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Use the Kuratowski ordered pairThe axiom of pairing first constructs $\{x\}$ and $\{x,y\}$, then constructs the displayed pair of those sets, proving that it is a set.

For any $a,b$, the intersection of the members of this pair is $\{a\}$, and their union is $\{a,b\}$. Thus equality of two ordered pairs first gives $\{a\}=\{c\}$, hence $a=c$, and then $\{a,b\}=\{a,d\}$. If $b=a$, this latter set is a singleton and forces $d=a$. If $b\ne a$, its unique member distinct from $a$ is $b$, forcing $d=b$. Hence equality of ordered pairs gives equality of both coordinates.

For a set $R$ of ordered pairs, both coordinates of every pair belong towhich is a set by two applications of the Axiom of union. Axiom of separation inside $U$ givesThese are sets, including when $R$ is empty.

The Axiom of foundation asserts that every nonempty set $A$ has an element $a$ with $a\cap A=\varnothing$. If a descending membership sequence $f:\omega \to V$ existed, replacement would give the nonempty range set $S=\{f(n):n\in \omega \}$. Foundation would give some $f(n)$ disjoint from $S$. But $f(n^{+})\in f(n)$ and $f(n^{+})\in S$, contradicting that disjointness. Thus foundation rules out every such sequence.

Conversely suppose there is no descending membership sequence, but foundation fails. There is then a nonempty set $A$ such that $a\cap A\ne \varnothing$ for every $a\in A$. The axiom of choice selects a function $h$ on $A$ with $h(a)\in a\cap A$. Choose $a_0\in A$ and use recursion to defineEvery value remains in $A$, and $f(n^{+})\in f(n)$ for every $n$. This is the forbidden sequence. Therefore, with the stated choice assumption, foundation is equivalent to the absence of descending membership sequences. The selection step explains exactly where choice is used; dependent choice would suffice.

Use the Kuratowski ordered pairThe axiom of pairing first produces $\{a\}$ by pairing $a$ with itself, then $\{a,b\}$, and then the pair of these two sets. The axiom of extensionality supplies the usual uniqueness of each specified set. In particular, no use of Axiom schema of replacement is needed to construct a single ordered pair from two given sets.

To recover the coordinates from $P=⟨a,b⟩$, observe thatThe first equality recovers $a$. Once $a$ is known, the second recovers $b$: if its union is the singleton $\{a\}$ then $b=a$, and otherwise $b$ is its unique element different from $a$. Hence equality of two encoded pairs forces equality of both coordinates, including the degenerate case $a=b$.

The first ordered triple encoding works. Ifinjectivity of the ordered pair construction first gives $a=a^{\prime }$ and $⟨b,c⟩=⟨b^{\prime },c^{\prime }⟩$, and a second application gives $b=b^{\prime }$ and $c=c^{\prime }$. The construction uses only two successive ordered pair constructions, so is a set as well.

Represent a function by its graph of Kuratowski ordered pairs. For $f:\omega \to {\omega }^2$,The bound is attained by $f(n)=\omega n$, which is a cofinal function into ${\omega }^2$. Taking the rank of the set of all such functions adds one at this attained maximum. Thus $\boxed{\mathrm{rk}(({\omega }^2{)}^{\omega })={\omega }^2+1}$. The extra successor is essential: unlike the previous product, this set actually contains a member of rank ${\omega }^2$.

Use the Kuratowski ordered pair $(m,n)=\{\{m\},\{m,n\}\}$. Its rank of a set is $\max (m,n)+2$. Every pair in the Cartesian product has finite rank, but those ranks are unbounded in $\omega$. Therefore $\mathrm{rk}(\omega \times \omega )={\sup }_{m,n<\omega }(\max (m,n)+3)=\boxed{\omega }$.

The essential fact is absoluteness of cardinalhood in limit ranks. Let $\alpha \le \beta$ be limit ordinals and $x\in V_{\alpha }$. Being an ordinal is absolute between these transitive sets, so a non-ordinal is a cardinal number in neither structure. For an ordinal $\delta \in V_{\alpha }$, we have $\delta <\alpha$.

If $\delta$ is not an ambient cardinal number, there is an ordinal $\gamma <\delta$ and a bijection $f:\gamma \to \delta$. With Kuratowski ordered pairs, its graph has rank of a set at most $\delta +3$. Since $\alpha$ is a limit ordinal, $\delta +3<\alpha$, so $f\in V_{\alpha }$. The assertion that $f$ is such a bijection is a bounded formula in set theory and is therefore evaluated correctly by each rank. Conversely, any internal bijection witnessing failure of cardinal number status is an actual one by the same set-theoretic absoluteness. ConsequentlyUnlike arbitrary transitive models, these rank-initial structures contain every sufficiently low-rank witness. That witness availability is what upgrades the usual downward absoluteness of cardinalhood to agreement in both directions.