On Halmos' proof that "there is no universe"

On pp. 6-7 of his Naive Set Theory, Paul Halmos proves a result that I know very well, but I have a hard time following the argument he gives here. This question is about Halmos' argument (and not about the result itself, which I understand well enough).

I'll quote Halmos at length to make sure I don't miss anything. (I'll be as faithful as possible, but I will change his notation slightly, since he uses $\epsilon$ for $\in$, and $\epsilon^\prime$ for $\notin$; I will stick to $\in$ and $\notin$). The excerpt below begins right after he has given the Axiom of specification. (In the excerpt, I'll use boldface to indicate what I'm having trouble with.)

...To indicate the way $B$ is obtained from $A$ and from $S(x)$ it is customary to write $$B = \{x \in A:S(x)\}$$

To obtain an amusing and instructive application of the axiom of specification, consider, in the role of $S(x)$, the sentence $$\mathrm{not}\;(x \in x).$$

It will be convenient, here and throughout, to write $``x \notin A"$ ... instead of $``\mathrm{not}\;(x \in A)"$; in this notation, the role of $S(x)$ is now played by $$x \notin x.$$

It follows that, whatever the set $A$ may be, if $B = \{x\in A:x\notin x\}$, then for all $y$, $$(*)\;\;\;\;\;y\in B\;\mathit{\;if\;and\;only\;if\;} \;(y\in A\;\;\mathit{and}\;\;y\notin y).$$

Can it be that $B \in A$? We proceed to prove that the answer is no. Indeed, if $B\in A$, then either $B\in B$ also (unlikely, but not obviously imposssible), or else $B\notin B$. If $B\in B$, then, by $(*)$, the assumption $B\in A$ yields $B\notin B$—a contradiction. If $B\notin B$, then, by $(*)$ again, the assumption $B\in A$ yields $B\in B$—a contradiction again. This completes the proof that $B\in A$ is impossible, so we must have $B\notin A$.

(In the subsequent discussion, Halmos argues that this result means that "there is no universe [of discourse]", etc., etc.)

Now, this is what I don't get. If I replace $y$ with $B$ in $(*)$, I get

$$B\in B\;\mathit{\;if\;and\;only\;if\;} \;(B\in A\;\;\mathit{and}\;\;B\notin B).$$

This means that $B\in B$ and $(*)$ together imply $B\notin B$. Contrary to what Halmos writes, "the assumption $B\in A$" is not needed to draw this conclusion.

What am I missing?

(By the way, I would have worded the second part of the proof a bit differently, namely:

If $B\notin B$, then, by $(*)$ again, the assumptions $B\in A$ and $B\notin B$ jointly imply that $B\in B$.

I mention this only in case this rewording gives a clue about where my confusion lies.)

Solution 1:

You are right that the appeal to $B\in A$ in the bolded part of your quote is not necessary for the conclusion it's trying to draw.

I'd chalk it up to sloppy editing and move on.

Solution 2:

One way to view "There is no universe" is: "Given any set A, then exist a set B such that $B\notin A$". What Halmos does is to show a contradiction from $B\in A$, in this sense $B\in A$ is necessary.