# On Set Convergence II

This post comes as a sequel of On Set Convergence I where we introduced some necessary notions that are useful for studying how sets converge in a topological space. Alongside, we provided two non-topological definitions of convergence, namely the limits inferior and superior. In this post we will delve a bit further into the topological properties of the inner and outer limits of sequences of sets.

We start with an important result which characterises the topological nature of the inner limit:

Proposition 4. Let $\{C_n\}_{n}$ be a sequence of sets in a Hausdorff topological space $\left(\mathcal{X},\tau\right)$. Then

$\liminf_n C_n = \left\{ x\left| {\begin{array}{c} \forall V\in\mho\left(x\right),\ \exists N\in \mathcal{N}_\infty\\ \forall n\in N: C_n\cap V\neq \varnothing \end{array} } \right.\right\}$

or equivalently:

$\liminf_n C_n = \left\{ x\left| {\begin{array}{c} \forall V\in\mho\left(x\right),\ \exists N_0\in \mathbb{N},\\ \forall n\geq N_0: C_n\cap V\neq \varnothing \end{array} }\right. \right\}$

Proof. (1). If $x\in\liminf_n C_n$ then we can find a sequence $\{x_k\}_{k}$ such that $x_k\to x$ while $x_k\in C_{n_k}$ and $\{n_k\}_{k}\subseteq \mathbb{N}$ is a strictly increasing sequence of indices. For any $V\in\mho\left(x\right)$ 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, $C_{n_i}\cap V\neq \varnothing$. Therefore $x$ is in the right-hand side set of equation the first equation in Proposition 4.

(2). For the reverse direction assume that $x$ belongs to the right-hand side set of the first equation in Proposition 4. Then, there is a strictly increasing sequence $\{n_k\}_{k}$ such that for every $V\in\mho\left(x\right)$ we can find a $x_k\in C_{n_k}\cap V.$ Hence, $x_k\to x$ (in the topology $\tau$). ♣

Or course for every result regarding the inner limit, there is a corresponding one for its complementary object: the outer limit. Therefore, following similar steps with the ones in the proof of Proposition 4 we can show that:

Proposition 5. Let $\{C_n\}_{n}$ be a sequence of sets in a Hausdorff
topological space $\left(\mathcal{X},\tau\right)$. Then

$\limsup_n C_n = \left\{ x\left|{\begin{array}{c} \forall V\in\mho\left(x\right),\ \exists N\in \mathcal{N}_\infty^\#,\\ \forall n\in N: C_n\cap V\neq \varnothing \end{array} } \right.\right\}$

or equivalenty:

$\limsup_n C_n = \left\{ x \left| {\begin{array}{c} \forall V\in\mho\left(x\right),\exists \{n_k\}_{k}\subseteq\mathbb{N}\\ \{n_k\}_{k}\uparrow,\forall k\in \mathbb{N}:\\C_{n_k}\cap V\neq \varnothing \end{array} } \right.\right\}$

Instead of arbitrary open sets, when $\mathcal{X}$ is a normed space we may use open balls, i.e. sets of the form $\mathcal{B}\left(\varepsilon\right)=\left\{x:\ \|x\|<\varepsilon\right\}$. We denote the unit ball of $\mathcal{X}$ by $\mathcal{B}:=\mathcal{B}\left(1\right)$. Then $\mathcal{B}\left(\varepsilon\right)=\varepsilon\mathcal{B}.$ This leads us to the following corollary:

Corollary 1. Let $\{C_n\}_{n}$ be a sequence of sets in a normed space
$\left(\mathcal{X},\|\cdot\|\right)$. Then,

$\liminf_n C_n = \left\{ {\begin{array}{c} x|\forall \varepsilon>0,\ \exists N\in \mathcal{N}_\infty,\\ \forall n\in N: x\in C_n+\varepsilon\mathcal{B} \end{array} } \right\}$

and

$\limsup_n C_n = \left\{ {\begin{array}{c} x|\forall \varepsilon>0,\ \exists N\in \mathcal{N}_\infty^\#,\\ \forall n\in N: x\in C_n+\varepsilon\mathcal{B} \end{array} } \right\}$

This corollary yields another equivalent characterization for the inner limit. The predicate “$\forall\varepsilon>0$” translates into the intersection over all $\varepsilon>0$, the requirement “$\exists N\in\mathcal{N}_\infty^\#$” will be written as the union over all $N\in\mathcal{N}_\infty$ and then “$\forall n\in N$” corresponds to an intersection. This allows us to restate the corollary
using set elementary operations as follows:

