Discrete Mathematics with Applications, 3rd ed., by Susanna Epp mentions that Russell's Paradox can be avoided by ensuring that any condition elements of a set are to satisfy must contain the restriction that they are elements of a known set. As for Russell's Paradox, the textbook gives this reason as to why this new definition/restriction of set theory no longer leads to a contradiction:
Let $U$ be a universal set and suppose that all sets under discussion are subsets of $U$. Let $S=\{A|A \subseteq U \text{ and } A \notin A \}$. … [This leads to the conclusion] that $S \not\subseteq U$.
But as far as I can understand, the argument above still leads to a contradiction, namely, that $S \subseteq U$ and $S \not\subseteq U$. Here is my understanding that I would like to have checked:
So the conclusion is that $S$ does not exist in our universe. But $S$ is implicitly in $U$ by virtue of being in our discussion (as every set in our discussion is by definition in $U$) [why would this sentence be erroneous, if it is?]. Thus a contradiction is derived. But what led to the contradiction? The set $S$ is a perfectly valid set under the new axioms of set theory (because the condition of $S$ is valid under the new definition), so it must be that the new axioms of set theory are wrong.
The above argument seems very similar to the argument of Russell's Paradox, but I can't pinpoint what is wrong with it.
Best Answer
Here is what the author means by a "universal set" (see p. 314):
Everything what is done in sequel only concerns subsets of such a universal set $U$ (union, intersection, ...).
This does no mean that one universal set suffices for all purposes. For example, if $U = \mathbb R$, then it is clear that there are also releavant sets which are no subsets of $U$ (for example $\mathbb R^n$ with $n > 1$).
When the author introduces the power set $\mathscr P(A)$ of a set $A$ and the Cartesian product $A_1 \times \ldots \times A_n$ of sets $A_1,\ldots, A_n$, it becomes very clear that certain natural constructions with sets produce new sets not belonging to a given universe $U$. For example, $\mathscr P(U)$ is never a subset of $U$.
What the author writes about Russell’s paradox is, in my opinion, misleading.
It is not at all clear what this "suppose" means. Anyway, one can define the set $$S = \{A | A ⊆ U \text{ and } A \notin A\}$$ This is a well-defined subset of $\mathscr P(U)$. Since $\mathscr P(U)$ is no subset of $U$, we cannot expect that a given subset of $\mathscr P(U)$ is a subset of $U$ - but it is not impossible. If it were true that the above $S$ is a subset of $U$, we would again end with the contradiction that $S \in S$ implies $S \notin S$ and $S \notin S$ implies $S \in S$ (Russell’s paradox). Therefore we conclude that $S$ is no subset of $U$, and there is no contradiction.
Thus "suppose that all sets under discussion are subsets of $U$" is certainly intended to say the following:
As long as we stay in $U$, it is impossible to reproduce Russell’s paradox. This paradox is a consequence of a "naive", but illegitimate definition of the "set" of all sets $A$ with $A \notin A$.