# liminf in terms of the point-to-set distance

Let $\mathcal{X}$ be a normed space and $C\subseteq \mathcal{X}$. We define the point-to-set distance for the set $C$ to be:

$$d_C:\mathcal{X}\ni x \mapsto d_c(x):= \inf_{y\in C}\|x-y\| \in [0,\infty]$$

Additionally, we define the inner limit of a sequence of sets $C_n$ in $\mathcal{X}$ to be:

$$\liminf_n C_n = \bigcup_{n=1}^\infty \bigcap_{m=n}^\infty C_m$$

This definition is equivalent to:

$$\liminf_n C_n = \left\{x \in \mathcal{X} | x\in C_k \text{ ultimately for all } k \right\}$$

My initial question was: I need to prove that: $$\liminf_n C_n = \left\{ x \in \mathcal{X} | \liminf_{n\to\infty} \left(d_{C_n}(x)\right)=0\right\}$$

But this is not true!

The following facts hold true:

1. $\liminf_n C_n \subseteq \left\{ x \in \mathcal{X} | \liminf_{n\to\infty} \left(d_{C_n}(x)\right)=0\right\} = \limsup_n C_n$
2. $\liminf_n C_n \subseteq \left\{ x \in \mathcal{X} | \lim_{n\to\infty} \left(d_{C_n}(x)\right)=0\right\}$

Edit 1. The meaning of "ultimately for all $k$" should be interpreted as follows:

$$\liminf_{n\to\infty} C_n = \{x | \forall k\in\mathbb{N} \exists x_k\in C_{n_k}:\ x_k\to x\}$$ where $n_k\in\mathbb{N}$ is a strictly increasing sequence of indices (i.e. $\{C_{n_k}\}_{k\in\mathbb{N}}$ is some subsequence of $\{C_n\}_{n\in\mathbb{N}}$).

Edit 2. I changed the formula with the closure which was wrong according to Tim (thanks to the answer to Tim's question by Matthias Klupsch). I'll post a new question for what I read in Rockafellar's book and gave me the confusion.

• (1.) I don't think $\lim\inf_n C_n = \bigcap_{n=1}^\infty \overline{\bigcup_{m=n}^\infty C_m}$ makes sense; the right hand side seems to correspond to $\lim\sup_n C_n$. (2.) Your equivalent characterization is not quite right. The second definition works if we ignore the closure operation in the first definition, but not the way the question is stated now. Nov 22, 2011 at 10:51
• @Srivatsan: See R.T. Rockafellar and R. J-B. Wets, "Variational Analysis", Grundlehren der mathematischen Wissenschaften, vol. 317, 1998, p. 110. However, you may omit the closure if it helps you answer my question. Nov 22, 2011 at 10:56
• @DavideGiraudo: Thanks for the hints... but I still don't see how I can construct this sequence. For $k=1$ I can find a sequence $N=\mathbb{N}_{\geq v_1}$ such that $x\in\lim\inf_n C_n$ implies that $x\in C_n + \mathcal{B}$ for $n\geq v_1$ (where $\mathcal{B}$ is the unit ball). Eventually, $x\in C_n + k^{-1}\mathcal{B}$ for $n\geq v_k$ (and $v_k\geq v_{k-1}\geq \ldots$ ). I feel I'm close... can you give me a hint... Nov 22, 2011 at 11:43
• @Srivatsan: Why is $\lim\sup_n C_n = \bigcap_{n=1}^\infty \overline{\bigcup_{m=n}^\infty C_m}$?
– Tim
Nov 22, 2011 at 20:35
• @Tim Actually, it isn't. But the right hand side looks more like limsup than liminf. See this chat discussion between tb and me: chat.stackexchange.com/transcript/message/2503305#2503305. Nov 22, 2011 at 20:36

Let $x\in \liminf C_n$. Then we construct a strictly increasing sequence of integers $\{n_k\}$ and a sequence $\{x_k\}$ such that $\lVert x-x_k\rVert\leq k^{-1}$ and $x_k\in C_{n_k}$. Since $x\in\overline{\bigcup_{m\geq 1}C_m}$, we can find $x_1\in\bigcup_{m\geq 1}C_m$ such that $\lVert x-x_1\rVert\leq 1$ and we choose $n_1$ an integer such that $x_1\in C_{n_1}$. If $x_1,\ldots,x_k$ and $n_1,\ldots,n_k$ are constructed, since $x\in \overline{\bigcup_{j\geq n_k+1}C_j}$, we can find $x_{k+1}$ such that $\lVert x-x_{k+1}\rVert\leq (k+1)^{-1}$ and we choose $n_{k+1}$ as an integer $\geq n_k+1$ such that $x_{k+1}\in C_{n_{k+1}}$. Since $d_{C_{n_k}}(x)\leq k^{-1}$, we get that $0\leq \liminf_n d_{C_n}(x)\leq \liminf_k d_{C_{n_k}}(x) =0$ and we showed $\subset$.

Conversely, if $\liminf_n d_{C_n}(x)=0$ then we can find a strictly increasing sequence of integers $\{n_k\}$ such that $\lim_k d_{C_{n_k}}(x) =0$. Now, taking $n\in\mathbb N$ and $\delta>0$, we can find $k$ such that $n_k>n$ and $d_{C_{n_k}}(x)\leq \frac{\delta}2$. By definition of infimum, we can choose $y\in C_{n_k}$ such that $\lVert x-y\rVert\leq\delta$. Therefore, $x\in\overline{\bigcup_{m\geq n}C_m}$ for all $n$ and $x\in\liminf_n C_n$.

