# Does Axiom of Power Set implies Axiom of Subset?

I am learning set theory based on Pinter textbook. In the textbook, the author suggests that the axiom of replacement implies the axiom of pairing and the axiom of subset. I was trying to deduce the axiom of subset based on the axiom of replacement and some other axioms. But I noticed that I don't even need the axiom of replacement to do it. This is my claim. For any set, the axiom of power set guarantees the existence of power set. Then, for a subset of the set, the subset is a member of the power set, so it is a set, due to the definition of set.

If this is true, why do we even need the axiom of power set? What is wrong? And how should I properly deduce the axiom of subset from the axiom of replacement?

• The question is, how do you guarantee that the power set of $X$ contains any elements besides $X$ and $\varnothing$? – Malice Vidrine Apr 15 '17 at 15:37
• @MaliceVidrine The main question is that what is the relationship between the two axioms. – Kang Apr 15 '17 at 16:19
• Yes, that's what I was trying to indicate.... Assuming by "axiom of subset" you mean "axiom schema of separation," you would need to show that merely the axiom of power set proved every instance of separation on $X$. If it did, then you could prove just from power set that there were elements other than the two I mentioned above. If by "axiom of subset" you mean something different, then you'll need to specify what. – Malice Vidrine Apr 15 '17 at 16:32

## 2 Answers

The power set axiom tells you that when $X$ exists, then a $\mathcal P(X)$ also exists, with the property that everything that exists and is a subset of $X$ is an element of $\mathcal P(X)$.

It doesn't guarantee that any particular subsets exist, only that whenever you find something in your model that is a subset of $X$, it will be in $\mathcal P(X)$.

For example, in a countable model of set theory (which has to exist due to Skolem-Löwenheim), not all the subsets of the model's natural numbers will actually exist as sets in the model, but some of them will -- and what the model considers to be $\mathcal P(\mathbb N)$ will consist of those and only those.

To prove that $\{x\in X\mid \phi(x)\}$ exists as a set, divide into cases depending on whether anything in $X$ satisfies $\phi$. If not, then you're looking for the empty set, which you should already know exists. Otherwise choose some $y\in X$ such that $\phi(y)$ and consider the range of $$F(x) = \begin{cases} x &\text{if }\phi(x) \\ y &\text{otherwise} \end{cases}$$

• I understand that the axiom of power set does not imply the existence of any particular subset. But I think it still impies the axiom of subset, which states that every subclass of a set is a set. Both don't state the existence of subset. They just talk about whether a given subclass of a set is a set or not. I still don't understand why we need the axiom of subset although we have the axiom of power set, which is more powerful. – Kang Apr 15 '17 at 16:17
• @Kang: Is this based on NBG, MK, or some other non-ZF axiom system? If so, please quote the exact formal statement of the version of the power set axiom you have. – Henning Makholm Apr 15 '17 at 16:19
• I don't know what axomatic sysyem is used in the Pinter textbook exactly, but in the book, it states "If A is a set then P(A) is a set" where the definition of P(A) is "the class of all subsets of A" – Kang Apr 15 '17 at 16:24
• Does the existence of P(A) is guaranteed by the axiom of subset? I just figured out this idea – Kang Apr 15 '17 at 16:27
• @Kang: Neither implies the other, in their usual formulations. (And if you have a book that doesn't bother to state the axioms formally, then it might be a good idea to find one that does instead). As it is, I think the best I can say is the if you have a subCLASS of $A$, then without the subset axiom you don't know that it is a subSET, and therefore nobody says it has to be an element of $\mathcal P(A)$. – Henning Makholm Apr 15 '17 at 16:31

Note that the class of ordinals satisfies the power set axiom (if $\alpha$ is an ordinal, then a subset of an ordinal which is also an ordinal is some $\beta\leq\alpha$, and therefore $\alpha+1$ is the power set of $\alpha$ with respect to the ordinals; or in simpler terms, if $\alpha$ is an ordinal, then $\mathcal P(\alpha)\cap\mathrm{Ord}=\alpha+1$).

But most certainly the ordinals do no satisfy Separation, since "the class of nonempty elements" (of some ordinal larger than $1$) is not an ordinal, and therefore does not make a set in this context.