Corollary 2. Let $\{C_n\}_{n}$ be a sequence of sets in a normed space $\left(\mathcal{X},\|\cdot\|\right)$. Then,

$\liminf_n C_n = \bigcap_{\varepsilon>0}\bigcup_{N\in\mathcal{N}_\infty}\bigcap_{v\in N} C_v + \varepsilon\mathcal{B}$

and

$\limsup_n C_n = \bigcap_{\varepsilon>0}\bigcup_{N\in\mathcal{N}_\infty^\#}\bigcap_{v\in N} C_v + \varepsilon\mathcal{B}$

Before we state and prove the following result (Proposition 7) we need to recall the following description of the closure of a set. Firstly, if $\left(\mathcal{X},\tau\right)$ is a topological space and $C\subseteq\mathcal{X}$, its closure is defined as:

$\mathrm{cl} C := \bigcap\left\{F:\ F\supseteq C,\ F^c \in \tau \right\}$

Then, we have the following useful proposition:

Proposition 6. $\mathrm{cl} C=\left\{x: \forall V\in \mho\left(x\right),\ V\cap C\neq \varnothing \right\}$

We may then show the following result:

Proposition 7. Let $\left(\mathcal{X},\tau\right)$ be a Hausdorff topological space and $\{C_n\}_{n}$ be a sequence of sets in $\mathcal{X}$. Then,

$\liminf_n C_n = \bigcap_{N\in\mathcal{N}_\infty^\#}\mathrm{cl}\bigcup_{n\in N} C_n$

Proof. (1). Let $x\in\liminf_n C_n$ and let $\Sigma\in\mathcal{N}_\infty^\#$. Let $W$ be a neighbourhood of $x$. There is a $N_0\in\mathbb{N}$ sucht that for all $n\geq N_0$ such that $n\in\Sigma$:

$W\cap C_n \neq \varnothing$

Thus,

$x\in \mathrm{cl} \bigcup_{n\in\Sigma}C_n$

(2). Assume that $x\notin \liminf_n C_n$. Then, there is an  open neighbourhood of $x$, let $W\ni x$, such that  $\Sigma_0:=\left\{n\in\mathbb{N}| W\cap C_n = \varnothing\right\}$. Therefore, $x\notin \mathrm{cl} \bigcup_{n\in\Sigma_0}C_n$. This completes the proof. ♣

Proposition 6 reveals a very important property of the inner limit:

Corollary 3. For any sequence of sets $\{C_n\}_{n}$ in a Hausdorff topological space $\left( \mathcal{X},\tau\right)$ the limit $\liminf_n C_n$ is closed.

Indeed, the inner limit is given as an arbitrary intersection of closed sets $\liminf_n C_n = \bigcap_{N\in\mathcal{N}_\infty^\#}\mathrm{cl}\bigcup_{n\in N} C_n$ which is closed.

The following corollary is another immediate consequence of Proposition 6.

Corollary 4. Let $\{C_n\}_{n}$ be a sequence of sets in $\mathcal{X}$. Then,

$\bigcap_{n=1}^\infty C_n \subseteq \mathrm{cl} \bigcap_{n=1}^\infty C_n \subseteq \liminf_n C_n$

The following result is preliminary to what follows. I put it here just because the proof has some interesting steps. It is expedient to know beforehand that the interior limit of a sequence of sets in normed spaces is described by:

$\liminf_n C_n = \left\{x\in \mathcal{X} | \lim_{n} d\left(x,C_n\right)=0\right\}$

Proposition 8. Let $\left(\mathcal{X},\|\cdot\|\right)$ be a normed space and $\{C_n\}_{n}$ be a sequence of sets in $\mathcal{X}$. If  $x\in\liminf_n C_n$ then $\liminf_{n} d\left(x,C_n\right)=0$.

Proof. Let $x\in\liminf_{n} C_n$. We will construct a sequence $\{x_k\}_{k}\subset X$ and a strictly increasing sequence of indices $\{n_k\}_{k}\subseteq \mathbb{N}$ such that $\|x-x_k\|\leq \frac{1}{k}$ while $x_k \in C_{n_k}$.

