Hartogs theorem says that for every set $X$ there is a least ordinal $h(X)$ for which no injection $h(X)\to X$ exists. This does not require the axiom of choice. Form the set $W$ of well-orderings of subsets of $X$. It can be defined by the axiom schema of specification insiderecording a subset and its well-order. Each member of $W$ has a unique ordinal order type. Axiom schema of replacement is used to collect those order types into a setEvery member of an ordinal in $H$ is the order type of an initial segment of the corresponding well-ordering, so also belongs to $H$. Thus $H$ is a transitive set of ordinals and is itself an ordinal. An ordinal $\alpha$ belongs to $H$ exactly when it injects into $X$: use a bijection to a well-ordered subset in one direction, and transport the ordinal order along the injection in the other. If $H$ injected into $X$, this characterization would give $H\in H$, impossible for an ordinal. All smaller ordinals lie in $H$ and do inject into $X$. Therefore $H=h(X)$. Apart from the standard order-type theorem, the explicit collection of all order types is the Axiom schema of replacement step; no axiom of choice of a representative ordering for each type is needed.

Zorn's lemma states that a nonempty partially ordered set in which every chain has an upper bound has a maximal element. Use the Hartogs bound on partial ordinal injections to obtain a set-sized partially ordered set. Consider the partial injections into $X$ whose domains are ordinals. Hartogs theorem bounds every such domain strictly below $h=h(X)$. Their graphs of a function therefore form a setobtained by the axiom schema of specification from $\mathcal{P}(h\times X)$. Order $P$ by extension. It is nonempty because it contains the empty function. A chain has as an upper bound the union of its graphs of a function: extension makes this a function, injectivity is preserved, and its domain is an union of ordinals and hence an ordinal. Its range is contained in $X$, so Hartogs theorem forces its domain to be strictly below $h$; thus it belongs to $P$. The empty chain also has the empty function as an upper bound.

By Zorn's lemma choose a maximal $f:\alpha \to X$. If an element $x\in X$ lay outside its range, appending $f(\alpha )=x$ would give a larger injection on $\alpha +1$. Such an injection itself guarantees $\alpha +1<h$ by Hartogs theorem, so it is a legitimate member of $P$, contradicting maximality in a partially ordered set. Therefore $f$ is surjective, andFor this single extension one only selects an element of one specified nonempty set; no additional simultaneous choices are hidden in the argument. Zorn's lemma is the form of the axiom of choice used to obtain maximality in a partially ordered set.

Finally work with cardinals, represented by initial ordinals, and put $\mu =\max (\kappa ,\lambda )$. It is infinite by hypothesis. Nonzeroness of each factor gives an injection of the larger cardinal into the Cartesian product by holding a coordinate of the other fixed. The supplied infinite-ordinal square result gives $|\mu \times \mu |=\mu$. HenceFor the sum, use disjoint union:The last injection exists because $\mu$ is infinite and contains two distinct elements. Cantor-Schröder-Bernstein theorem turns both pairs of bounds into equalities, provingThese final operations are cardinal sum and multiplication, measuring disjoint unions and Cartesian products. They are not the noncommutative ordinal arithmetic considered in question 5.