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I'm just learning a bit about axiomatic set theory, and I'm kind of confused as to why we need/want this axiom?

Does not accepting it imply that there exists some set which doesn't have a power set?

If that was the case, why would we consider it a problem?

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    $\begingroup$ No, it does not imply that, otherwise ZFC would be obviously inconsistent. Not provable does not mean false. $\endgroup$ – Stefan Perko Jan 20 '16 at 7:43
  • $\begingroup$ Right, that's true. Thanks! $\endgroup$ – YoTengoUnLCD Jan 20 '16 at 7:44
  • $\begingroup$ What happens if we partially negate it, for example assuming $\mathbb{N}$ has no power set? $\endgroup$ – Dan Brumleve Jan 20 '16 at 7:45
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    $\begingroup$ Is there some reason you are asking this specifically about the Power Set axiom? Or do you have a similar question about any of the other axioms? $\endgroup$ – Ted Jan 20 '16 at 7:54
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    $\begingroup$ @esotechnica: Just because you can prove the existence of a set, does not mean that you can prove the existence of its power set. That's the whole point of the power set axiom. $\endgroup$ – Asaf Karagila♦ Oct 25 '17 at 9:37
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The power in the power set axiom is the ability to create larger sets than any other axiom is capable of. At least we want it because we probably want $\mathbb R$ (to be a set).

The other axioms doesn't seem to be strong enough to guarantee the existence such large set (larger than $\mathbb N$).

Note that dropping an axiom would not make it (automatically) false, because if it could be proven to be false when dropped we could use the same proof in a set theory where we don't drop it - and thereby get a contradiction.

In order to get a set theory where we're guaranteed to have a set without a power set we would actually have to replace the axiom of power set with it's negation (saying that there's a set without a power set). The possibility to do so without contradiction requires that the axiom is independent of the others (that is it can't be proven from the other).

To see what the axioms are able to produce out of the set we supply (that is what we already have) let's pick them one by one:

Axiom of Extensionality: Doesn't guarantee any existence at all (it's only stating a relation between supplied sets).

Axiom of Regularity: It hardly produce anything new, it only guarantee properties of the sets inside any set.

Axiom schema of Specification: It will only produce smaller sets, the produced set is a subset (with elements having the property) of the supplied set.

Axiom of Pairing: It will produce sets of size one or two given a supplied set.

Axiom of Union: It will produce the union of supplied sets, this will at most have cardinality of the outer set multiplied with the highest cardinality of the contained set. At most we can square cardinalities by this.

Axiom schema of Replacement: It will produce a set of lower cardinality, however the cardinality of it's element can be any cardinality we can construct. This means we can't increase cardinality in any way here.

Axiom of infinity: This is allowed to produce an infinite set out of nothing. Not only can we create an infinite cardinality, but we also can create sets which can be supplied to the other axioms. No particular infinity are not specified in the axiom, just that it has at least cardinality $\aleph_0$.

Well ordering axiom: This can construct a total order on a set. The cardinality of a total order on $A$ is $|A|(|A|+1)/2$. At infinite cardinalities this means squaring the cardinality.

Axiom of Power set: This will create power sets of any set $A$ which has the cardinality $2^{|A|}$.

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We want sets to model the idea of "mathematical objects which are collections of mathematical objects". And our naive understanding is that if $X$ is a set, then there is a collection of all the subsets of $X$. And even in set theories without the power set axiom, we can still talk about such a collection, since it defines a class.

Now we go back to finite sets, and we notice that every finite set has a power set, since we can literally write down this power set for very small sets, and prove inductively that if a set with $n$ elements has a power set, then under very mild assumptions, a set with $n+1$ elements will have a power set as well (these assumptions include union of two sets is a set; basic separation axioms; and very basic Replacement axioms in the finitary case (i.e., if we defined a function on a finite set, the image is a set as well)).

So if finite sets have a power set, why shouldn't infinite sets have a power set? I mean, it's a collection of objects and we can refer to this collection.

Now, if we consider the fact that the power set axiom is a powerful and useful axiom which provides us with the set of real numbers, and with Cartesian products, and with the set of functions from $\Bbb N$ to $\Bbb R$, which itself gives us all sort of useful sets like $\ell_2$... well, this is a good thing, then.

