# limsup and liminf of a sequence of points in a set

My ways to define/write limsup and liminf of a sequence of points in a set $X$:

They come from what I have understood. If you have other ways of understanding, really appreciate if you can reply here.

1. When $X$ is a complete lattice, then $$\limsup \, x_n := \inf_{n \geq 0} \left(\sup_{m \geq n} \, x_m\right) = \mathop{\wedge}\limits_{n \geq 0}\left( \mathop{\vee}\limits_{m\ \geq n} \, x_m\right).$$
2. When $X$ is a topological space and also a partially ordered set (can be strengthened to be complete in some sense, such as a complete lattice if needed), then there are different ways to define/write:

• (1) $\displaystyle\limsup \, x_n := \sup \{\text{limit of any subsequence of }\{x_n\}\text{ that converges}\}$
• (2) $\displaystyle\limsup \, x_n := \lim_{n \rightarrow \infty} ( \sup_{m \geq n}x_m )$.

$\liminf x_n$ is defined similarly.

Questions:

1. Are there some conditions on $X$ that can make 2(1) and 2(2) equivalent?
2. Are there some conditions on $X$ that can make 1 and 2(1) equivalent? For example, when $X$ is both a topological space and a complete lattice, I can define limsup in both 1 and 2(1) ways, so when will they agree?
3. Are there other definitions/forms of limsup of a sequence of points in a set $X$ possibly with some structures? Can they be related/shown to be equivalent to the definitions/forms given above?
4. Is 2(1) exactly the specialization from limsup of a filter base in $X$ to limsup of a sequence of points in $X$: $$\limsup \, x_n := \sup \cap \{\bar{B}_0 | B_0 \in B \}$$ where $B:=\{ \{x_n | n \geq n_0 \} \forall n_0 \in \mathbb{N} \}$, and $\bar{B}_0$ is the closure of $B_0$?
5. When $X$ is a topological space and ordered, if $\limsup \, x_n = \liminf \, x_n$, both of which are defined in the same way as 2(1), what is the condition for $\limsup \, x_n = \liminf \, x_n$ to be the same as $\lim \, x_n$ defined with respect to the topology?

Thanks and regards!

As to Question 4, I think it is. $\forall B_0 \in B$, it is a tail sequence of $\{ x_n \}$, so any accumulation point of $B_0$ will be an accumulation point of any other tail sequence, and an accumulation point of the original sequence $\{ x_n \}$. Therefore \begin{align*} \cap \{\bar{B}_0 | B_0 \in B \} &= \{\text{accumulation point of }\{ x_n \} \}\\ &= \{\text{limit of any subsequence of }\{x_n\}\text{ that converges}\}. \end{align*} Please correct me if I am wrong.

Arturo, really appreciate your reply! It can take me a while to fully understand what you wrote.

I think I will probably put the following into a new post if it is not proper of adding questions here given the length of this post. But anyway, here they are:

1. In definition 1, when will $\limsup x_n = \liminf x_n$?
2. In definition 2(1), when will $\limsup x_n = \liminf x_n$?

Basically, I would like to know when one can define "limit" of a sequence in a set, not solely depending on or even independent of topology.

• I think that if you have the order topology on a totally ordered set, then 2(1) and 2(2) are equivalent (in that one exists if and only if the other does, in which case they are equal). I suspect the same answer holds for 2, or at least if there is sufficient relation between the lattice order and the topology (it may be enough for one to be finer than the other, but they should be related somehow). – Arturo Magidin Jan 13 '11 at 16:29

I suspect that if you have a complete lattice $X$ and you endow it with the order topology (where a subbasis of the topology is given by the sets of the form $(a,\infty) = \{x\in X\mid a\lt x\}$ and $(-\infty,b) = \{x\in X\mid x\lt b\}$ with $a,b\in X$ arbitrary), then 1, 2(1), and 2(2) will be equivalent in the sense that one exists if and only if the others exists and in that case they will be equal (the only caveat is that we have to be careful with 2(1) in the case of unbounded sequences; technically, for instance, if you take $1,2,1,3,1,4,1,5,1,6,\ldots$, the only converging subsequences converge to $1$, but the sups are never defined, so you could interpret 2(1) as saying that the limit superior is $1$, and 2(2) as saying it does not exists).

