# $\limsup$ and $\liminf$ of a sequence of subsets relative to a topology

From Wikipedia

if $\{A_n\}$ is a sequence of subsets of a topological space $X$, then:

$\limsup A_n$, which is also called the outer limit, consists of those elements which are limits of points in $A_n$ taken from (countably) infinitely many n. That is, $x \in \limsup A_n$ if and only if there exists a sequence of points $\{x_k\}$ and a subsequence $\{A_{n_k}\}$ of $\{A_n\}$ such that $x_k \in A_{n_k}$ and $x_k \rightarrow x$ as $k \rightarrow \infty$.

$\liminf A_n$, which is also called the inner limit, consists of those elements which are limits of points in $A_n$ for all but finitely many n (i.e., cofinitely many n). That is, $x \in \liminf A_n$ if and only if there exists a sequence of points $\{x_k\}$ such that $x_k \in A_k$ and $x_k \rightarrow x$ as $k \rightarrow \infty$.

According to the above definitions (or what you think is right), my questions are:

1. Is $\liminf_{n} A_n \subseteq \limsup_{n} A_n$?
2. Is $(\liminf_{n} A_n)^c = \limsup_{n} A_n^c$?
3. Is $\liminf_{n} A_n = \bigcup_{n=1}^\infty\overline{\bigcap_{m=n}^\infty A_m}$? This is based on the comment by Pantelis Sopasakis following my previous question.
4. Is $\limsup_{n} A_n = \bigcap_{n=1}^\infty\overline{\bigcup_{m=n}^\infty A_m}$, or $\limsup_{n} A_n = \bigcap_{n=1}^\infty\operatorname{interior}(\bigcup_{m=n}^\infty A_m)$, or ...?

Thanks and regards!

• I didn't know that definition but beware of the fact that there is another (commonly used, in particular in probability theory) definition of limsup and liminf that does not require any topology. also 1. is clearly true. Nov 22, 2011 at 16:21
• @Glougloubarbaki: Thanks! What is the other definition of limsup and of liminf you referred to, particularly in probability theory? I just want to see if it is the same as in my previous question linked.
– Tim
Nov 22, 2011 at 16:27
• Wikipedia article on Kuratowski convergence might be interesting for you, too. Nov 22, 2011 at 16:38
• @tim the definition i'm talking about is mentionned in the answer from artura magidin to your previous post : $\limsup_{n \in \mathbb{N}} X_n = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} X_m$ and similarly for the liminf. the main use I know is the Borel Cantelli lemma, which is a very useful tool. Nov 22, 2011 at 18:20
• @MartinSleziak: Thanks! Now I have the same exact question here math.stackexchange.com/questions/120885/….
– Tim
Mar 16, 2012 at 12:31

I must admit that I did not know these definitions, either.

1. Yes, because if $x \in \liminf A_n$ you have a sequence $\{x_k\}$ with $x_k \in A_k$ and $x_k \rightarrow x$ and you can choose your subsequence $\{A_{n_k}\}$ to be your whole sequence $\{A_n\}$.

2. For $X = \mathbb{R}$ take $A_n = \{0\}$ for all $n$. Then you have $(\liminf A_n)^c = \{0\}^c = \mathbb{R}\setminus{\{0\}}$, but $\limsup A_n^c = \mathbb{R}$. So in general you do not have equality.

3. Of course you have "$\supseteq$". But if you take $A_n := [-1,1- \frac{1}{n}) \subseteq \mathbb{R}$ you have the sequence $\{x_k\}$ defined by $x_k := 1 - \frac{2}{k}$ which converges to $1$ which is not in $\overline{\bigcap_{n = m}^{\infty} A_n} = [-1,1-1/m]$ for any $m \in \mathbb{N}$. But it works of course, if $\{A_k\}$ is decreasing.

4. In your first assumption you have "$\subseteq$". A sequence $\{x_k\}$ with $x_k \in A_{n_k}$ with $\{A_{n_k}\}$ some subsequence of $\{A_n\}$ lies eventually in $\bigcup_{n = m}^\infty A_n$ for all $m$ by the definition of subsequence. Now for metric spaces you have "$\supset$", too, as you can start by choosing $x_1 \in A_{n_1}$ with $d(x_1,x) < 1$ and proceed by induction choosing $x_{n_k} \in A_{n_k}$ with $n_k > n_{k-1}$ and $d(x_{n_k},x) < \frac{1}{k}$. I doubt that this holds in general (at least this way of proving it does not).