You could argue, though, I don't care about these sets which are mentioned above. I just care about these collections. But mathematical foundations using set theory, which is a fairly common and fairly useful form of mathematical foundation, deals with objects in the universe and these objects are sets. So when you want to assert that $\ell_2$ exists, you essentially assert that it is a set. So in that case, you should care about this.

Of course, there are alternative set theories, e.g. Pocket set theory, which do not have the power set axiom and in which the real numbers form a proper class. And sure, you can still talk about the real numbers, and develop a bunch of analysis just fine. But there is a problem now. Every time you want to talk about theorems about the real numbers, you need to refer to your meta-meta-theory (since Pocket set theory is now your meta-theory), and you need to start paying much more attention to that one. And that is a big hindrance in the common workflow of most mathematicians.

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Ultimately, this is one of the basic elements of the arithmetic of sets and of higher order reasoning.

Subsets of a set $X$ are a thing we like to talk about. Therefore, we'd like a set of all of them, so that we can most naturally use set theory to talk about them.

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The power set axiom is used to construct Cartesian products, which are used for relations, which are used for defining "function."

See https://en.wikipedia.org/wiki/Axiom_of_power_set at least for Cartesian products.

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  • $\begingroup$ Can't Cartesian products be defined without this axiom? $\endgroup$ – YoTengoUnLCD Jan 20 '16 at 8:07
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    $\begingroup$ If you only want Cartesian product of finite number of factors (as the example on wikipedia) you should be able to do that with the other axioms (using replacement and union axioms for example). $\endgroup$ – skyking Jan 20 '16 at 8:10
  • $\begingroup$ Hmm... given a set $x$, we can create $\{x\}$ by separation, union, and pairing, and given $x,y$, we can create $(x,y):=\{\{x\},\{x,y\}\}$ by the above and pairing again. Then with replacement, for $x\in X$ and a set $Y$, we can get the set of all $(x,y)$ for $y\in Y$, and with replacement again the set of all $(x,y)$. I didn't realize you didn't need powersets for this. (Anyway, another justification for powersets: we want to be able to write things like $\{f(U) : U\subset X\text{ and } P(U)\}$ and claim it is a set.) $\endgroup$ – Kyle Miller Jan 20 '16 at 8:19
  • $\begingroup$ It just says that it allows you to create a simple definition, not that it is necessary. $\endgroup$ – Brandon Thomas Van Over Jan 20 '16 at 8:19
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If you consider the axioms of separation pairing and union, you can apply them to a finite set, $A$, to create a set of all subsets of $A$, called the powerset of $A$, notated as $\mathscr P(A)$.

E.g. for $A = \{1, 2\}$ you can use separation to create the subsets $\emptyset, \{1\}, \{2\}, \{1, 2\}$. Then pairing to combine $\emptyset, \{1\}$ into $\{ \emptyset, \{1\} \}$ and $\{2\}, \{1, 2\}$ into $\{ \{2\}, \{1, 2\}\}$. Finally, union to combine $\{ \emptyset, \{1\} \}$ and $\{ \{2\}, \{1, 2\}\}$ into $\{ \emptyset, \{1\}, \{2\}, \{1, 2\}\} = \mathscr P(A)$.

What is rarely stated explicitly though is that the expression used in to separate a subset of a set must be finite. So if $A$ is an infinite set you can't create the powerset this way. Since we would like to be able to do this it requires a new axiom to allow it - the axiom of the power set.

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  • $\begingroup$ You are not using union, but pairing. The union of $\emptyset,\{1\},\{2\},\{1,2\}$ is $\{1,2\}$. In any case, the usual formulation of the axiom of union would require the existence of $\{A,B\}$ prior to forming $A\cup B$. $\endgroup$ – Andrés E. Caicedo Jan 20 '16 at 13:29
  • $\begingroup$ @AndrésCaicedo. Thanks: hopefully corrected now. $\endgroup$ – Tom Collinge Jan 21 '16 at 6:45
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Yes, cartesian products do not require power sets. But once power sets are accepted one obtains the existence of all functions between any pair of sets since this corresponds to a subset of the power set of the product consisting of all graphs. This then proves the existence of the set of all sequences of any set. I think this is a justification for power sets which is not often considered.

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  • $\begingroup$ The downvote seems a bit tough. Power set is indeed necessary for the existence of $X^\omega$. For example, HC is a model of ZFC - Power Set + $2^\omega$ does not exist. $\endgroup$ – Reveillark Oct 30 at 3:10

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