# Kuratowski $\limsup$ and $\liminf$ of a sequence of subsets of a topological space?

From Wikipedia:

Let $(X, d)$ be a metric space. For any point $x ∈ X$ and any non-empty compact subset $A ⊆ X$, let $$d(x, A) = \inf \{ d(x, a) \mid a \in A \}.$$ For any sequence of such subsets $A_n ⊆ X, n ∈ \mathbb{N}$, the Kuratowski limit inferior (or lower closed limit) of $A_n$ as $n → ∞$ is \begin{align*} \mathop{\mathrm{Li}}_{n \to \infty} A_{n} &= \left\{ x \in X \;\left|\; \limsup_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}\\ &= \left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset \mbox{ for large enough } n \end{matrix} \right. \right\}; \end{align*} and the Kuratowski limit superior (or upper closed limit) of $A_n$ as $n → ∞$ is \begin{align*} \mathop{\mathrm{Ls}}_{n \to \infty} A_{n} &= \left\{ x \in X \;\left|\; \liminf_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}\\ &= \left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset \mbox{ for infinitely many } n \end{matrix} \right. \right\}.\end{align*}

The second part of each definition is purely topological and has nothing to do with metric. So I try to represent them using operations on subsets, such as union, intersection, closure, interior, .... Questions:

1. Is it correct that:

• "For all open neighbourhoods $U$ of $x$, $U \cap A_{n} \neq \emptyset$" is equivalent to $x \in \mathrm{closure}(A_n)$?

• So $$\left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset \mbox{ for large enough } n \end{matrix} \right. \right\} = \cup_{i=1}^\infty \cap_{n=i}^\infty \mathrm{closure}(A_n),$$ and $$\left\{ x \in X \;\left|\; \begin{matrix} \mbox{for all open neighbourhoods } U \mbox{ of } x, \\ U \cap A_{n} \neq \emptyset \mbox{ for infinitely many } n \end{matrix} \right. \right\} = \cap_{i=1}^\infty \cup_{n=i}^\infty \mathrm{closure}(A_n)?$$

2. I don't think they are the same as those provided by Andre, since here there is no taking closure after taking union to make the results to be closed. But this is contrary to the linked Wikipedia article:

$\mathop{\mathrm{Li}}_{n \to \infty} A_{n}$ and $\mathop{\mathrm{Ls}}_{n \to \infty} A_{n}$ are always closed sets in the metric topology on $(X, d)$

Which one is right, Wikipedia or I?

3. How can it be shown that the first one in part 1 is $\left\{ x \in X \left| \limsup_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}$, and the second is $\left\{ x \in X \left| \liminf_{n \to \infty} d(x, A_{n}) = 0 \right. \right\}$?

Thanks and regards!

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Didn't you already ask this? – t.b. Mar 16 '12 at 12:02
@t.b.: I doubt if the questions here and there are the same, are they? Here the limsup and liminf are always closed subset as the same link says, but they seem not the same as those in my post linked by you, does it? If they do, I am glad to see how they are the same. – Tim Mar 16 '12 at 12:07
I don't know if it is exactly the same but you should be able to come up with the answer yourself. It's twice a yes, of course. – t.b. Mar 16 '12 at 12:28
@t.b.: I added my thought. Could you let me know what you think? – Tim Mar 16 '12 at 18:00
@t.b.: Indeed, it turns out not the same as my previous one linked by you. See the reply I selected. I believe I was not able to figure it out on my own either. – Tim Mar 16 '12 at 21:58

(1) "For all open neighbourhoods $U$ of $x$, for $n$ large enough (resp. infinitely many $n$), $U \cap A_n \neq \emptyset$"

is not equivalent to

(2) "For $n$ large enough (resp. infinitely many $n$), for all open neighbourhoods $U$ of $x$, $U \cap A_n \neq \emptyset$"

(1) is the one present in the definition of the Kuratowski limit, and (2) is the one present in your definition using closure(An)

When you switch the order of quantifiers from a $\forall x \exists y P(x,y)$ into a $\exists y \forall x P(x,y)$, the numbers of things your choice of $y$ can depend on gets smaller, so you almost never get equivalent statements. However, there is an easy implication between the two : $\exists y \forall x P(x,y) \implies \forall x \exists y P(x,y)$. So $\cup_{i\ge 1} \cap_{n\ge i} A_n \subset \liminf A_n$ and $\cap_{i\ge 1} \cup_{n\ge i} A_n \subset \limsup A_n$

$\limsup A_n$ and $\liminf A_n$ are always closed : $x \notin \limsup A_n$ means that there is an open neighboorhood $U$ of $x$ such that $P(U)$ for some property $P$, and so for every $y \in U$ it is again the case that there is a neighboorhood of $y$, namely $U$, such that $P(U)$.

So in fact, we have $\cup_{i\ge 1} \cap_{n\ge i} A_n \subset \overline{\cup_{i\ge 1} \cap_{n\ge i} A_n} \subset \liminf A_n$ and $\cap_{i\ge 1} \cup_{n\ge i} A_n \subset \overline{\cap_{i\ge 1} \cup_{n\ge i} A_n} \subset \cap_{i\ge 1} \overline{\cup_{n\ge i} A_n} = \limsup A_n$.

