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I'm little confused about the whole concept of limits of sets. I'm working with the definition:

$$\liminf_{n\to\infty}(A_n) := \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}(A_k).$$

As an example consider the sequence of sets $A_n=\{2k | k\ge n\}$, for all $n \in \mathbb{N}$. I'm assuming all the sets $A_n$ include $+\infty$ (I'm not sure my notation is correct for that).

Then, for all $m \in \mathbb{N}$, the $\sup_{n\ge m}(A_n) = +\infty$, and so: $$\limsup_{n\to\infty}(A_n) = \{+\infty\}.$$

However $\inf_{n\ge m}(A_n) \ne {+\infty}$, because for every $n$, $$\inf (A_n) = 2n < +\infty,$$ which would lead to a contradiction. Given that, it seems $$\liminf_{n\to\infty}(A_n)=\varnothing.$$

But this is wrong because $\{-\infty\} \notin \varnothing$. Could someone please help me understand where I went wrong in my argument?

Davide Giraudo
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Numik
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    If we are talking of limsup and liminf of sets, then $\limsup A_n = \cap_{n\ge 1}\cup_{m\ge n} A_m$. You need to consider unions and intersections of sets. This is not related to the order of the numbers contained in these sets. – Gribouillis May 22 '23 at 12:25
  • Please include (in the body of the question) the definition of what $\liminf_{n\to\infty}(S_n)$ means for a sequence of sets $S_n$. (It would help if you say where you got the definition.) When the definition is known, we can talk about how to apply it. – David K May 22 '23 at 12:30
  • Edited now @DavidK. I'm confused by the other comment because if liminf(An) is the event that all but finitely many $A_n$ occur then this should be null but it seems to be $+\infty$ (or is it ${+\infty}$?) – Numik May 22 '23 at 12:43
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    The two "definitions" that you wrote are not at all equivalent. – Gribouillis May 22 '23 at 12:45
  • That would explain my confusion @Gribouillis. These were stated in my lectures as equivalent. What's the difference then? – Numik May 22 '23 at 12:50
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    I think you are jumbling the definition of liminf for sets with the definition of liminf for sequences of numbers. In this case, only the first definition (for sets) is relevant, so work with this. – Gribouillis May 22 '23 at 12:53
  • It sounds like you're still trying to apply the wrong notion of limits. This problem has nothing to do with $\pm\infty$, because we are not comparing numbers using $<$. Forget about the relative sizes of individual numbers and use the union/intersection definition. – Karl May 22 '23 at 14:17
  • Perhaps of interest: https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Hans Lundmark May 22 '23 at 15:02

1 Answers1

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It may be useful to point out that the definitions "supremum" and "infimum" make sense in any partially ordered set $P$: If $S \subset P$ is any subset of the poset $P$, then $\mathrm{sup}(S)$ (if it exists) is the unique element $t\in P$ such that

(i) $s \leq t$ for all $s \in S$ (i.e. it is an upper bound for S) and

(ii) if $t'$ is any element of $P$ satisfying $s\leq t'$ for all $s\in S$ then $t\leq t'$ (i.e. it is the least upper bound).

You can define the infimum in the same way. If, as the poset, you take the power set $\mathcal P(X)$ of a set $X$, i.e. the set of all subsets of $X$, where the partial order is given by containment, then if $\{S_i:i \in I\}$ is a subset of $\mathcal P(X)$, clearly $\mathrm{sup}\{S_i: i\in I\} = \bigcup_{i \in I} S_i$, and similarly $\mathrm{inf}\{S_i: i \in I\} = \bigcap_{i \in I} S_i$.

If we have a sequence $(A_n)$ of subsets of $X$, then $$ \mathrm{limsup}_n A_n = \mathrm{inf}_n(\mathrm{sup}\{A_k: k \geq n\}) = \bigcap_{n \geq 0}\bigcup_{k \geq n}A_k $$ and similarly $$ \mathrm{liminf}_n (A_n) = \mathrm{sup}_n(\mathrm{inf}\{A_k: k \geq n\}) = \bigcup_{n \geq 0}\bigcap_{k \geq n}A_k $$

The $\mathrm{limsup}_n(S_n)$ are elements of $X$ which occur in infinitely many of the $S_n$s, while $\mathrm{liminf}_n(A_n)$ are the elements of $X$ which occur in all but finitely many of the $A_n$s.

If you take $X = \mathbb N$ and $A_n = \{2k: k \geq n\}$ then $\mathrm{limsup}_n(A_n) = \mathrm{liminf}_n(A_n) = \emptyset$, since any even integer $2k$ lies in only finitely many of the subsets $\{A_n\}$.

The $\mathrm{limsup}$ and $\mathrm{liminf}$ of sequences of real numbers is a completely different notion, using the total order on the real numbers rather than partial order of containment.

krm2233
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  • Thank you, that's a very useful way to put it. Now I see why the notation is similar even though the notion is quite different. – Numik May 22 '23 at 15:04
  • There is one way to relate the limsup/liminf of subsets of $X$ to the limsup/liminf of sequences: Given a sequence $(A_n){n \geq 0}$ of subsets of a set $X$, let $f_n=1{A_n}$ be the indicator function of $A_k$, so that $f_n(x)=1$ if $x \in A_n$ and is $0$ otherwise. Then for any $x\in X$ we obtain a sequence $(f_n(x))$, and $$\mathrm{liminf}n f_n(x) = 1{\mathrm{liminf}n(A_n)}(x), \quad \mathrm{limsup}_n f_n(x) = 1{\mathrm{limsup}_n(A_n)}(x)$$. – krm2233 May 23 '23 at 22:19