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For my statistics class we had elementary set theory.

It was stated that:

$$\inf_{k\geq n } A_k = \bigcap\limits_{k=n}^{\infty} A_k$$

and

$$\sup_{k\geq n } A_k = \bigcup\limits_{k=n}^{\infty} A_k$$

From this was deduced that:

$$\lim\limits_{n\to\infty} \inf A_k = \bigcup\limits_{n=1}^{\infty} \bigcap\limits_{k=n}^{\infty} A_k$$

and

$$\lim\limits_{n\to\infty} \sup A_k = \bigcap\limits_{n=1}^{\infty} \bigcup\limits_{k=n}^{\infty} A_k$$

I absolutely have no idea why. Could someone explain it to me in the least technical way possible? I neither get why the intersection of Ak from n onwards should be the infimum nor why the union of all intersections should be the limit of that infimum.

M. Winter
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Carol Eisen
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2 Answers2

14

It is clear that for real numbers, e.g. $3$ and $8$, that the minimum is the smaller one, i.e. $3$. There is a way to talk about smaller sets: a set $A$ is "smaller" than a set $B$ if it is completely contained in it, i.e. $A\subseteq B$. This is natural somehow, as e.g. $\{1,2,3\}$ is smaller than $\{1,2,3,4,5,6\}$. In this case, check that the intersection gives

$$A\cap B=A.$$

This means that the intersection somehow acts like the minimum function on sets. By the same reasoning, you can argue that the union gives $A\cup B=B$ and acts like a maximum function. It is a bit less natural if neither $A\subseteq B$ nor $B\subseteq A$, but above should deal as a motivation.

Because $\inf$ and $\sup$ generalizes minimum and maximum for infinitely many numbers (and sets in our case), it is natural to define

$$\inf \,\{A_k\}=\bigcap A_k,\qquad \sup \,\{A_k\}=\bigcup A_k.$$

Now remember that $\liminf_{n\to\infty}$ was defined as $\sup_{n\in\Bbb N}\inf_{k\ge n}$. This hopefully explains why the formula for $\liminf$ contains both a union and an intersection.

M. Winter
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  • Thanks for the explanation. It makes it much clearer to me. thanks! One small thing: shouldn't A intersection B equal A and not B if A is fully contained in B? – Carol Eisen Oct 12 '17 at 13:36
  • @CarolEisen Of course! Sorry for this avoidable mistake. And thank you to the editor, too! – M. Winter Oct 12 '17 at 15:01
  • Thanks a lot for the answer. It was very helpful. My apologies for the duplicate ! – Carol Eisen Oct 16 '17 at 10:33
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If $(x_n)$ is an increasing sequence of real numbers, then it converges (possibly to $\infty$) and $\lim_{n\to \infty} x_n = \sup_{x\in \mathbb N} x_n$. Now given any sequence $(x_n)$, the sequence $(\inf_{k\geq n} x_k)_{n\in \mathbb N}$ is increasing, so $\lim_{n\to \infty} \inf_{k\geq n} x_k = \sup_{n\in \mathbb N} \inf_{k\geq n} x_k$.

Now you apply the same ideas to subsets of a set substituting $\inf$ with $\bigcap$ and $\sup$ with $\bigcup$.

Stefan Perko
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