First we introduce the following notation: $$ \mathcal{N}_\infty:= \{N\subset \mathbb{N}| \mathbb{N} \text{\ }N \text{ is finite}\} $$ and $$ \mathcal{N}_\infty^\#:= \{N\subset \mathbb{N}| N \text{ is infinite}\} $$ In most textbooks of real analysis, the limit inferior is defined in one of the following two ways: $$ \liminf_n C_n = \bigcup_{n=1}^\infty \bigcap_{m=n}^\infty C_m $$ or $$ \liminf_n C_n = \left\{x \in \mathcal{X} | x\in C_k \text{ ultimately for all } k \right\} $$ We need to show that: $$ \liminf_n C_n = \bigcap_{N\in \mathcal{N}_\infty^\#} \overline{\bigcup_{n\in N}C_n} $$ where the overline denotes the set closure in the respective topology. For the limit superior we need to show that: $$ \limsup_n C_n = \bigcap_{N\in \mathcal{N}_\infty} \overline{\bigcup_{n\in N}C_n} $$ These properties appear in [p.110, 1] as exercises.

[1] R.T. Rockafellar and R. J-B. Wets, "Variational Analysis", Grundlehren der mathematischen Wissenschaften, vol. 317, 1998.

  • $\begingroup$ there is already a very similar question. maybe they should be merged $\endgroup$
    – Albert
    Nov 22, 2011 at 21:41
  • $\begingroup$ @Glougloubarbaki : If you know a very similar question perhaps you should post it here. $\endgroup$ Nov 22, 2011 at 22:00
  • $\begingroup$ @Pantelis : What have you tried? $\endgroup$ Nov 22, 2011 at 22:01
  • $\begingroup$ @PatrickDaSilva I have tried showing that the one set is included in the other. But I don't see how I should proceed. $\endgroup$ Nov 22, 2011 at 22:43
  • 1
    $\begingroup$ I've posted a similar comment to several questions related to this one, but I'll add it here anyway, since it might be relevant. Definition 5.2.1 and Proposition 5.2.2 in Beer: Topologies on closed and closed convex sets seems to be a more general version of this. (He works with nets instead of sequences.) $\endgroup$ Nov 23, 2011 at 7:48

2 Answers 2


Thanks to Martin Sleziak (for pointing at the book of G. Beer [p.145, Prop. 5.2.2. in 1]) the proof is as follows:

Proposition 1. Let $(\mathcal{X},\mathcal{T})$ be a Hausdorff topological space. Then: $$ \liminf_n C_n = \bigcup_{N\in\mathcal{N}_\infty^\#}\overline{\bigcap_{v\in N}C_v} $$


(1). Let $x\in\liminf_n C_n$ and let $\Sigma\in\mathcal{N}_\infty^\#$. Let $W$ be a neighborhood of $x$. There is a $N_0\in\mathbb{N}$ sucht that for all $n\geq N_0$ such that $n\in\Sigma$: $$ W\cap C_n \neq \emptyset $$ Thus, $$ x\in\overline{\bigcup_{n\in\Sigma}C_n} $$ (2). Assume that $x\notin \liminf_n C_n$. Then, there is an open neighborhood of $x$, let $W\ni x$, such that $\Sigma_0:=\{n\in\mathbb{N}| W\cap C_n = \emptyset\}$. Therefore, $x\notin \overline{\bigcup_{n\in\Sigma_0}C_n}$. This completes the proof. $\square$

Note 1: Characterization of the closure of a set: Let $C\subset \mathcal{X}$ and $\bar{C}$ denote its closure which is defined as:

$$ \bar{C}=\bigcap\{F\supset C| F^c\in \mathcal{T}\} $$ Then:

$$ x\in\bar{C} \Leftrightarrow \forall V\in\mathcal{T},\ V\ni x:\ V\cap C \neq \emptyset $$

Note 2: This result is stated in [1] for nets of sets in $\mathcal{X}$, $\{C_n\}_{n\in\Lambda}$ where $\Lambda$ is a partially ordered set. Then the class $\mathcal{N}_\infty^\#$ is replaced by the family of cofinal sets of $\Lambda$. Set set $\Sigma$ is called a cofinal subset of $\Lambda$ if for all $\lambda\in\Lambda,\ \exists\sigma\in\Sigma:\ \sigma\geq\lambda$.

