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For an infinite cardinal number $\kappa$, use cardinal arithmetic with the axiom of choice. If $\lambda =2^{\kappa }$ had cofinality $\mu \le \kappa$, choose cofinal ordinals ${\alpha }_i<\lambda$, $i<\mu$. Their cardinalities $a_i$ satisfy ${\sum }_i{a}_i=\lambda$. The König theorem for cardinal numbers gives $\lambda <{\prod }_i{a}_i^{+}\le {\lambda }^{\mu }\le (2^{\kappa }{)}^{\kappa }=\lambda$, a contradiction.

Choose cofinal ordinals ${\alpha }_i<\kappa$, indexed by $\mu =\mathrm{cf}(\kappa )$, and set $a_i=|{\alpha }_i|$. Then ${\sum }_i{a}_i=\kappa$ and $a_i^{+}\le \kappa$. The König theorem for cardinal numbers gives $\kappa <{\prod }_i{a}_i^{+}\le {\kappa }^{\mu }$. The estimate holds for every infinite cardinal number, regular or singular, under the usual axiom of choice convention for cardinal arithmetic.

Constant functions give the lower bound ${\aleph }_2\le {\aleph }_2^{{\aleph }_0}$. By the Generalized continuum hypothesis, ${\aleph }_2=2^{{\aleph }_1}$. The currying law for cardinal exponentiation and infinite cardinal arithmetic therefore giveHence $\boxed{{\aleph }_2^{{\aleph }_0}={\aleph }_2}$.

Use the usual cardinal arithmetic with the axiom of choice. The König theorem for cardinal numbers states that for any index set $I$ and cardinal numbers ${\kappa }_i<{\lambda }_i$,Represent the cardinal numbers as initial ordinals. Let $A=\{(i,\alpha ):i\in I,\alpha <{\kappa }_i\}$ and $B={\prod }_{i\in I}{\lambda }_i$. There is an injection $A\to B$: send $(i,\alpha )$ to the function whose $i$th coordinate is $\alpha +1$ and whose other coordinates are zero. The unique nonzero coordinate identifies $i$, and its value identifies $\alpha$. All values are allowed because $\alpha +1\le {\kappa }_i<{\lambda }_i$.

For any proposed surjection $F:A\to B$, its $i$th row has at most ${\kappa }_i$ elements, soChoose $b_i$ to be the least omitted ordinal. The function $b=(b_i{)}_{i\in I}$ belongs to $B$ and differs from $F(i,\alpha )$ at coordinate $i$, for every $(i,\alpha )\in A$. Thus $F$ is not surjective. Cardinal comparison now gives the strict inequality. If $I=\varnothing$, the conclusion is $0<1$, also valid.

The Cantor-Schröder-Bernstein theorem says that injections $X\to Y$ and $Y\to X$ imply a bijection $X\to Y$. Thus mutual cardinal inequalities imply equal cardinalities, without a choice assumption.

Write $\mathfrak{c}=2^{{\aleph }_0}=|\mathbb{R}|$. Standard cardinal arithmetic gives $\mathfrak{c}\cdot \mathfrak{c}=\mathfrak{c}$ and ${\mathfrak{c}}^{{\aleph }_0}=\mathfrak{c}$. The correspondence $x+iy\leftrightarrow (x,y)$ is a bijection $\mathbb{C}\leftrightarrow {\mathbb{R}}^2$. HenceEquivalently, the real axis supplies the lower inequality and a coding of two real coordinates supplies the upper one; Cantor-Schröder-Bernstein then gives the equality.

The ordinal operations are defined by transfinite recursion in the right argument. Addition satisfies $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ at a nonzero limit. Multiplication satisfies $\alpha \cdot 0=0$, $\alpha \cdot (\beta +1)=\alpha \cdot \beta +\alpha$, and $\alpha \cdot \lambda ={\sup }_{\beta <\lambda }\alpha \cdot \beta$. Exponentiation satisfies ${\alpha }^0=1$, ${\alpha }^{\beta +1}={\alpha }^{\beta }\cdot \alpha$, and ${\alpha }^{\lambda }={\sup }_{\beta <\lambda }{\alpha }^{\beta }$ at nonzero limits. These are ordinal operations, not commutative cardinal arithmetic.

