1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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 $X$? $\endgroup$
    – Taladris
    Commented Oct 20, 2022 at 4:29
  • $\begingroup$ $\max()$ is defined on $X$ if there is an order relation on $X$, so if $X$ is arbitrary, you can't use $\max()$. $\endgroup$
    – Taladris
    Commented Oct 20, 2022 at 4:30
  • $\begingroup$ @Taladris Understood. $X$ is only arbitrary. $\endgroup$
    – Y.T.
    Commented Oct 20, 2022 at 4:33
  • $\begingroup$ 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. $\endgroup$
    – dmtri
    Commented Oct 20, 2022 at 4:36
  • $\begingroup$ @dmtri There is no other restriction. $\endgroup$
    – Y.T.
    Commented Oct 20, 2022 at 4:40

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .