# What's an example of a function with $\mathcal{P}(\mathcal{P}(X)) \rightarrow \mathcal{P}(X)$ and $X \rightarrow \mathcal{P}(X)$

What's an example of a function with $$\mathcal{P}(\mathcal{P}(X)) \rightarrow \mathcal{P}(X)$$ and $$X \rightarrow \mathcal{P}(X)?$$

Absolute beginner here.

The first confusion I have is whether I have to provide two functions or one function satisfying both constraint. I will assume the first interpretation is true.

So $$\mathcal{P}(\mathcal{P}(X))$$ refers to all subsets of the set containing all subsets of X. For example if $$X = \{0\}$$, then $$\mathcal{P}(\mathcal{P}(X))$$ is $$\{\emptyset, \{0\}, \{\emptyset\}, \{0, \emptyset\} \}$$.

A function with $$\mathcal{P}(\mathcal{P}(X)) \rightarrow \mathcal{P}(X)$$ means that the domain is $$\mathcal{P}(\mathcal{P}(X))$$ and the codomain is $$\mathcal{P}(X)$$. An example I can think of is the $$\max()$$ function after I flatten out the element of a set(i.e. remove nested $$\{$$ and $$\}$$), but is it correct?

• How could one function have two different domains? Except if there is something I am missing, you need to provide two functions. Also, is $X$ given, or do you need a definition that would work for every choice of set $X$? Commented Oct 20, 2022 at 4:29
• $\max()$ is defined on $X$ if there is an order relation on $X$, so if $X$ is arbitrary, you can't use $\max()$. Commented Oct 20, 2022 at 4:30
• @Taladris Understood. $X$ is only arbitrary.
– Y.T.
Commented Oct 20, 2022 at 4:33
• Usually a function is called a map from any set to $\mathbb{R}$. So here we are interested in finding two maps, given only their sets. Any restrictions? Otherwise you can find easily such maps. Commented Oct 20, 2022 at 4:36
• @dmtri There is no other restriction.
– Y.T.
Commented Oct 20, 2022 at 4:40

There are many functions $$X\to \mathcal P(X)$$: for example,

1. $$x\mapsto \emptyset$$
2. More generally, if $$A$$ is a subset of $$X$$, $$x\mapsto A$$ is a constant function.
3. $$x\mapsto \{x\}$$ is an example of non constant function.
4. You can play with set operations and consider $$x\mapsto \{x\}\cup A$$ or $$x\mapsto \{x\}\cap A$$ or $$x\mapsto A\setminus\{x\}$$.

Similarly, since $$\emptyset$$ is a subset of every set, $$A\in\mathcal P(\mathcal P(X)) \mapsto \emptyset$$ is a constant function from $$\mathcal P(\mathcal P(X))$$ into $$\mathcal P(X)$$. If $$X$$ is not empty, you can construct many more constant functions.

Interesting examples of functions $$\mathcal P(\mathcal P(X))\to\mathcal P(X)$$ includes

1. the topology induced by a family of subsets of $$X$$ (as a subbasis)
2. the $$\sigma$$-algebra induced by a family of subsets of $$X$$

(thanks to @Dave L. Renfro for these examples and links).

As @dmtri mentioned in comments, it is easy to construct functions if there is no further restriction. More precisely, if $$A$$ and $$B$$ are two sets, there is always at least one function $$A\to B$$, the only exception being when $$B$$ is empty but $$A$$ isn't.