For a normal function on an ordinal, prove domination by transfinite induction. At zero, $0\le \Phi (0)$ automatically. If $\alpha \le \Phi (\alpha )$, strict increase at successors gives $\alpha +1\le \Phi (\alpha )+1\le \Phi (\alpha +1)$. At a nonzero limit $\lambda$, continuity and the induction hypotheses giveTherefore $\boxed{\alpha \le \Phi (\alpha )\text{ for every ordinal }\alpha }$. No axiom of choice is needed for this ordinal induction.

Call a set function $f$ with ordinal domain $\gamma$ good if $f(\xi )=G(\xi ,\mathrm{ran}(f{|}_{\xi }))$ for every $\xi <\gamma$. Any two good functions agree on their common domain: otherwise their first disagreement has identical earlier ranges and hence identical values under $G$. This is transfinite induction, not an assumption of the desired recursion.

Good functions exist with every ordinal domain. At a successor stage extend the unique previous function by its prescribed value under $G$. At a limit stage, Axiom schema of replacement collects the unique good functions on all smaller domains into a set, and their compatible union is a good function on the limit domain. Each earlier range is a set, so $G$ is applied only to sets. Transfinite induction now establishes existence at all stages. Define $F(\alpha )$ to be the value at $\alpha$ of the unique good function on $\alpha +1$. This is a definable class function because goodness and the uniqueness assertion are formulas in the defining formula for $G$. Compatibility gives

An initial ordinal is an ordinal not bijective with any smaller ordinal. It is the canonical ordinal representative of a well-orderable cardinal, so cardinal arithmetic can be defined by taking the initial ordinal equinumerous with a disjoint union, Cartesian product or function set. In the usual setting with the axiom of choice, every cardinal has such a representative. Hartogs theorem says that for every set $A$ there is an ordinal $h(A)$ which does not inject into $A$, and there is a least such ordinal. For an initial ordinal $\kappa$ it is the next larger initial ordinal: every smaller ordinal injects into $\kappa$, and if $h(\kappa )$ were equinumerous with a smaller ordinal it too would inject into $\kappa$.

Use the recursion just proved to set ${\omega }_0=\omega$, ${\omega }_{\beta +1}=h({\omega }_{\beta })$ and ${\omega }_{\lambda }={\sup }_{\beta <\lambda }{\omega }_{\beta }$ for nonzero limit $\lambda$. These instructions use the earlier value set alone: at a successor it has a maximum, and at a limit its supremum is taken. They are therefore implemented by an appropriate definable $G$. The limit supremum is initial: if it were bijective with $\gamma$ below it, choose an earlier ${\omega }_{\beta }>\gamma$; then ${\omega }_{\beta }$ would inject into $\gamma$, contradicting initiality. The sequence is strictly increasing and continuous. There are no omitted infinite initial ordinals, since each successor chooses the least next one and each limit is the supremum; more formally, a least omitted initial ordinal cannot lie between successive terms, and the least term above it cannot be a limit term. Terms eventually exceed any prescribed ordinal, since a strictly increasing ordinal sequence satisfies ${\omega }_{\beta }\ge \beta$.

There is a fixed point of a normal ordinal function. Explicitly let ${\alpha }_0=0$, ${\alpha }_{n+1}={\omega }_{{\alpha }_n}$ and $\alpha ={\sup }_{n<\omega }{\alpha }_n$. The sequence is strictly increasing, by induction from $0<{\omega }_0$ and strict increase of the enumeration. It is cofinal in the limit $\alpha$, so continuity gives

Work with well-ordered cardinals, as usual with the axiom of choice. We first prove the infinite cardinal arithmetic identity ${\kappa }^2=\kappa$. If a least infinite counterexample $\kappa$ existed, order pairs of ordinals below $\kappa$ first by their maximum coordinate, then lexicographically within each level. Every initial segment ending at $(\xi ,\eta )$ lies inside $(\max (\xi ,\eta )+1{)}^2$. Its cardinal is less than $\kappa$, by the smaller-cardinal identity or finite counting. This well-order cannot have an element with $\kappa$ predecessors, so its order type is at most the initial ordinal $\kappa$. Thus $|\kappa \times \kappa |\le \kappa$, contrary to its being a counterexample; the reverse injection is $\xi \mapsto (\xi ,0)$. HenceA sum of finitely many finite cardinals is finite, so ${\aleph }_0$ is regular. A sum of fewer than ${\aleph }_1$ smaller cardinals is a countable union of countable sets and has size at most ${\aleph }_0^2={\aleph }_0$, so ${\aleph }_1$ is regular.

