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The cardinal product is the cardinality of the set of functions $f$ on the index set with $f(i)<{\kappa }_i$ at every coordinate. The empty product is one, because it contains the empty function. A zero factor makes a nonempty-index product zero. In ZFC, a family of nonzero cardinals has nonzero product by the axiom of choice.

The cardinal sum is the cardinality of a disjoint union, and the cardinal product is the cardinality of the set of choice functions:Here the cardinals are represented by initial ordinals. Tagging by $\alpha$ makes the summands disjoint even when their underlying sets overlap. For an empty index set the sum is zero and the product is one, because there is exactly one empty function. If any factor in a nonempty product is zero, that product is empty.

Zorn's lemma states that a nonempty partially ordered set in which every chain has an upper bound contains a maximal element. Here a chain means a subset totally ordered by the inherited order; it need not have a greatest member.

View every partial function as its graph, a subset of $\mathbb{R}\times \mathbb{R}$. The family $\mathcal{F}$ is a set by axiom of separation from the power set of this product. Let $\mathcal{C}\subseteq \mathcal{F}$ be a chain and put $h=\bigcup \mathcal{C}$. If two members of the union assign values to the same input, the chain property makes one graph contain the other, so those values agree. Thus $h$ is a partial function.

If $x\in \mathrm{dom}h$ and $q\in \mathbb{Q}$, choose a chain member containing $x$ in its domain. Its rational-scalar closure gives $qx\in \mathrm{dom}h$. If $x,y\in \mathrm{dom}h$, choose two chain members containing them; the larger one contains both, and therefore contains $x+y$ with value $h(x)+h(y)$. Hence $h$ satisfies the required additive identity and domain closure. Consequentlyand it is an upper bound for the chain. For the empty chain its union is the empty function, which satisfies the defining logical implications vacuously, so this case is covered too.

A nonempty weakly linear function has a domain $D$ closed under rational scalars and addition, hence a vector subspace of $\mathbb{R}$ over $\mathbb{Q}$. Additivity first gives $f(0)=0$ and $f(-x)=-f(x)$, then $f(nx)=nf(x)$ for integers $n$. If $q=m/n$ with $n>0$, rational closure puts $qx$ in $D$ andThus $f(qx)=qf(x)$: it is a $\mathbb{Q}$-linear map. The empty function is not maximal because it extends to the function on $\{0\}$ with value zero.

Suppose a maximal $f$ had proper domain $D\ne \mathbb{R}$. Choose $t\notin D$ and any prescribed real value $a$, for instance $a=0$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$: if two expressions coincide and the coefficients differ, then $t$ would belong to $D$. DefineUniqueness makes this well-defined. The displayed logical formula and the $\mathbb{Q}$-linearity of $f$ show directly that $h$ is additive and has a domain closed under rational scalars. Hence it is a weakly linear function and extends $f$. It is a proper extension because $t$ is now in its domain, contradicting maximality. ThereforePart (a) and Zorn's lemma guarantee the existence of maximal extensions as well as this description of their domains.

True in ZF. A function from $M$ into the functions $B\to A$ corresponds bijectively to a function from $B\times M$ into $A$:The inverse sends $f$ to the function $m\mapsto (b\mapsto f(b,m))$. These two explicit transformations are inverses, proving the currying law for cardinal exponentiation. Empty domains and codomains cause no exception: the same correspondence still applies, including the unique empty function when its domain is empty. No selection of representatives or choice functions is involved.

Hartogs theorem says that for every set $A$ there is a least ordinal $h(A)$ that does not inject into $A$. This theorem requires no axiom of choice.

To prove it, collect all well-orderings of subsets of $A$. They are set-sized data, coded by pairs from $\mathcal{P}(A)$ and $\mathcal{P}(A\times A)$. Each has a unique ordinal order type, so replacement gives a set $H$ of their order types. If $\eta \in H$ and $\xi <\eta$, restricting the corresponding well-order to an initial segment shows $\xi \in H$. Hence $H$ is an ordinal. If this ordinal itself injected into $A$, its transported well-order would put $H$ among its own elements, impossible for an ordinal. Every smaller ordinal does inject by definition. Thus $h(A)=H$ has the asserted property.

The recursion printed in the source needs two boundary restrictions. Its new-row successor clause allows $\beta =\alpha$, which would assign a value at $(\alpha ,\alpha +1)$ outside the declared domain. Also, at $\alpha =0$, its row-start clause gives $g_1(0,0)=0$ while its column-start clause gives $g_1(0,0)=g_1(0,0)+1$. Taken literally, these clauses do not define a function.

The intended extension fills the new row first and the remaining new column second. Restrict the row successor clause to $\beta <\alpha$, and apply the column-start clause only when $\alpha >0$. With these two restrictions the domains are disjoint:All successors here are ordinal successors. Simultaneous transfinite induction proves a stronger result: $g_{\alpha }$ is a bijection of $\alpha \times \alpha$ onto an ordinal ${\gamma }_{\alpha }$, and the functions extend one another.

At zero this is the empty function. Assuming it at $\alpha$, the least missing ordinal is ${\gamma }_{\alpha }$. Recursion down the new row gives values ${\gamma }_{\alpha }+\beta$ for $\beta \le \alpha$; recursion down the remaining column gives ${\gamma }_{\alpha }+(\alpha +1)+\beta$ for $\beta <\alpha$. These enumerate two disjoint consecutive ordinal intervals above the old range, so the extension is well defined and bijective, withAt a limit $\lambda$, take the union of the earlier functions. Compatibility makes it a function, and every pair of coordinates below $\lambda$ is already present at some earlier stage. Its range is the ordinal ${\bigcup }_{\alpha <\lambda }{\gamma }_{\alpha }$, and injectivity is preserved by the compatible union. This completes the construction of the row-and-column enumeration of an ordinal square.

For finite stages ${\gamma }_n=n^2$. For example the table of $g_3$, with the first coordinate indexing rows, isThe remaining parts refer to this domain-consistent construction. The source's literal boundary claims are preserved in the archive, rather than silently asserted as valid.

A weakly linear partial function on the real numbers has a domain closed under rational scalar multiplication and addition, and it preserves addition on that domain. A nonempty domain is a vector space over the rational numbers, and additivity implies rational homogeneity of an additive function by integer multiplication and division. The empty function also satisfies the defining conditions vacuously. This terminology does not assert homogeneity under arbitrary real scalars.