By Proposition 6, if $x\in\liminf_n C_n$, then for any $\Sigma_1\in \mathcal{N}_\infty^\#$:

$x\in \mathrm{cl}\bigcup_{i\in\Sigma_1}C_i \Rightarrow \exists x_1 \in\bigcup_{i\in\Sigma_1}C_i : \|x-x_1\|<1$

Choose $n_1\in\Sigma_1$ to be the smallest integer such that $x\in C_{n_1}$. Set $\Sigma_2=\left\{\sigma\in\Sigma_1,\ \sigma> n_1\right\}$.
Note: $\Sigma_2\subseteq \Sigma_1$ is cofinal for $\Sigma_1$. Then,

$x\in\mathrm\bigcup_{i\in\Sigma_2}C_i \Rightarrow \exists x_2 \in\bigcup_{i\in\Sigma_2}C_i : \|x-x_2\|<\frac{1}{2}$

and take $n_2\in\Sigma_2$ to the be smallest such that $x_2\in C_{n_2}$.
Eventually, we construct the aforementioned sequences. We now have that:

$\|x_k-x\|<\frac{1}{k},\ x_k \in C_{n_k}\Rightarrow 0\leq d\left(x,C_{n_k}\right)<\frac{1}{k}$

Therefore we have proven that $\lim_k d\left(x,C_{n_k}\right)=0$, i.e. $\liminf_n d\left(x,C_n\right)=0$. ♣

The previous result is extended to provide a particularly important description of the notion of the inner limit on normed spaces. We use the point-to-set distance function to describe the inner limit of a sequence of sets:

Proposition 9. Let $\left(\mathcal{X},\|\cdot\|\right)$ be a normed space and $\{C_n\}_{n}$ be a sequence of sets in $\mathcal{X}$. The inner limit of a sequence of sets is:

$\liminf_n C_n = \left\{x\in \mathcal{X} | \lim_{n} d\left(x,C_n\right)=0\right\}$

Proof. (1).  We firstly need to show that for any $x\in\mathcal{X}$ it is $\limsup_n d\left(x,C_n\right)=0$. Let us assume that $\limsup_n d\left(x,C_n\right)>0$, i.e. there exists an increasing sequence of indices $\{n_k\}_{k}$ so that $d\left(x,C_{n_k}\right)\to_k a > 0$. This suggests that there is a $\varepsilon_0>0$ such that for all $k\in\mathbb{N}$ one has that $d\left(x,C_{n_k}\right)>\varepsilon_0$. However, according to Proposition 6,

$x\in\mathrm{cl}\bigcup_{k\in\mathbb{N}}C_{n_k}$

whereas

$d\left(x,\mathrm{cl}\bigcup_{k\in\mathbb{N}}C_{n_k}\right)\geq\varepsilon_0$

which contradicts our initial assumption. Hence, $\limsup_n d\left(x,C_n\right)=0$, i.e. $\lim_n d\left(x,C_n\right)=0$ and this way we have proven that $x$ is in the right-hand side set.

(2). Conversely, assume that $x$ in the right-hand side of the equation in the statement of Proposition 8, that is $\lim_n d\left(x,C_n\right)=0$. For any $\varepsilon>0$; we can find $n_0\in\mathbb{N}$ such that $d\left(x,C_{n}\right)\leq \frac{\varepsilon}{2}$ for all $n\geq n_0$. By definition, we have that

$d\left(x,C_{n}\right)=\inf\left\{\|x-y\|,\ y\in C_{n}\right\},$

thus we can find a $y_n\in C_{n}$ such that

$\|y_n-x\|

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 $x\in\liminf_n C_n$ (see Corollary 1). ♣

Note. For a sequence $\{a_n\}_{n}$, in order to show that $\liminf_n a_n=a$, it suffices to find a subsequence of it that converges to $a$, i.e. it suffices to determine a strictly increasing sequence $\{n_k\}_{k}\subseteq \mathbb{N}$ such that $a_{n_k} \to_k a$, i.e. $\lim_k a_{n_k}=a$. Using this fact we can prove the following:

Proposition 10.  Let $\left(\mathcal{X},\|\cdot\|\right)$ be a normed space and $\{C_n\}_{n}$ be a sequence of sets in $\mathcal{X}$. The outer limit of a sequence of sets is:

$\limsup_n C_n = \left\{x\in \mathcal{X} | \liminf_{n} d\left(x,C_n\right)=0\right\}$

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