• Thanks! I wonder if $\liminf_n A_n$ and $\limsup_n A_n$ as defined in Wikipedia can be equivalently defined as something as clean as $\bigcup_{n=1}^\infty\overline{\bigcap_{m=n}^\infty A_m}$, in other words, in some formulas using union, intersection, taking closure, and/or taking interior, ...?
– Tim
Nov 22, 2011 at 17:04
• (1) Based on your counterexample in 3, is $\liminf_{n} A_n = \overline{\bigcup_{n=1}^\infty\overline{\bigcap_{m=n}^\infty A_m}}$? (2) I don't know how to modify 4 for $\limsup_{n} A_n$. Perhaps $\limsup_{n} A_n = \operatorname{interior}(\bigcap_{n=1}^\infty\operatorname{interior}(\bigcup_{m=n}^\infty A_m))$?
– Tim
Nov 22, 2011 at 17:14
• (1) I just realized that taking $A_k = \emptyset$ for some $k \in \mathbb{N}$ always gives you $\liminf A_n = \emptyset$. (2) Take singelton sets in $\mathbb{R}$. Nov 22, 2011 at 17:34
• If $A_n=A$ is constant, then $\limsup A_n^c = A^c$. So (2) is true Jul 4, 2018 at 16:10

First of all, I'd like to notice that the closure of a set $A \subset X$ is NOT, in general, the same as the set of all limit points (for sequences) of $A$. Nevertheless, given the topology $\tau$ on $X$, you can always define another topology where the closed sets are those that contains all its limit points (for sequences). This new topology is stronger (or equal) the original one, since the closed sets contains all its limit points. Let's assume our topology is of this type!

Regarding the notions for $\limsup$ and $\liminf$, notice that whenever you have a countably complete lattice you have the $\limsup$ and $\liminf$ for sequences. That is, if you have an order where every "growing" sequence of elements $x_n$ have a limit: $\sup x_n$. Similarly, every "decreasing" sequence of elements $x_n$ have a limit: $\inf x_n$. If you notice that $y_n = \inf_{j \geq n} x_n$ is a "growing" sequence, then, you will realise that it does have a limit: $\lim y_n = \sup_n \inf_{j \geq n} x_n$ This is the $\liminf x_n$. That is, $$\liminf x_n = \sup_{n}\, \inf_{j \geq n} x_j.$$ Analogously, $$\limsup x_n = \inf_{n}\, \sup_{j \geq n} x_j.$$ You say that $x_n$ has a limit when $\limsup x_n = \liminf x_n$.

One of the lattices one might be interested is the power set of $X$: $\mathcal{P} = \mathcal{P}(X)$. That is, the family of all subsets of $X$. This is a lattice where $\inf A_\lambda = \bigcap A_\lambda$ and $\sup A_\lambda = \bigcup A_\lambda$. Here, I use $\lambda$ to emphasise that the family might not be countable.

If you have a $\mathcal{L} \subset \mathcal{P}$, it might happen that for $A_n \in \mathcal{L}$, $\sup A_n \not \in \mathcal{L}$, or $\inf A_n \not \in \mathcal{L}$. However, if $\inf A_\lambda \in \mathcal{L}$ for every family $A_\lambda$, then we still have a supremum in $\mathcal{L}$ for $A_n$. To avoid confusion, let's denote the supremum and infimum in $\mathcal{L}$ by $\bigvee$ and $\bigwedge$. That is, $$\bigwedge A_n = \inf A_n,$$ and $$\bigvee A_n = \inf_{\substack{A \in \mathcal{L}\\A \supset \sup A_n}} A_n.$$ This is what we do if we have a family of groups, or a family of topologies, for example. If $X = G$ is a group, then the intersection of any family of subgroups is a group. But the union is not. The group generated by $H_\lambda \subset G$ is not $\sup H_\lambda$, but rather, $\bigvee H_\lambda$; that is, the smallest subgroup that contains all $H_\lambda$. Notice that the "generated group" could be determined for an arbitrary family of sets $S_\lambda$ instead of $H_\lambda$. But notice that $\bigvee S_\lambda$ would be just the same as $\bigvee \langle S_\lambda \rangle$, where $\langle S_\lambda \rangle = \bigwedge_{H \supset S_\lambda} H$. Nevertheless, there is a problem if we want to define $\bigwedge S_\lambda$. We could say that it is the smallest group that contains $\bigcap S_\lambda$, that is, $$\bigwedge S_\lambda = \bigwedge_{H \supset \bigcap S_\lambda} H;$$ or, we could say that it is $$\bigwedge S_\lambda = \bigwedge \langle S_\lambda \rangle.$$ Those sets might not be the same! (why?) But they will be the same if $S_\lambda$ is a subgroup. That is, if $S_\lambda = \langle S_\lambda \rangle$. This ambiguity also makes it ambiguous to talk about the $\liminf$, but not about $\limsup$ for $S_\lambda$ in the lattice of subgroups of $G$. (why?)