However, I am having trouble proving it in full generality; the equivalence of 2(1) and 2(2) is not difficult if you assume the set is totally ordered and complete, so that suprema exist for any bounded set (throw in an über-sup and an über-inf like $\infty$ and $-\infty$ to the reals if necessary). I'm somewhat unhappy about this, but I figure something is better than nothing here. I've been toying with this and going down a fair number of silly alleys along the way, so maybe after it is written down I'll see that it works in any lattice; however, right now I am using the fact that any sequence contains a monotone subsequence, and this requires total order, not just lattice order. (I'm also using the fact that the order topology is Hausdorff for totally ordered sets; I don't know if that works for lattices).

Note that a totally ordered set that has the order topology is necessarily Hausdorff: given $x\neq y$, say $x\lt y$. If there exists $z$ such that $x\lt z\lt y$, then the sets $(-\infty,z)$ and $(z,\infty)$ provide disjoint open neighborhoods of $x$ and $y$, respectively; if there is no element of $X$ strictly between $x$ and $y$, then $(-\infty,y)$ and $(x,\infty)$ provide the necessary disjoint neighborhoods.

Theorem. Suppose $X$ is a totally ordered set such that every nonempty subset has a supremum. If we endow $X$ with the order topology, then definitions 2(1) and 2(2) are equivalent.

Proof. Say $\lim\limits_{n\to\infty}\sup\{x_m\mid m\geq n\}$ is equal to $L$. This means that for every open neighborhood $U$ of $L$, there exists a $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$, and in particular for every $k\geq K$ there exists $m\geq k$ such that $x_m\in U$.

First, I claim that the set of limit points of subsequences of $x_n$ is bounded above by $L$. Indeed, suppose that you have a subsequence that converges to $M$, with $M\gt L$. Let $U$ and $V$ be neighborhoods of $L$ and $M$ that are disjoint; intersect $U$ with $(-\infty,M)$ and $V$ with $(L,\infty)$, if necessary. Then there exists $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$; in particular, for all $k\geq K$, we have $x_k \leq\sup\{x_m\mid m\geq k\}\notin V$; therefore, since the sequence is eventually not in $V$, it is not infinitely often in $V$, and therefore there can be no subsequence that converges to $M$. This contradicts the choice of $M$, so we conclude that $M\leq L$, as claimed.

Since every sequence contains a monotone subsequence, and monotone subsequences in complete sets have limits (their sup if they are increasing, their inf if they are decreasing), the set of limit points of subsequences is nonempty; therefore, since it is bounded above by $L$, we know that $$\sup\{\text{limit of any subsequence of }\{x_n\}\text{ that converges}\} \leq L=\lim_{n\to\infty}\sup\{x_m\mid m\geq n\}.$$

Suppose that the supremum of the limits is $M$, and is strictly smaller than $L$; let $V$ be an open neighborhood of $M$ contained in $(-\infty,L)$, $U$ an open neighborhood of $L$ contained in $(M,\infty)$, with $U\cap V=\emptyset$. We know that there exists $K$ such that for all $k\geq K$, $\sup\{x_m\mid m\geq k\}\in U$; in particular, for all $k\geq K$ there exists $m\geq k$ such that $x_m\in U$. Thus, we can find a subsequence of $\{x_n\}$ that is contained in $U$; this subsequence contains a monotone subsequence, which therefore coverges to some point in the closure of $V$; the closure of $V$ is contained in the complement of $U$ and of $(-\infty,M)$, and therefore the closure of $V$ does not include $M$ and consists only of points strictly larger than $M$; that is, the limit of this monotone subsequence is strictly larger than $M$, contradicting the choice of $M$. Thus, $M\geq L$, giving equality. QED

I don't know about 3 and 4, I'll think about 5, though again I expect you'll need the topology to be related to the order.

In any lattice, with definition 1, you always have $\liminf x_n\leq \limsup x_n$. To see this, note that for every $k,m$, you have $$\mathop{\wedge}\limits_{n\geq k} x_n \leq \mathop{\vee}\limits_{n\geq m}x_n$$ because if $r\geq \max\{k,m\}$, then you have $$\mathop{\wedge}\limits_{n\geq k}x_n \leq x_r \leq \mathop{\vee}\limits_{n\geq m}x_n.$$ Therefore, each $\mathop{\vee}\limits_{n\geq m}x_n$ is an upper bound for all of $\mathop{\wedge}\limits_{n\geq k}x_n$, hence for each $m$ we have $$\liminf x_n = \sup\mathop{\wedge}\limits_{n\geq k}x_n \leq \mathop{\vee}\limits_{n\geq m}x_n.$$ Therefore, $\liminf x_n$ is a lower bound for each of the joins, hence is less than or equal to their inf; that is, $$\liminf x_n \leq \inf\mathop{\vee}\limits_{n\geq m}x_n = \limsup x_n.$$ (The same is also true for definition 2(1), for silly reasons: the limit superior is the supremum of a particular nonempty set, the limit inferior is the infimum of that same set, and since the set is nonempty, then the infimum is less than or equal to the maximum).