• When you say that $x\in\overline{\bigcup_{m\geq 1 }C_m}$ did you use the fact that $\liminf_n C_n=\bigcap_\Sigma\{\overline{\bigcup_{i\in\Sigma}}; \Sigma \text{ is infinite in } \mathbb{N}\}$? Do you think there is a way to show it based only on the fact that $\liminf_n C_n = \bigcup_{n=1}^{\infty}\bigcap_{m=n}^{\infty} C_m$? Thanks a lot! Nov 23, 2011 at 9:04
• I don't think it's even true with this definition: consider the case $\mathcal X=\mathbb R$ with the usual metric and $C_n=\left[0,1-\frac 1n\right]$, and $x_0=1$. Then $d_{C_n}(x)=\frac 1n$ and converges to $0$, but $1$ is never in $C_n$. Nov 23, 2011 at 9:20

Important note: According to @Did 's comments, I need to clarify that in my original question I confused two different definitions of limit for sequences of sets. The limit defined as $\liminf_n C_n = \left\{x \in \mathcal{X} | x\in C_k \text{ ultimately for all } k \right\}$ is different from the inner limit of Rockafellar and Wets. In what follows I use the definition given by Rockafellar and Wets.

Notation: Let $\left( \mathcal{X},\mathcal{T}\ \right)$ be a topological space. We denote the family of open neighbourhoods of $x\in\mathcal{X}$ by $\mho(x):=\{V\in\mathcal{T}:\ x\in V\}$.

Proposition 1: Let $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in a Hausdorff topological space $\left( \mathcal{X},\mathcal{T}\ \right)$. Then, $$\liminf_n C_n = \{x|\forall V\in\mho(x),\ \exists N\in \mathcal{N}_\infty, \forall n\in N: C_n\cap V\neq \emptyset\}$$ or equivalenty: $$\liminf_n C_n = \{ x|\forall V\in\mho(x),\ \exists N_0\in \mathbb{N}, \forall n\geq N_0: C_n\cap V\neq \emptyset \}$$

Proof.

(1). If $x\in\liminf_n C_n$ then we can find a sequence $\{x_k\}_{k\in\mathbb{N}}$ such that $x_k\to x$ while $x_k\in C_{n_k}$ and $\{n_k\}_{k\in\mathbb{N}}\subseteq \mathbb{N}$ is a strictly increasing sequence of indices. For any $V\in\mho(x)$ there is a $N_0\in\mathbb{N}$ such that for all $i\geq N_0$ it is: $x_i\in V$; but also $x_i\in C_{n_i}$. Thus $C_{n_i}\cap V\neq \emptyset$. Therefore $x$ is in the right-hand side set of the equation.

(2). For the reverse direction assume that $x$ belongs to the right-hand side set of given equation. Then, there is a strictly increasing sequence $\{n_k\}_{k\in\mathbb{N}}$. Then, for every $V\in\mho(x)$ we can find a $x_k\in C_{n_k}\cap V$. Hence, $x_k\to x$ ( in the topology $\mathcal{T}$ ).

$\square$

Proposition 2: Let $(\mathcal{X},\|\cdot\|)$ be a normed space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. The inner limit of a sequence of sets is: $$\liminf_n C_n = \{ x\in X | \lim_n d(x,C_n)=0 \}$$

Proof.

(1). We now need to show that $\limsup_n d(x,C_n)=0$. Let us assume that $\limsup_n d(x,C_n)>0$, i.e. there exists an increasing sequence of indices $\{n_k\}_{k\in\mathbb{N}}$ so that $d(x,C_{n_k})\to_k a > 0$. This suggests that there is a $\varepsilon_0>0$ such that for all $k\in\mathbb{N}$ one has that $d(x,C_{n_k})>\varepsilon_0$. However, according to proposition \ref{propo:un_int}, $x\in\text{cl}\bigcup_{k\in\mathbb{N}}C_{n_k}$ while $d(x,\text{cl}\bigcup_{k\in\mathbb{N}}C_{n_k})\geq\varepsilon_0$ which is a contradiction. Hence, $\limsup_n d(x,C_n)=0$, i.e. $\lim_n d(x,C_n)=0$ and this way we have proven that $x$ is in the right-hand side set.

(2). Assume that $x$ in the right-hand side of the given equation. This is $\lim_n d(x,C_n)=0$. For any $\varepsilon>0$, we can find $n_0\in\mathbb{N}$ such that $d(x,C_{n})\leq \frac{\varepsilon}{2}$ for all $n\geq n_0$. By definition, we have that $d(x,C_{n})=\inf\{\|x-y\|,\ y\in C_{n}\}$, thus we can find a $y_n\in C_{n}$ such that

$$\|y_n-x\|<d(x,C_{n})+\frac{\varepsilon}{2}=\varepsilon$$

we can do that following the steps pointed out by Davide Giraudo in his answer to this question. That is:

$$\exists\ y_n\in C_{n}:\ \|y_n-x\|<\varepsilon$$ Therefore, $x\in C_{n} + \varepsilon \mathcal{B}$ from which it follows that $x\in\liminf_n C_n$ (According to proposition 1).

$\square$

Proposition 3: Let $(\mathcal{X},\|\cdot\|)$ be a normed space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. The inner limit of a sequence of sets is: $$\limsup_n C_n = \{ x\in X | \liminf_n d(x,C_n)=0 \}$$

Note: We know that $\liminf_n C_n \subseteq \limsup_n C_n$. We may therefore prove that :

$$\liminf_n C_n \subseteq \{ x\in X | \liminf_n d(x,C_n)=0 \}$$

• Pantelis: This (accepted) answer of yours and the text of the question itself both contain several serious inaccuracies based on a confusion between some distinct definitions (roughly speaking, set theoretic limits vs topological limits). This was pointed to you here and you acknowledged it, if I understand you correctly. I suggest you mention this fact somewhere in the present question and answer, instead of letting people believe otherwise.
– Did
Jan 9, 2012 at 8:42