lim sup and lim inf of sequence of sets.

Solution 1:

A member of $$ \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n $$ is a member of at least one of the sets $$ \bigcap_{n\ge N} A_n, $$ meaning it's a member of either $A_1\cap A_2 \cap A_3 \cap \cdots$ or $A_2\cap A_3 \cap A_4 \cap \cdots$ or $A_3\cap A_4 \cap A_5 \cap \cdots$ or $A_4\cap A_5 \cap A_6 \cap \cdots$ or $\ldots$ etc. That means it's a member of all except finitely many of the $A$.

A member of $$ \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n $$ is a member of all of the sets $$ \bigcup_{n\ge N} A_n, $$ so it's a member of $A_1\cup A_2 \cup A_3 \cup \cdots$ and of $A_2\cup A_3 \cup A_4 \cup \cdots$ and of $A_3\cup A_4 \cup A_5 \cup \cdots$ and of $A_4\cup A_5 \cup A_6 \cup \cdots$ and of $\ldots$ etc. That means no matter how far down the sequence you go, it's a member of at least one of the sets that come later. That means it's a member of infinitely many of them, but there might also be infinitely many that it does not belong to.

Solution 2:

I just came up with this mnemonic story:

There is a company with employes and one day a whole bunch $\mathcal A$ of them get fired. They become beggars and have to live on the street. One day, the local church decides to start to give out free food for them every week. In the $n$th week, $A_n$ are the people who show up (this is a sequence of $\mathcal A$-people: $\forall n\ (A_n\subseteq \mathcal A)$. Yes, beggars never die).

1. Some of the people eventually get a new job and never show up at the church again.

2. Others are too proud and try not to be seen around all the time, but they need to eat so they always come back eventually.

3. Lastly there are the people who have low self-esteem; they feel inferior, and at one point, they don't care anymore and start to get their food from the church each week.

$\lim \sup A_n$ are all the people who don't get another job. (Categories 2 and 3). Thus, $\lim \inf A_n^C$ are all the people who eventually get a new job. (Category 1)

$\lim \inf A_n$ are the people who become weekly regulars. (Category 3)

Clearly $\lim \inf A_n \subseteq \lim \sup A_n$.

A series converges implies all the people who can't get another job eventually swallow their pride and become regulars too: $\lim \inf A_n$ = $\lim \sup A_n$. We call this the limit of $A_n$.

Solution 3:

(This is a late answer, but hopefully it may add another perspective to the discussion.)

When one thinks about the problem of defining the limit of a sequence of sets, there are two easy cases: if the sequence is increasing, or if it's decreasing.

For example, when defining improper integrals $\iint_D$ in multivariable calculus, one looks at sequences $D_1 \subseteq D_2 \subseteq D_3 \subseteq \dotsb$ which exhaust $D$, meaning that

• each $D_i$ is a subset of $D$, and
• each element of $D$ is contained in $D_n$ for all sufficiently large $n$ (in other words, $D=\bigcup_{n=1}^\infty D_n$).

In this situation, students often ask if they can write $D_n \to D$, and the answer is usually “well, we haven't really defined limits of sets in this course, but if we had...”. So it's pretty intuitive that for an increasing sequence of sets, the limit should be defined as the union of all sets in the sequence.

Similarly, for a decreasing sequence $E_1 \supseteq E_2 \supseteq E_3 \supseteq \dotsb$, it's natural to define the limit as the intersection : $E_n \to E$ as $n\to\infty$, where $E=\bigcap_{n=1}^\infty E_n$.

Now, for an arbitrary sequence of sets $(A_1,A_2,A_3,\dots)$, we can squeeze it between an increasing sequence $(D_n)$ and a decreasing sequence $(E_n)$, like this: $$ \begin{array}{lcl} D_1 = A_1 \cap A_2 \cap A_3 \cap \dotsb &\quad\subseteq\qquad A_1 \qquad &\subseteq \qquad E_1 = A_1 \cup A_2 \cup A_3 \cup \dotsb \\ D_2 = \phantom{A_1 \cap{}} A_2 \cap A_3 \cap \dotsb &\quad\subseteq\qquad A_2 \qquad &\subseteq \qquad E_2 = \phantom{A_1 \cup{}} A_2 \cup A_3 \cup \dotsb \\ D_3 = \phantom{A_1 \cap A_2 \cap{}} A_3 \cap \dotsb &\quad\subseteq\qquad A_3 \qquad &\subseteq \qquad E_3 = \phantom{A_1 \cup A_2 \cup{}} A_3 \cup \dotsb \\ & \qquad\vdots & \end{array} $$ Moreover, $(D_n)$ is the largest increasing sequence such that $D_n \subseteq A_n$ for all $n$, and $(E_n)$ is the smallest decreasing sequence such that $A_n \subseteq E_n$ for all $n$, so it makes sense to define $$ \liminf_{n\to\infty} A_n = \lim_{n\to\infty} D_n ,\qquad \limsup_{n\to\infty} A_n = \lim_{n\to\infty} E_n . $$ And if these lower and upper limits are equal, then we say that that's $\lim\limits_{n\to\infty} A_n$.

(What Willie Wong's answer says is that the same construction makes sense in any partially ordered set where you can take the least upper bound and greatest lower bound of infinitely many elements, if you replace $(\subseteq,\cap,\cup)$ by $(\le,\wedge,\vee)$.)