# Further clarification on the connection between limsup and liminf for sequences of sets and real numbers

Big picture goal: I am trying to reconcile the difference between between the definition of $\liminf,\limsup$ of a sequence of real numbers and a sequence of sets for the special case of the discrete metric (as per wikipedia).

liminf: I know that liminf can be thought of as a union of a sequence of intersections: $\liminf = \bigcup_{N \geq 1} ( \underbrace{\bigcap_{n\geq N} A_n}_{I_n} )$, where $I_n$ is non-decreasing: $I_n \subseteq I_{n+1}$, and so we can easily see that it can be written as: $\liminf A_n = \{A_n :$ for all except finitely many $\}$.

limsup: I also know that the limsup can be thought of as: $\limsup = \bigcap (\underbrace{\bigcup A_n}_{J_n})$, where $J_n$ is non-increasing: $J_n \supseteq J_{n+1}$.

Question: Why can we write this as $\limsup A_n = \{A_n:$ infinitely often $\}$? giving us a possibly infinite set where it doesnt belong? In other words, I see it at this:

$\liminf A_n = \{A_n: \text{true infinitely often + finitely false} \}$
$\limsup A_n = \{A_n: \text{true infinitely often} \}$

or rather:

$\limsup A_n = \underbrace{\{A_n: \text{i.o + finitely false} \}}_{\liminf} \bigcup \underbrace{\{A_n: \text{i.o + infinitely false} \}}_{???}$

but am unsure as to how thats the case for the $\limsup$? Wikipedia says: "So the limit supremum is contained in all subsets which are upper bounds for all except finitely many sets of the sequence", so it seems we should also be able to make a finite-type statement for limsups as well?

I think the source of my confusion is trying to start from this general definition on sets under the discrete metric, where we can make two statements for $\liminf$ (i.o and finitely false) and only one statement for $\limsup$ (i.o), to the case of sequences of real numbers where we can make two for each! For example, as in this "alternate definition":

A number $t$ is the limit superior of a sequence $\langle a_n\rangle$ if the following two conditions are both satisfied:

• For every $s<t$ we have $s<a_n$ for infinitely many $n$'s.

• For every $s>t$ we have $s<a_n$ for only finitely many $n$'s (possibly none).

Similarly, a number $t$ is the limit inferior of a sequence $\langle > a_n\rangle$ if the following two conditions are both satisfied:

• For every $s>t$ we have $s>a_n$ for infinitely many $n$'s.

• For every $s<t$ we have $s>a_n$ for only finitely many $n$'s (possibly none).

Some great questions clarifying the definition in either specific domains (too many to link)

## 1 Answer

One key remark is that it is equivalent to say of a set of natural numbers that it is unbounded or that it is infinite (or dually, that it is bounded is equivalent to saying that it is finite).

For any fixed $x$, consider the truth value $[x\in A_n]$, as a function of $n$, of the statement "$x \in A_n$" (with the value $1$ standing for "true" and $0$ for "false). There are three exclusive possibilities (in classical logic...):

• either $[x\in A_n]$ is eventually $0$ (meaning $\exists N\, \forall n\geq N\,(x \not\in A_n)$), or equivalently, takes the value $1$ only finitely many times (meaning that $x$ belongs to only finitely many $A_n$'s),

• or $[x\in A_n]$ takes both the values $0$ and $1$ for arbitrarily large $n$, or equivalently, for infinitely many $n$'s,

• or $[x\in A_n]$ is eventually $1$ (meaning $\exists N\, \forall n\geq N\, (x \in A_n)$), or equivalently, takes the value $0$ only finitely many times (meaning that $x$ fails to belong to only finitely many $A_n$'s).

In the last case, we say that $x$ belongs to the lim inf of the $A_n$'s, and $\exists N\, \forall n\geq N\,(x \in A_n)$ translates exactly into $x \in \bigcup_{N\geq 0}\bigcap_{n\geq N} A_n$. In the last two cases, we say that $x$ belongs to the lim sup of the $A_n$, and the negation of the first case ($\exists N\, \forall n\geq N\,(x \not\in A_n)$), i.e. $\forall N\, \exists n\geq N\,(x\in A_n)$ translates exactly into $x \in \bigcap_{N\geq 0}\bigcup_{n\geq N} A_n$.

