Let $(E_n)$ be a sequence of sets. I was giving the following definitions:

$$ \limsup_{n \to \infty} E_n = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k} E_n $$

$$ \liminf_{n \to \infty} E_n = \bigcup_{k=1}^{\infty} \bigcap_{n \geq k} E_n $$

I am having hard time trying to understand this definitions. I was thinking on a concrete example and see how it works. For instance, let $(E_n) = \left( \dfrac{1}{n} \right) $ be sequence of sets. Then,

$$ \limsup E_n = \bigcap_{k=1}^{\infty} \bigcup_{n \geq k} \left( \frac{1}{n}\right) = \bigcap_{k} \left( \frac{1}{k}\right) = \{1 \}$$

Is this correct? I am still kind of puzzled with this definition.


Look at the intervals $$A_n = \left[-\frac{1}{n},\,\,\,\,\, 1+\frac{1}{n}\right]$$ Well, they clearly "converge" to $$A=\mathrm{''lim''}A_n=\left[0,\,1\right]$$ Unfortunately, not all sequences of sets converge so "cleanly" to a "limit". Look at this example:

$$ B_n=\begin{cases} \left[0,\,\,1-\frac{1}{n}\right] & \mbox{if $n$ is odd} \\ \left[0,\,\,2+\frac{1}{n}\right] & \mbox{if $n$ is even} \\ \end{cases} $$

In this case, there are two subsequences "converging" to $\left[0,1\right]$ and $[0,2]$. These intervals $\left[0,1\right]$ and $\left[0,2\right]$ are the $\liminf$ and $\limsup$.

Definition of $\limsup$ and $\liminf$

One way to motivate this definition of $\liminf$ and $\limsup$ is the following. When a sequence of sets is increasing (that is, $A_{n+1}\supset A_n$), the "limit" of the sequence is intuitively the union of all sets: $\mbox{''lim''}A_n=\bigcup_n A_n$ (because if an element is in at lease one set of the sequence, it is also in all subsequent sets). When the sequence is decreasing ($A_{n+1}\subset A_n$), the limit is the intersection of all sets: $\bigcap_n A_n$ (if one elements fails to be inside one set, it will fail to be in all subsequent sets).

Now, the sequence $S_k=\bigcap_{n\geq k} A_n$ is always increasing. Is is some kind of "reverse disacumulation": $S_k$ is the set of "all elements that are inside $A_k$ and $A_{k+1}$ and ...". What we gain by considering the $S_k$ sequence is that if an element fails to be, say, in $A_3$ and $A_7$, but it is in all other $A_n$ for $n\neq3 \mbox{ or } 7$, then this element will make it into $S_8$ (and into all other $S_k$ for $k\geq 8$, since the $S_k$ are increasing).

Since the $S_k$ are increasing, we want to take its union: $$S = \bigcup_k S_k = \bigcup_k \bigcap_{n\geq k} A_n$$ and call it some kind of limit. We call it the inferior limit because it is very restrictive: for an element to be in it, it has to be in all of the $A_k$ (except at most some finite number, like the $A_3$ and $A_7$ in the example above)

As for the superior limit, we can define the always-decreasing sequence $T_k=\bigcup_{n\geq k} A_n$. Here, $T_k$ is the set of all elements that belong to some $A_n$ with $n\geq k$. If an element belongs to only a few (finite number of) $A_n$, say, $A_5$ and $A_9$, it won't make it into $T_{10}$. Why? Well, because $T_{10}$ is the set of elements that appear somewhere from $A_{10}$ onwards. This element will never appear again, in any $T_k$ with $k\geq 10$ -- that's why the $T_k$ are decreasing. Since the $T_k$ are decreasing, we can take its' intersection: $$\bigcap_k T_k = \bigcap_k \bigcup_{n\geq k} A_n$$ and call it the superior limit.