Now, instead of a group, let's consider the family given by the closed sets mentioned at the beginning of this answer. Given $A \subset X$, consider $\langle A \rangle = \overline{A}$. That is, $\langle \cdot \rangle$ is simply the closure operation, or, the set of all limit points of sequences in $A$. Then, for this lattice of closed sets, we can define \begin{align*} \liminf(1) A_n &= \liminf \overline{A}_n \\&= \bigvee_{n} \bigwedge_{j \geq n} \overline{A}_j \\&= \bigvee_{n} \bigcap_{j \geq n} \overline{A}_j \\&= \overline{\bigcup_{n} \bigcap_{j \geq n} \overline{A}_j}. \end{align*} But we could also define \begin{align*} \liminf(2) A_n &= \bigvee_{n} \bigwedge_{F \supset \cap_{j \geq n} A_j} F \\&= \bigvee_{n} \overline{\bigcap_{j \geq n} A_j} \\&= \overline{\bigcup_{n} \overline{\bigcap_{j \geq n} A_j}} \\&= \overline{\bigcup_{n} \bigcap_{j \geq n} A_j}. \end{align*} These two are not the same! But it is clear that $\liminf(2) A_n \subset \liminf(1) A_n$.

Now, notice that in the definition of $\liminf$ in your question, $x \in \liminf A_n$ iff for any neighbourhood $V$ of $x$, you have that $V \cap A_j$ for all $j \geq n$ for a certain $n$. That happens iff for some $n$, $V \cap \bigcap_{j \geq n} \overline{A}_j \neq \emptyset$. And this just means that $$x \in \overline{\bigcup_{n} \bigcap_{j \geq n} \overline{A}_j}.$$ That is, the $\liminf$ in the question is exactly $\liminf(1)$.

Using the same argument, one can prove that the $\limsup$ in your question is the same as the $\limsup$ in this answer.

Notice that your statement "2" is not correct because while $\limsup$ is a closed set, the complement of $\liminf$ is an open set! Also, you could do the same construction with the "open sets". This would yield a not ambiguous $\liminf$ and an ambiguous $\limsup$. I guess that taking $\limsup$ for the closed sets and $\liminf$ for the open sets would yield the validity of statement "2". I think this is what you want in statement "4", when you use the "interior" of a set.

Edit: Fixed definition of $\langle \cdot \rangle$.

• Thank you for that insight, I totally did not take into account that closure and sequential closure were different. But your $\liminf (1)$ seems still to be different from the one in the question for me if you take the empty set for one of the $A_k$s (you fail to get $x_k \in A_k$). Any idea? Nov 23, 2011 at 6:31
• @MatthiasKlupsch: Interesting. It seems that what comes after the "that is" in the definition of $\liminf$ is not the same as the definition stated before the "that is": - $\liminf$ [...] consists of those elements which are limits of points in $A_n$ for all but finitely many $n$ [...] Nov 23, 2011 at 13:00
• @MatthiasKlupsch: and the same happens for $\limsup$! Nov 23, 2011 at 13:02
• @AndréCaldas: Thanks! Do you mean $\langle S_\lambda \rangle = \bigwedge_{H \supset S_\lambda} H$ instead of $\langle S_\lambda \rangle = \bigwedge_{H \supset \bigcup S_\lambda} H$?
– Tim
Nov 23, 2011 at 14:24
• @Tim: Yes. I will edit it. I am happy you read it! :-) Nov 23, 2011 at 14:31