Is there a topology induced in $\mathcal P (X)$ for an infinite set $X$?

Intuitively, if $s_1 \supseteq s_2 \supseteq s_3 ...$ and $\bigcap_i s_i = s$ then we could say $s$ is the limit of $\{s_i\}$. But I don't exactly know how to define the open sets to make that formal.

Even better: Is there a metric I could impose on the subsets to get such limits to be meaningful? If not, is there a restriction to $X$ such that I can impose a metric on $\mathcal P (X) $?

Here is why I'm interested: I want to find a fixpoint for a function $F:\mathcal P (X) \rightarrow \mathcal P (X)$. Interestingly, in my application, I can find approximations of the fixpoint such that $F( s_{i+1}) = s_{i}$ and $s_i \supseteq s_{i+1}$. Now I just need to require $F$ to be continuous, to get $F( \lim s_i ) = \lim s_i$. So I need to be able to define continuous.

  • $\begingroup$ If I am not mistaken, this is special case of this limit of sets which is special case of Kuratowski convergence. Wikipedia article mentions that it convergence in Fell topology. $\endgroup$ – Martin Sleziak Dec 4 '16 at 16:53
  • $\begingroup$ Any idea what the Fell topology looks like with the discrete topology on $X$? $\endgroup$ – Skuge Dec 4 '16 at 18:09
  • $\begingroup$ It seems to that when $X$ is discrete, the Fell topology, on $\mathcal P (X)$ is also discrete. So only constant sequences converge... Am I missing something? $\endgroup$ – Skuge Dec 4 '16 at 18:22
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    $\begingroup$ There is a well known fixed point theorem for functions $F:\mathcal P(X)\to\mathcal P(X)$ such that $s\subseteq t\implies F(s)\subseteq F(t).$ $\endgroup$ – bof Dec 4 '16 at 19:18
  • $\begingroup$ Thank's bof. My function is not covariant in the entire space. Only in the sequence $\{ s_i\}$ does it satisfy the property. The sequence is not a lattice, without the limmit, right? $\endgroup$ – Skuge Dec 5 '16 at 20:07

There are a number of reasonably well-known topologies on $\wp(X)$ and subsets thereof, but it’s not clear that any of them will be of much use in your setting.

Let $X$ be a topological space. For each closed $F\subseteq X$ and open $U\subseteq X$ such that $F\subseteq U$ let

$$\mathscr{B}(F,U)=\{A\subseteq X:F\subseteq A\subseteq U\}\;;\tag{1}$$

the sets $\mathscr{B}(F,U)$ are a base for the hyperspace topology on $\wp(X)$. If we consider only finite subsets of $X$ (and restrict $F$ in $(1)$ to being finite), we get the Pixley-Roy topology on the space of finite subsets of $X$.

However, if $X$ has the discrete topology, the hyperspace topology is also discrete, as $\mathscr{B}(A,A)=\{A\}$ is open for each $A\subseteq X$. Thus, the only convergent sequences are those that are eventually constant.

It is more common to consider the space of closed subsets of $X$. This is usually given the Vietoris topology, which has a base consisting of all sets of the form

$$\mathscr{B}(\mathscr{U})=\left\{F\subseteq\bigcup\mathscr{U}:F\text{ is closed, and }F\cap U\ne\varnothing\text{ for each }U\in\mathscr{U}\right\}\;,$$

where $\mathscr{U}$ is a finite family of open sets in $X$. It is also common to look at the subspace of this space whose elements are the compact subsets of $X$. If $X$ is a compact metric space, the compact subsets are precisely the closed subsets, and this topology is induced by the Hausdorff metric.