Corollary 2. Let $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of subsets of $\mathcal{X}$. Then: $$ \bigcap_{n\in\mathbb{N}}C_n \subseteq \overline{\bigcap_{n\in\mathbb{N}}C_n} \subseteq \liminf_n C_n $$

Proof. It follows from Proposition 1 taking $N=\mathbb{N}\in\mathcal{N}_\infty^\#$. $\square$

[1] G. Beer, "Topologies on Closed and Closed Convex Sets", Kluwer Academic Publishers, ISBN: 0-7923-2531-1.

  • $\begingroup$ You should notice that the definition from Beer is - in the case of sequences - the following: $x\in\liminf A_n$ $\Leftrightarrow$ every neighborhood $U$ of $x$ intersects $A_n$ for all but finitely many $n$'s\\ $x\in\limsup A_n$ $\Leftrightarrow$ every neighborhood $U$ of $x$ intersects $A_n$ for infinitely many $n$'s\\ This is the same as your definition if you work with discerete topology. $\endgroup$ Nov 23, 2011 at 13:50
  • $\begingroup$ @MartinSleziak This is easy to prove the two definitions to be equivalent I mentioned (by the way, I found the book), i.e. in a topological space $(\mathcal{X},\mathcal{T})$ let $\mho(x)$ be the family of open neighborhoods of $x$. Then $\liminf_n C_n = \{x|\forall V\in\mho(x),\ \exists N\in \mathcal{N}_\infty, \forall n\in N: C_n\cap V\neq \emptyset\}$. Equivalently: $\liminf_n C_n = \{x|\forall V\in\mho(x),\ \exists N_0\in \mathbb{N}, \forall n\geq N_0: C_n\cap V\neq \emptyset\}$. $\endgroup$ Nov 23, 2011 at 14:09

I believe the overline you put shouldn't be there ; it has no reason to be. I'll prove what you want to prove without putting it there to prove my point.

I'll do the $\liminf$ case for you, the $\limsup$ case is symmetric. You want to show that $$ \bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) = \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right). $$ You were right wanting to say that you want to prove $(\subseteq)$ and $(\supseteq)$, but you can make your life more easier a little. To show that $$ \bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) \subseteq \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right), $$ you only need to show that $$ \forall N \in \mathcal N_{\infty}^{\#}, \qquad \bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) \subseteq \bigcup_{n \in N} C_n $$ since being in the intersection means being in all of the things you intersect over. Now suppose $$ x \in \bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) \qquad \Longrightarrow \qquad \exists n \, \text{ s.t. } x \in \bigcap_{m=n}^{\infty} C_m \qquad \Longrightarrow \qquad \forall m \ge n, \quad x \in C_m. $$ Since $N$ is an infinite subset of $\mathbb N$, there exists $m \ge n$ with $m \in N$. Therefore there exists $m \in N$ such that $$ x \in C_m \subseteq \bigcup_{n \in N} C_n, $$ hence we are done with this part.

It's just manipulations of large symbols, don't be astonished by the length of the proof. To show that $$ \bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) \supseteq \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right), $$ You can restrict yourself to show that $$ x \in \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right) \quad \Longrightarrow \quad \exists n \, \text{ s.t. } \, x \in \bigcap_{m=n}^{\infty} C_m. $$ Consider the set $N_X = \{ m \in \mathbb N \, | \, x \notin C_m \}$. If $N_X \in \mathcal N_{\infty}^{\#}$, we have $$ x \in \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{m \in N} C_m \right) \qquad \Longrightarrow \qquad x \in \bigcup_{m \in N_X} C_m, $$ but this is a contradiction because $x$ is (by definition of $N_X$) not in any of those $C_m$'s. Thus $N_X \in N_{\infty}$, and therefore there exists $n \in \mathbb N$ such that for all $m \ge n$, $x \in C_m$. This completes the argument.

Hope that helps,


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