Let $L=\liminf x_n$ and $M=\limsup x_n$.

Suppose that $\liminf x_n \lt \limsup x_n$, and that you endow $X$ with the order topology inherited from the complete lattice structure. Because $\liminf x_n$ and $\limsup x_n$ are comparable, then we can prove the existence of disjoint open neighborhoods that separate them the same way I did for a totally ordered set: if there is an element $z\in X$ such that $L\lt z\lt M$, then the open sets $(-\infty,z)$ and $(z,\infty)$ are disjoint open neighborhoods of $L$ and $M$, respectively; and if there is no such element, then $(-\infty,M)$ and $(L,\infty)$ are disjoint open neighborhoods that separate them. Taking disjoint neighborhoods of $L$ and $M$ you can obtain subsequences that are eventually not in the corresponding neighborhood, so if $L\lt M$ then there are subsequences that do not converge.

If $L=M$, on the other hand, then let $U$ be an open neighborhood of $L=M$ that is an interval; because the sequence $w_k = \wedge\{ x_m\mid m\geq k\}$ is nondecreasing, and $U$ is an interval containing $L$, there exists $K$ such that for all $k\geq K$ you have $w_k\in U$. Likewise, by considering the nonincreasing sequence $z_k\vee\{x_m\mid m\geq k\}$ which converges to $M$, there exists $H$ such that for all $h\geq H$ we have $z_h\in U$. Therefore, for every sufficiently large $k$ we have that both the meet and the join of $\{x_m\mid m\geq k\}$ lie in the interval $U$, then all the terms lie in the interval $U$; that is, the sequence is eventually in $U$. That is, given any open neighborhood of $L=M$, then the sequence is eventually in $U$; this means that it converges to $L=M$; and from here it is easy to verify that every subsequence also converges to $L$.

Therefore, $\limsup x_n = \liminf x_n$ if and only if there exists $L$ such that every subsequence converges to $L$, if and only if there exists $L$ such that the sequence converges (in the topological sense) to $L$.

I don't know if the limit needs to be unique (in a topology, you can have a sequence converge to two different things if the topology does not separate the points enough).

For 2(1), assuming the topology is the order topology, and $X$ is totally ordered, if $L\lt M$, then we can find intervals $(-\infty,a)$ and $(b,\infty)$ that are disjoint, $L\lt a$ and $b\lt M$, as above. Then there is a subsequence that converges to an element of $(-\infty,a)$ (in fact, of $[L,a)$) and a subsequence that converges to an element of $(b,\infty)$ (in fact, of $(b,M]$). Therefore, the original sequence is infinitely often in $(-\infty,a)$, and infinitely often in $(b,\infty)$; since these two are disjoint, the sequence cannot converge (if it converges to $x$, then for every open neighborhood of $x$ the sequence is eventually in the neighborhood; but separate $x$ from either $L$ or $M$ and get that the sequence is eventually not in the corresponding neighborhood of $x$). Thus, if $L\lt M$ then the sequence does not converge.

If $L=M$, then every converging subsequence converges to $L$; but if the set is not complete, then you'll run into trouble: take for example $X=\mathbb{Q}$, a sequence of rationals that converges to $\sqrt{2}$, and interweave it with the constant sequence $1$. If you take a subsequence, then either it contains infinitely many terms of the sequence that converges to $\sqrt{2}$, and therefore does not converge in $X$, or else it contains only finitely many terms of the sequence, and therefore converges to $1$; so the set of limits of converging subsequences is $\{1\}$, hence the limits inferior and superior are equal under definition (1), but the sequence does not converge; if the space $X$ is complete relative to the order topology (so that every set has a supremum and every set has an infimum), then we have the equivalence between definition 2(1) and 2(2) by the above, and then we are essentially in the situation of 1 that I just dealt with to show that the sequence converges to $L=M$ in the sense of 2(2), and hence in the sense of 2(1).