It may seem oddly asymmetric that lim sup comprises the two last cases whereas lim inf comprises only one. But this is necessarily the case if we want lim sup to be a superset of lim inf (so it can be broken down into the disjoint union of lim inf, viz. the last case, and the relative difference, viz. the middle case). The reason for this asymmetry is that we are dealing with $A_n$ rather than their complements (assume everything lives into some fixed large set): if we replace the $A_n$ by their complements, then lim sup and lim inf are interchanged (more precisely, the lim sup of the complement is the complement of the lim inf and vice versa). The point of using $0$ and $1$ truth values instead is that the symmetry is less obscured.

In the case of the lim sup and lim inf of a sequence of real numbers, the symmetry is more apparent because we can replace an inequality such as $s<t$ by $s>t$ (inverting the order on the reals, i.e., passing to the negative interchanges lim sup with lim inf), which is more transparent than if we replace $x\in A_n$ by $x\not\in A_n$.

But here are some ways the two notions (lim inf and lim sup of sets and reals) can be connected which perhaps makes the symmetry more apparent:

• $[x \in \limsup A_n] = \limsup [x\in A_n]$ and $[x \in \liminf A_n] = \liminf [x\in A_n]$. This makes it possible to define the lim sup and lim inf of a sequence of sets using the lim sup and lim inf of a sequence of reals, here the truth values $[x\in A_n]$.

• $t = \limsup a_n$ (with $a_n$ a sequence of reals) iff $(-\infty,t) \subseteq \limsup (-\infty,a_n) \subseteq (-\infty,t]$ (or equivalently $(-\infty,t) \subseteq \limsup (-\infty,a_n] \subseteq (-\infty,t]$, or, passing to complements: $(t,+\infty) \subseteq \liminf (a_n,+\infty) \subseteq [t,+\infty)$, or again: $(t,+\infty) \subseteq \liminf [a_n,+\infty) \subseteq [t,+\infty)$). This makes it possible to define the lim sup of a sequence of reals using the lim sup or lim inf of a sequence of sets (half-lines).

• $t = \liminf a_n$ (with $a_n$ a sequence of reals) iff $(-\infty,t) \subseteq \liminf (-\infty,a_n) \subseteq (-\infty,t]$ (again rewriteable in many different forms: $(-\infty,t) \subseteq \liminf (-\infty,a_n] \subseteq (-\infty,t]$, etc.).

• im sorry Im still confused by your answer – user79950 Jan 13 '16 at 5:20
• OK, well if you can't explain what's confusing you, let me try a very short answer instead: "true infinitely often and false finitely often" is redundant, you might as well just say "false finitely often" or equivalently "eventually true" (my third bullet point), which implies "true infinitely often". It's one statement, not two (notwithstanding the fact that counting statements doesn't mean anything…). – Gro-Tsen Jan 13 '16 at 6:51
• I am just trying to disambiguate when were dealing with sets, the way limsup can be (true i.o + false +i.o) or (fininately false), while liminf can only be finitely false, and how that situation changes/reduces when were dealing with real numbers – user79950 Jan 13 '16 at 18:18
• Are you bothered by the lack of symmetry? Since lim.sup ("true i.o.") is a superset of lim.inf, of course you can break it in two, namely lim.inf ("false f.o.") and the relative difference ("true i.o. and false i.o."). To restore the symmetry, you have to switch to complements (assume everything lives in some fixed large set): the complement of lim.sup is lim.inf of the complements, and vice versa. So yes, you're right in that there's a lack of symmetry, which comes from focusing on the sets rather than their complements, or from trying to write "true i.o." as a disjunction rather than conj. – Gro-Tsen Jan 13 '16 at 19:37
• I guess I'm bothered by it becasue I dont see where it comes from. Could you elaborate on that in your answer and Ill mark it accepted? Thanks for the help! – user79950 Jan 13 '16 at 19:52