0
$\begingroup$

I recently obtained an old textbook on the theme Mathematical Structures (in my mother language, Slovenian). The content is truly fascinating, so naturally, after I dug through some of the main concepts, I sought universal applicability, or in simpler words, how the notation ultimately corresponds, regardless of the language in which the different textbooks are written.

To my horror, I quickly found a damming difference between my textbook and the ZFC axioms as presented in this Wikipedia website. The Definition of Power Set in my textbook is such:

enter image description here

To put it in LaTeX context:

$$\mathcal{P}A = \{ X; \; X \subseteq A \}$$

Which we could read as: A Power Set of a chosen set A is a set which consists of all the subsets of set A. As you can see, an example is given in the photo I provided as well. Up to this point, everything is normal. But for the existence of a Power Set, we have to propose a special axiom, which ensures a non-contradictory framework (so Cantor sets won't be a problem):

enter image description here

Once again, a LaTeX equivalence:

$$(\forall A)(\exists ! C)(C = \mathcal{P}A)$$

We read: For a given set A, there exists precisely one set C such that set C is the Power Set of set A. I find the little detail in this axiom that is $\exists !$ absolutely necessary, as there mustn't exist multiple modules of one set that derive from one definition. However, the Wikipedia page about ZFC axioms proposes this:

$$\forall X \exists Y \forall Z[Z \subseteq X \Rightarrow Z \in Y]$$

Which, put in words, gives: For every set X, there exists a set Y, such that for every set Z follows that if Z is a subset of X, then Z is an element of Y. At first glance, this axiom is even more elegant than the one found in my textbook, as it uses three sets and doesn't immediately assume their origin. But I am very bothered by the fact that the operand is not a $\exists !$, rather a $\exists$, thus not excluding a possibility of multiple Power Sets, at least it seems so.

Please help me. Though I have a hypothesis as to why this is so, it is NOT straight-forward: My initial thought was that the operand $\exists !$ was not needed in the Wikipedia page, because the definition of a power set was wider, therefore making the specification obsolete. Answers, criticisms and remarks are welcome. Cheers!

EDIT: Please ignore the fact that the author used old mathematical notation (e.g.: the old equivalence of $\exists !$ is $\mathsf{E} !$). Rest assured that all of the symbols mean the same as new notation.

$\endgroup$
4
$\begingroup$

There are two reasons:

  1. $\exists!$ is not a standard quantifier. It is a shorthand, $\exists!x\varphi(x)$ is really $\exists x(\varphi(x)\land\forall y(\varphi(y)\rightarrow x=y))$. This means that any statement where $\exists!$ is used, is inherently more complicated than a statement where only $\exists$ is used.

  2. The axiom of extensionality ensures that definitions give us unique sets. In particular, if there are two sets whose elements are exactly the subsets of $x$, then the two sets are equal. Therefore, we can prove that $\exists y$ is really $\exists!y$, when we define $y$ to be the power set of $x$.

$\endgroup$
  • $\begingroup$ Thank you for the answer! Since you have proved that $\exists !$ is more composite than $\exists$, how would you comment on the fact that the author chose the former operand to construct the axiom. Is the proposition therefore faulty/could be done better? Do I have good mathematical literature? $\endgroup$ – Gregor Perčič Sep 30 '17 at 18:21
  • 1
    $\begingroup$ The zfc axiom says there is a set containing the power set. The axiom of separation then gives the actual power set from this. $\endgroup$ – Kyle Miller Sep 30 '17 at 18:22
  • $\begingroup$ @Gregor: I don't know why the author chose this formulation. This can be pedagogical, where one does not care at all about logical complexity; or it could be that whoever wrote this document is not actually a working set theorist, and wasn't paying attention to this kind of detail. Or it could be anything else. I don't know. $\endgroup$ – Asaf Karagila Sep 30 '17 at 18:22
  • $\begingroup$ @Kyle: Yes, one can formulate the power set axiom in a broader sense, but one can also require that there is a set which is exactly the power set. $\endgroup$ – Asaf Karagila Sep 30 '17 at 18:23
  • $\begingroup$ @AsafKaragila Okay. I wanted to make sure that I am reading legit literature and that after digesting it, I will have a deep-enough understanding of Set Theory. Cheers! $\endgroup$ – Gregor Perčič Sep 30 '17 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.