The ZFC axioms

Self-membership is exactly the kind of self-reference that powers Russell's paradox; Foundation rules it out not by deriving a contradiction each time, but once and for all, as a structural rule on how sets can be built. This is what lets mathematicians safely define things by induction/recursion on the membership relation, and it is why the sets that appear throughout ordinary mathematics never contain themselves.

Suppose, for contradiction, that some set $x$ satisfies $x \in x$.

By the Axiom of Pairing, the set $\{x\}$ exists (pair $x$ with itself). This set is nonempty, since it contains $x$.

Apply the Axiom of Foundation to the nonempty set $\{x\}$: there must exist $y \in \{x\}$ with $\{x\} \cap y = \emptyset$. But the only element of $\{x\}$ is $x$ itself, so $y = x$, and the conclusion becomes $\{x\} \cap x = \emptyset$.

Now recall the contradiction hypothesis $x \in x$. Since $x \in \{x\}$ (by definition of the pair) and $x \in x$ (our assumption), the element $x$ belongs to both $\{x\}$ and $x$, so $x \in \{x\} \cap x$. This means $\{x\} \cap x$ is nonempty, directly contradicting $\{x\} \cap x = \emptyset$ obtained above.

This contradiction shows the assumption $x \in x$ is impossible for any set $x$; hence $x \notin x$ for every set $x$, as claimed.

Zorn's Lemma, Choice and Well-Ordering look completely different — one is about order and maximal elements, one is about choosing simultaneously from many sets, one is about a total order with no infinite descent — yet each encodes exactly the same underlying power to make infinitely many unconstrained choices at once. This is why algebraists reach for Zorn's Lemma to prove existence statements (a maximal ideal, a basis, an algebraic closure) that Choice alone would state less conveniently.

We prove Choice $\implies$ Zorn's Lemma (the other equivalences are standard but longer; this direction is the one used constantly in algebra). Let $(P, \le)$ be a nonempty poset in which every chain has an upper bound in $P$, and suppose toward contradiction that $P$ has no maximal element.

Since $P$ has no maximal element, every $x \in P$ has some strict upper bound in $P$ (an element $y$ with $x < y$): otherwise that $x$ would itself be maximal. In particular, every chain $C \subseteq P$ has an upper bound (by hypothesis) which itself has a strict upper bound, so every chain has a strict upper bound in $P$.

By the Axiom of Choice, fix a choice function that selects, for every chain $C \subseteq P$, some strict upper bound $g(C)$ of $C$. Using $g$, build a transfinite sequence $(a_\alpha)$ indexed by all ordinals $\alpha$: let $a_0 = g(\emptyset)$, and for each ordinal $\alpha$, once $a_\beta$ has been defined for every $\beta < \alpha$, the set $\{a_\beta : \beta<\alpha\}$ is a chain (by construction each new term is a strict upper bound of all earlier ones), so define $a_\alpha = g(\{a_\beta : \beta<\alpha\})$, a strict upper bound of every earlier term.

This produces a strictly increasing map from the ordinals into $P$: $\alpha < \beta \implies a_\alpha < a_\beta$. But the ordinals do not form a set (there is no set of all ordinals), while $P$ is an ordinary set; a strictly increasing map from the ordinals into $P$ would make the ordinals no larger in cardinality than $P$, contradicting the fact (a consequence of Replacement) that no set can be mapped injectively onto a collection as large as all the ordinals. This contradiction shows the assumption "$P$ has no maximal element" is false, so $P$ has a maximal element, proving Zorn's Lemma from Choice.