim \sup A_i$ is "the set of elements that will be visited infinitely often".

From these intuitive interpretations, we can see that for your example, only $\emptyset$ is contained in every set while $1$ occurs infinitely often.

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A more general example may help to answer this question. Let us construct a set sequence $\{A_n\}$ where $A_{2j}=X,\ A_{2j-1}=Y,\ j=1,2,\dots$ (notice both $X$ and $Y$ are sets). Now we specify

  • $\lim \inf A_n=\bigcup_{k \ge 1}\bigcap_{n\ge k}A_n=\bigcup_{k\ge 1}(X \bigcap Y)=X\bigcap Y$,

  • $\lim \sup A_n=\bigcap_{k \ge 1}\bigcup_{n\ge k}A_n=\bigcap_{k\ge 1}(X \bigcup Y)=X\bigcup Y$.

JonathanV
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