How to determine the existence of all subsets of a set?
Given
- The definition of subset;
- The axiom of power set: for any set $S$, there exists a set $\wp$ such that $X \in \wp$ if and only if $X\subseteq S$
we know what a subset is and what a power set contains.
In a simple case where a set $A$ is supposed to exist, with $A=\{a, b, c\}$, we know what is and what is not a subset of $A$:
$\{a\}, \{b\}, \{c\}, \{a, b\}, \{a,c\}, \emptyset$ and $A$ are subsets of $A$ and anything different is not.
$\wp(A)=\big\{\{a\}, \{b\}, \{c\}, \{a, b\}, \{b,c\}, \{a,c\}, \{a, b, c\}, \emptyset\big\}$.
However the mere definition of something (and consequently it's recognition as such) does not guarantee its existence. $\emptyset$ and $A$ seem like the only subsets whose existence is immediate.
In other words, I know what a power set contains, but how do I know that the things it contains exist in the first place?
Because such a well-defined and existent set such as $\wp(A)$ should not contain nonexistent elements, to prove the existence of its elements is important. It seems that two alternatives arise:
- Being a member of $\wp(A)$ automatically makes this thing to exist;
or
- There should be an alternative to prove the existence of all subsets of $A$ without the axiom of power set.
The power set axiom just tells you what it says: for every $A$, there exists a set $\mathcal{P}(A)$ such that $$ \text{for all $B$, $B\in\mathcal{P}(A)$ if and only if $B\subseteq A$} $$ There is no claim of “existence” of any particular subset of $A$. In $\mathsf{ZFC}$ one can show that $|\mathcal{P}(A)|>|A|$, so there is plenty of subsets.
It should be noted that, if $A$ is infinite, there is no hope to find, for each subset of $A$, a formula “describing it”, because $\mathcal{P}(A)$ is uncountable. This is however not a problem: the axiom tells you that you have a “container” for all subsets of $A$; when you prove that a set $B$ is a subset of $A$, then you know it belongs to $\mathcal{P}(A)$; and conversely, if you pick $B\in\mathcal{P}(A)$, you know $B\subseteq A$.
The real purpose of the axiom is that the subsets of a set form a set. In particular, for instance, the equivalence relations on a set form a set that can be isolated from $\mathcal{P}(A\times A)$ using a suitable predicate and the axiom of separation.
I remember some good notes about this in Paul J. Cohen's “Set theory and the continuum hypothesis”.
I don't think you can actually derive what $P(S)$ is for any random $S$, from the axioms. Try to describe $P(\mathbb N)$ for instance.
If you have described a set $P'$ and want to know if $P'=P(S)$, then you can't do much more than try to prove that $$X\in P'\implies X\subset S\\ X\notin P'\implies X\not\subset S.$$ But that's only if you have found a way to describe $P'$, which will not always be possible.
Lastly, $\emptyset$ and $S$ are not the only subsets whose existence is immediate, for any set $S$. It is not very hard to prove the existence of sets with more than one element. This means we can do the following:
Suppose we have a set nonempty $S$ which is not a singleton, i.e. $(\forall x)(\emptyset\neq S\neq\{x\})$. Now, assume that $P(S)=\{\emptyset, S\}$. $$x\in S\implies \{x\}\subset S\implies \{x\}=\emptyset \vee \{x\}=S$$ Both $\{x\}=\emptyset$ and $\{x\}=S$ give a contradiction, so there have to be subsets of $S$, other than $\emptyset$ or $S$.