There should be examples where all the inequalities are strict. Also, I've not thought about cases where we don't necessarily pick compact sets for $A_n$.

Since we are dealing with positive sequences, and since the open balls form a basis for the topology, $$\limsup (d(x,A_n)) = 0 \Leftrightarrow \forall \varepsilon > 0 \exists n \forall m \ge n, d(x,A_m) < \varepsilon \\ \Leftrightarrow \forall \varepsilon > 0 \exists n \forall m \ge n, B(x,\varepsilon) \cap A_m \neq \emptyset \\ \Leftrightarrow \forall U \exists n \forall m \ge n, U \cap A_m \neq \emptyset$$ and $$\liminf (d(x,A_n)) = 0 \Leftrightarrow \forall \varepsilon > 0 \forall n \exists m \ge n, d(x,A_m) < \varepsilon \\ \Leftrightarrow \forall \varepsilon > 0 \forall n \exists m \ge n, B(x,\varepsilon) \cap A_m \neq \emptyset \\ \Leftrightarrow \forall U \forall n \exists m \ge n, U \cap A_m \neq \emptyset \\ \Leftrightarrow \forall n \forall U, U \cap (\cup_{m\ge n} A_m) \neq \emptyset$$

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+1. Thanks, although confused. (1) How are they not equivalent? (2) How are they equivalent to Wiki definition and mine respectively? – Tim Mar 16 '12 at 18:53
well because you switch the order of the quantifiers, you should NOT expect them to be equivalent. Arthur Fischer gives an explicit example where (1) is true and (2) is false. More precisely there is an "there exists n" somewhere in there, in (1), you are allowed to choose it depending on U, and in (2), you can't, you have to make a choice that works for every U. – mercio Mar 16 '12 at 19:02
Thanks! (1) Is there some relation between Wiki's definition and mine (the one using closure)? For example, is one subset of the other? (2) Is there some relation between Wiki's definition and Andre's (it is a modification of mine by taking another closure after taking union to ensure closedness)? – Tim Mar 16 '12 at 21:55
Thanks! I wonder if in your reply $\limsup A_n$ and $\liminf A_n$ are the same as $\mathop{\mathrm{Ls}}_{n \to \infty} A_{n}$ and $\mathop{\mathrm{Li}}_{n \to \infty} A_{n}$ respectively? If not, what definitions did you use for $\limsup A_n$ and $\liminf A_n$? – Tim Mar 20 '12 at 18:58
yes, sorry they are the Ls and Li defined as in the text you quoted. – mercio Mar 20 '12 at 19:00

(1) is false, even in the context of metric spaces (in fact, even for the real line $\mathbb{R}$). For $n \in \mathbb{N}$ let $A_n = [2^{-(n+1)} , 2^{-n} ]$. It is easy to show that $$\mathrm{Li}_{n \rightarrow \infty} A_n = \{ 0 \} = \mathrm{Ls}_{n \rightarrow \infty} A_n.$$ It is also quite easy to establish that $$\bigcup_{i=1}^\infty \bigcap_{n=i}^\infty \overline{A_n} = \bigcup_{i=1}^\infty \bigcap_{n=i}^\infty A_n = \bigcup_{i=1}^\infty \emptyset = \emptyset,$$ and also that $$\bigcap_{i=1}^\infty \bigcup_{n=i}^\infty \overline{A_n} = \bigcap_{i=1}^\infty \bigcup_{n=i}^\infty A_n = \bigcap_{i=1}^\infty ( 0 , 2^{-i} ] = \emptyset.$$

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+1 Thanks! (1) How do you show that the in the definition of $Li_{n} A_n$, the two expressions are equal? I.e. the first is based on metric and the second on topology? Similarly for $Ls_n A_n$. (2) How would you represent the definition in terms of subset operations, such as union, intersection, closure, interior, ...? – Tim Mar 16 '12 at 18:47
@Tim: (1) Using the fact that $\mathrm{Li}_n A_n \subseteq \mathrm{Ls}_n A_n$, we need only show that $0 \in \mathrm{Li}_n A_n$ and $\mathrm{Ls}_n A_n \subseteq \{ 0 \}$. The first is easy, since $0$ is the limit point of the sequence $\{ 2^{-n} \}_{n=1}^\infty$. The second isn't must more difficult since if $x \neq 0$, then $x$ has a neighbourhood which meets only finitely many of the $A_n$'s (the $\epsilon$-ball around $x$ where $\epsilon = \frac{|x|}{2}$ should work). – arjafi Mar 16 '12 at 19:09
@Tim: As for (2), I have doubts that there is a simple formula expressing these operations in terms of more basic topological operations. My reasoning is not based on anything mathematical, but rather because the only places I have seen these operations introduced has been in exactly this form. If these operations were reducible to more basic operations, this should have been mentioned somewhere. (Like I said, not a mathematical reason, but more of a psychological reason.) – arjafi Mar 16 '12 at 19:13