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I am particularly confused with alternative formulas describing the inner and outer limits of a sequence of sets in topological spaces. The inner limit of a sequence of sets $\left\{C_n\right\}_{n\in\mathbb{N}}$ is defined as:

$$ \liminf_n C_n := \left\{x|\ x\in C_n \text{ ultimately for all } n\right\} $$

or equivalently

$$ \liminf_n C_n = \left\{x|\exists N_0\in\mathbb{N} \text{ such that }x\in C_k,\ \forall k\geq N_0\right\} $$

Then on one hand we can prove the following well known result:

Proposition 1 : The inner limit of a sequence of sets is: $$ \liminf_n C_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty} C_m \ \ \text{(1)} $$ Proof. The proof can be found in http://planetmath.org/encyclopedia/LimitOfSequenceOfSets.html (Most books of analysis coping with sequences of sets pertain to this result).

$\square$

But on the other hand we can prove that:

Proposition 2 : Let $\left(\mathcal{X},\mathcal{T}\right)$ be a Hausdorff topological space and $\left\{C_n\right\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. Then, $$ \liminf_n C_n = \bigcap\left\{ \overline{\bigcup_{n\in N} C_n},\ N \text{ is a cofinal subset of } \mathbb{N} \right\} $$ Note (Notation) : $\bar{C}$ stands for the closure of the set $C$.

Proof. This result can be found in the Book of R.T. Rockafellar and R.J-B. Wets, Variational Analysis (Springer, Dordrecht 2000, ISBN: 978-3-540-62772-2) as an exercise while the proof can be found in the book of G.Beer, Topologies on Closed and Convex Closed Sets (Kluwer Academic Publishers, Dordecht 1993, ISBN: 0-7923-2531-1).

$\square$

Using Proposition 1, we arrive at the following result:

Proposition 3A : Let $\left\{C_n\right\}_n$ be a decreasing sequence of sets (i.e. $C_k \subseteq C_{k+1}$ for all $k\in\mathbb{N}$). Then:

$$ \liminf_n C_n = \bigcap_{n\in\mathbb{N}}C_n $$

Let $\left\{C_n\right\}_n$ be an increasing sequence of sets (i.e. $C_k \supseteq C_{k+1}$ for all $k\in\mathbb{N}$). Then:

$$ \liminf_n C_n = \bigcup_{n\in\mathbb{N}}C_n $$

Proof. The proof can be found at http://planetmath.org/encyclopedia/LimitOfSequenceOfSets.html.

$\square$

But then, using Proposition 2 we can see that:

Proposition 3B : Let $\left\{C_n\right\}_n$ be a decreasing sequence of sets. Then:

$$ \liminf_n C_n = \bigcap_{n\in\mathbb{N}}\overline{C_n} $$

Let $\left\{C_n\right\}_n$ be a decreasing sequence of sets. Then:

$$ \liminf_n C_n = \overline{\bigcap_{n\in\mathbb{N}} C_n} $$

Note : Rockafellar and Wets give the following example which to me seems paradoxical: Consider the constant sequence $C_v=\mathbb{Q}^p$ for $v\in\mathbb{N}$ in the normed space $\left(\mathbb{R}^p,\left\| \cdot \right\| \right)$ (with the Euclidean norm). Then $\liminf_v C_v = \limsup_v C_v = \mathbb{R}^p$ (NOT $\mathbb{Q}^p$; and they underline this fact). However if we consider the element $y=\left( \pi,\pi,\ldots,\pi\right)\in \mathbb{R}^p\setminus \mathbb{Q}^p$ then for every $n\in\mathbb{N}$, $y\notin C_n$. Therefore, I would say that $y\notin \limsup_v C_v$ and $y\notin \liminf_v C_v$.

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The definition of liminf used by Rockafellar and Wets (which you do not indicate) seems different from the definition you quote at the beginning of your post (which does not require any structure on the reference space, topological or anything else) hence the results are different. –  Did Nov 25 '11 at 12:17
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The definition at the beginning of your post does not require any topology. R&W's definition does, since it involves convergent sequences (in your comment, $x_v\to x$). Propositions 1 and 3A on the one hand, and propositions 2 and 3B on the other hand, are based on two different definitions of liminf. –  Did Nov 25 '11 at 12:44
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Of course it requires topology: it talks about convergence of sequences. –  Brian M. Scott Nov 25 '11 at 12:46
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The definition at the beginning of your post is the special case of R&W's when the underlying set is given the discrete topology. –  Brian M. Scott Nov 25 '11 at 12:50
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@Didier: I guess that we’ll have to agree to disagree on the pædagogy involved here, since I think that pointing out exactly how the two notions are related goes a long way towards explaining why they have the same name $-$ which, it seems to me, was the real source of the confusion. I certainly don’t see it as being in any way evangelism for the topological concept or for the pervasiveness of such concepts; that never even occurred to me. But I’ll let it rest here, since the discussion-warden is getting grumpy and I need to get to bed. –  Brian M. Scott Jan 9 '12 at 12:59

1 Answer 1

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The definition of liminf used by Rockafellar and Wets is different from the definition you quote at the beginning of your post (which does not require any structure on the reference space, topological or anything else) hence the results are different.

The notion of liminf defined at the beginning of your post, which I will call set theoretical and denote $\mathrm{set}\!\!-\!\!\liminf$, does not require any structure on the reference space, and in particular no topology is involved. R&W's notion of liminf, which I will call topological and denote $\mathrm{top}\!\!-\!\!\liminf$, does require a topology since, according to a comment of yours, it involves convergent sequences (in your comment, $x_v\to x$) and to define convergence one needs a topology. Hence Propositions 1 and 3A on the one hand, and Propositions 2 and 3B on the other hand, are based on two different definitions of liminf. For every $(C_n)$, $$ \mathrm{set}\!\!-\!\!\liminf C_n\subseteq\mathrm{top}\!\!-\!\!\liminf C_n, $$ and these can be quite different. For example, consider the real line $\mathbb R$ equipped with the usual topology, and $C_{2n}=D$ and $C_{2n+1}=\mathbb R\setminus D$ for every $n$, where $D$ is dense and co-dense, for example $D=\mathbb Q$. Then $\mathrm{set}\!\!-\!\!\liminf C_n$ is the empty set and $\mathrm{top}\!\!-\!\!\liminf C_n$ is $\mathbb R$.

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Pantelis: You asked three different questions that are all based on the same confusion between two different notions of liminf (which I call set theoretical and topological in my answer). You even accepted your own answer to one of the questions, which wrongly proved that these two notions are equivalent (they are not). At this point it seems that all that could be said on the subject has been said, hence I suggest that you try to understand the answers to your three questions, or that you address them in comments to their authors, before embarking on a new duplicate question. –  Did Nov 26 '11 at 9:16
    
My confusion sources from the use of the same symbol for two completely different notions. Thanks a lot for clarifying this and I admit it should be obvious! Actually it is all clear to me now. –  Pantelis Sopasakis Nov 26 '11 at 12:47

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