If you are wondering why the sequence $S_k=\bigcap_{n\geq k} A_n$ is increasing (it's a sequence of intersections, shouldn't it be decreasing??), note that every time we take the next $k$, we are removing one $A_n$ from the intersection, hence we are taking away one necessary condition for an element to belong the next $S_k$. We are making it easier for an element to be in this next $S_k$. The dual of this argument explains why the sequence $T_k=\bigcup_{n\geq k} A_n$ is decreasing, despite being a sequence of unions. Every time we take the next $k$ we are removing one sufficient condition for an element to belong to the next $T_k$, thus making it harder to make it into this next $T_k$.

I hope this explains the definitions of $\limsup$ and $\liminf$

Your example

Your example is incorrect. We have $$T_k = \bigcup_{n\geq k} E_n = \bigcup_{n\geq k} \left\{\frac{1}{n}\right\} = \left\{\frac{1}{k},\frac{1}{k+1},\cdots\right\}$$ The intersection: $$\limsup E_n = \bigcap_k \left\{\frac{1}{k},\frac{1}{k+1},\cdots\right\}=\emptyset$$ is the empty set, because no element belongs to all of the $T_k$. The sequence $T_k$ loses one element at a time, and "in the limit" is has no more elements at all.

The inferior limit is also empty (obviously, because it must be contained in the superior, which is empty, but let's calculate it): $$S_k = \bigcap_{n\geq k} E_n = \bigcap_{n\geq k} \left\{\frac{1}{n}\right\} = \emptyset$$ This time, the $S_k$ themselves are empty, so their union is will be empty: $$\liminf E_n = \bigcup_k \emptyset=\emptyset$$

Now, if you define this sequence: $$F_n = \left\{\frac{1}{1}, \frac{1}{2}, \cdots, \frac{1}{n}\right\}$$ you can show that both limits (superior and inferior) are: $$F=\liminf F_n = \limsup F_n = \left\{\frac{1}{1}, \frac{1}{2}, \cdots\right\}$$

  • $\begingroup$ In your last line, isn't $\limsup F_n = \varnothing $ ? $\endgroup$ – user203867 Dec 31 '14 at 7:04
  • $\begingroup$ No! For the $F_n$ sequence, we have $T_k=\bigcup_{n\geq k} F_n = \left\{1/1, 1/2, 1/3, \ldots\right\}$. Since $T_k$ doesn't depend on $k$, the limsup is equal to all of the $T_k$. But if you were talking about the lines in which I had written "$\limsup T_k$" and "$\limsup S_k$", I have just corrected them :) $\endgroup$ – fonini Dec 31 '14 at 7:23
  • $\begingroup$ $limsup F_n = T_1 \cap T_2 \cap T_3 \cap ... = \{1, 1/2,1/3,...\} \cap \{1/2,1/3,...\} \cap \{1/3,1/4,.. \} \cap ... = \varnothing ?$ $\endgroup$ – user203867 Dec 31 '14 at 7:33
  • $\begingroup$ You are assuming $T_k$ is $\left\{\frac{1}{k},\frac{1}{k+1},\frac{1}{k+2},\cdots\right\}$. This is true for the original $E_n$ sequence, but not for the new $F_n$ sequence. For this new $F_n$ sequence, we have $T_1=T_2=\cdots=\left\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\cdots\right\}$. Note that $\left(F_n\right)_{n\in\mathbb{N}}$ is increasing, while $E_n$ is neither decreasing nor increasing. $\endgroup$ – fonini Dec 31 '14 at 8:26
  • $\begingroup$ isnt the original $E_n = \{ 1, 1/2,..,1/n\} $ ? $\endgroup$ – user203867 Dec 31 '14 at 9:00

This may help:

$x\in \lim\sup E_n$ if and only if $x$ is in infinitely many of the $E_n$;

$x\in \lim\inf E_n$ if and only if $x$ is in $E_n$ for all sufficiently large $n$ (i.e., it is in all but finitely many of the $E_n$).

  • $\begingroup$ Good way to explain it. $\endgroup$ – Mathemagician1234 Dec 30 '14 at 4:45

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