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  • $\begingroup$ Very informative but, as you mentioned, probably not of much use for my setting. Is it also true that the Vietoris topology is discrete if $X$ is discrete? $\endgroup$ – Skuge Dec 4 '16 at 19:28
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    $\begingroup$ @Skuge: Yes. The only compact sets are the finite ones, and if $F$ is finite, we can let $\mathscr{U}=\big\{\{x\}:x\in F\big\}$ and observe that $\mathscr{B}(\mathscr{U})=\{F\}$. $\endgroup$ – Brian M. Scott Dec 4 '16 at 19:41

It is not entirely clear from the wording of the question whether this is what was intended, but I will read the question like this:

Let $X$ be a set. Is there a topology $\tau$ on $\newcommand{\powerset}[1]{\mathcal P(#1)}\powerset{X}$ such that for any non-increasing sequence of sets $s_1 \supseteq s_2 \supseteq \dots \supseteq s_n \supseteq s_{n+1} \supseteq\ \dots$ we have $$s=\bigcap_{i=1}^n s_i\qquad\text{if and only if}\qquad \lim_{i\to\infty} s_i=s,$$ i.e. $s$ is the intersection of this system if and only if the sequence $(s_i)$ converges to $s$ in the topology $(\powerset X,\tau)$?

In fact, we can take a very simple topology which fulfills this. There is natural bijection between $\powerset X$ and $\{0,1\}^X$ given by mapping $A\mapsto \chi_A$, i.e., every set is mapped to its characteristic function.

The we simply take discrete topology on $\{0,1\}$ and we endow $\{0,1\}^X$ with product topology. We get the space which is usually called Cantor cube.

In product topology, convergence is precisely pointwise convergence. So what is convergence in our topology? When a sequence $A_i$ of subsets of $X$ converges to $A$? This is equivalent to $$(\forall x\in X)\lim_{i\to\infty}\chi_{A_i}(x)=\chi_A(x),$$ where the convergence of $\chi_{A_i}(x)$ to $\chi_A(x)$ is convergence in the discrete space $\{0,1\}^X$.

In the other words, $\lim\limits_{i\to\infty} A_i=A$ iff for every $x\in X$:

  • If $x\notin A$, then there is $i_0$ such that $x\notin A$ for $i\ge i_0$.
  • If $x\in A$, then there is $i_0$ such that $x\in A$ for $i\ge i_0$.

Claim. If we take Cantor space topology on $\powerset X$, then for any non-increasing sequence of sets $s_1 \supseteq s_2 \supseteq \dots \supseteq s_n \supseteq s_{n+1} \supseteq \dots$ the equivalence $$s=\bigcap_{i=1}^n s_i\qquad\Leftrightarrow\qquad \lim_{i\to\infty} s_i=s$$ holds.

$\boxed{\Rightarrow}$ Let $s=\bigcap\limits_{i=1}^n s_i$. If $x\in s$, then clearly $x\in s_i$ for each $i$. If $x\notin s$, then there is and $i_0$ such that $x\notin s_{i_0}$. And since the sequence is non-increasing, we have $x\in s_i$ for each $i\ge i_0$.

$\boxed{\Leftarrow}$ Let $s_i\to s$. If $x\in s$, then $x\in s_i$ for every large enough $i$. But since the system is non-increasing, we get $s\in s_i$ for each $i$. This means that $$s\subseteq \bigcap\limits_{i=1}^n s_i.$$ We can show the opposite inclusion by a similar argument. (Or we can simply rephrase the above argument to get equivalences instead of implications.)

While the Cantor topology on $\powerset X$ is useful in many contexts and it is often used, when I read the first part of your question, it seemed to me that this might be closer to limit superior and limit inferior of sets. (For the second part, the choice of the topology would probably depend on the nature of the function $F$.)

You can read more about limit superior and limit inferior of sequence of sets in this Wikipedia article. Clearly, if they coincide, we can talk about limit. Similar convergence for for closed (or compact) subsets of a topological space is known as Kuratowski convergence.

If this seems like something which would be interesting you, you can find some other questions on this site about this type of convergence (and about limit superior/inferior of sets). You could find such questions, for example, in the . To find the ones which are about this type of extreme limits, perhaps you could add general-topology tag or elementary-set-theory tag.

For example, you can find this question which is somewhat similar to yours: Do limits of sequences of sets come from a topology? (And the answer given there is Cantor cube.)

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