I'm attempting to learn Set Theory and I'm currently working through Halmos' Naive Set Theory. I will say that I completely understand the essence of the Axiom of Extensionality. However, where I'm stumbling is grasping its formalization in set-builder notation. So, looking at: $$\forall x\forall y \, \left(x=y \leftrightarrow \forall z \, \left(z\in x \leftrightarrow z\in y\right)\right)$$

The part I do not follow is the $\left(z\in x\leftrightarrow z\in y\right)$

So, in my head, if I imagine that I have a set $x$ that is $\{ “a”, “b”, “c”\}$ and I have a set $y$ that is $\{“a”, “b”, “c”, “d” \}$, I intuitively understand that by the Axiom of Extension, sets $x$ and $y$ are not equal. However, I don't follow how the formal statement would yield this. So for example, suppose we have a set $z$ that is $\{ “a”, “b”, “c”\}$.

So, in my mind - here's what I do to mentally work through the formal statement of the axiom. First, I'll ask myself: Is $z$ in $y$? Well, since $y$ is $a, b, c, d$ and $z$ is $a, b, c$ - so it would certainly seem that $z$ is in fact in $y$. So then I'll ask myself: Is $z$ in $x$? Well, $x$ is $a, b, c$ and $z$ is $a, b, c$ - so it would certainly seem that $z$ is also in fact in $x$! Therefore, the expression $\left(z \in x \leftrightarrow z \in y\right)$ should evaluate to True and therefore $x = y$ (which of course I know is wrong).

So, hopefully you can see why I am stuck on this. I assume the root of my misunderstanding is a flaw in how I'm interpreting the formal statement; specifically the iff biconditional connective; but I'm just not seeing how/why I am.

Any help in providing clarity on where / how I'm stumbling on this would be greatly appreciated.

Thank you for you time.

  • $\begingroup$ This question may be useful for you. $\endgroup$ – Asaf Karagila Dec 27 '11 at 5:08
  • $\begingroup$ Knowing a bit about the philosophical background of the notion of "extensionality" provided insight for me when learning this. Try this. $\endgroup$ – Dan M. Katz Dec 27 '11 at 7:04
  • $\begingroup$ You're getting confused with $z\in x$ versus $z\subseteq x$. In the former, $z$ is an element of $x$, which means if you could list all the elements inside of the set $x$, you would (eventually) find $z$. In the latter, you're saying $z$ is a subset of $x$, which means every element of $z$ is also an element of $x$. That is, if you opened up the sets $z$ and $x$ and looked at each of their elements, you would find that everything inside $z$ is also inside $x$ (and there might even be more elements inside $x$ that are not inside $z$). $\endgroup$ – chharvey Jan 26 '14 at 6:32

You are confusing $z\in y$ ($z$ is an element of $y$) with $z\subseteq y$ ($z$ is a subset of $y$). There is also possible misunderstanding of the meaning of $\forall z$.

Let's look at $x$ and $y$ as in your example, and let $z=d$. Then, on the assumption that $d$ is not equal to any of $a$, $b$, or $c$, we have $z\in y$ but $z \notin x$. So by Extensionality, we conclude that $\lnot(x=y)$. Remember that $x=y$ if and only if for all $z$, $z\in x \iff z\in y$. You don't get to pick the $z$ that you will consider. Also, if you pick $z$ as you did, namely $z=\{a,b,c\}$, then we do not have $z\in x$, though we do have $z \subseteq x$. That is why I conjecture that you are having some trouble distinguishing between $z \in x$ and $z \subseteq x$.

Informally, Extensionality says that two sets $x$ and $y$ are equal if and only if $x$ and $y$ have the same elements.

Comment: The book you are going through is a classic work, very well written. Despite the passage of many years, it remains I think the best introduction to Set theory. Good choice!


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