Likewise fewer than ${\aleph }_2$ summands means at most ${\aleph }_1$ summands, each of size at most ${\aleph }_1$. Their sum is at most ${\aleph }_1^2={\aleph }_1<{\aleph }_2$. Thus ${\aleph }_2$ is regular. In contrast,is a countable sum of smaller cardinals, with ${\aleph }_0<{\aleph }_{\omega }$. Therefore ${\aleph }_{\omega }$ is singular, of cofinality $\omega$.

Take $\kappa ={\omega }_1$ and choose a club sequence on ${\omega }_2$, each $C_{\delta }$ of order type at most ${\omega }_1$. For a successor $\delta$ use its predecessor as a singleton; at a limit use a cofinal sequence of minimal length. Form the minimal-walk treeordered by proper extension. Its height is ${\omega }_2$.

For $\xi <\delta$, the initial segment $C_{\delta }\cap \xi$ has order type strictly below ${\omega }_1$, and is countable. The strict inequality follows because a point of $C_{\delta }$ at or above $\xi$ occurs later in its enumeration. Hence every entry of every trace is countable. Under the Continuum hypothesis, for $|\alpha |\le {\aleph }_1$,There are at most ${\aleph }_1$ finite sequences of such sets. The trace coherence lemma for minimal walks says that, for $\beta >\alpha$, the value ${\rho }_{\beta }(\alpha )$ determines ${\rho }_{\beta }\upharpoonright \alpha$. The case $\beta =\alpha$ adds at most one node. Thus $|T_{\alpha }|\le {\aleph }_1$ for every level.

Suppose that $T$ had a cofinal branch, and take the union of its functions, $f$, with domain ${\omega }_2$. Every ${\rho }_{\beta }$ is injective by the proper-initial-segment argument, so $f$ is injective too. On the stationary setthis set is stationary because the supremum of a strictly increasing ${\omega }_1$-sequence from any club set has cofinality ${\omega }_1$ and lies in that club. The union of the finitely many countable entries of $f(\alpha )$ is bounded below $\alpha$. Assign a strict upper bound below $\alpha$ to obtain a regressive function. By Fodor lemma, there is a stationary $S^{\prime }\subseteq S$ and a single $\eta <{\omega }_2$ such that every entry of $f(\alpha )$ lies inside $\eta$ for $\alpha \in S^{\prime }$. There are at most ${\aleph }_1$ such finite sequences by the same cardinal arithmetic, but $|S^{\prime }|={\aleph }_2$, contradicting injectivity.

Therefore $T$ is an aleph-two Aronszajn tree:

Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.

Choose sets $K,L,M$ of cardinalities $\kappa ,\lambda ,\mu$. Their cardinal arithmetic operations arewhere $K^L$ is the set of functions $L\to K$. The relation $\kappa \le \lambda$ means that there is an injective function $K\to L$.

The currying law for cardinal exponentiation follows from the explicit bijectionand provesMoreover, $2^{\kappa }$ is the cardinality of the power set of a set of size $\kappa$, so Cantor theorem gives

For completeness, identify each cardinal with its initial ordinal. Suppose that some infinite $\kappa$ violates ${\kappa }^2=\kappa$, and choose the least such cardinal. Well-order the pairs $(\alpha ,\beta )\in {\kappa }^2$ first by $\max \{\alpha ,\beta \}$ and then lexicographically. Every proper initial segment is contained in ${\gamma }^2$ together with finitely many boundary pieces for some $\gamma <\kappa$, and has cardinality below $\kappa$ by minimality. The resulting well-order therefore has cardinality at most $\kappa$. The reverse inequality is immediate from $\alpha \mapsto (\alpha ,0)$, contradicting the choice of $\kappa$. Thus the square of an infinite cardinal satisfies ${\kappa }^2=\kappa$.

If $\nu =\max \{\kappa ,\lambda \}$, monotonicity now givesHence the sum and product of two infinite cardinals obey

Assertion (i) can be false. For any infinite $\lambda$, take $\kappa =(2^{\lambda }{)}^{+}$. Then ${\kappa }^{\lambda }\ge \kappa >2^{\lambda }$, so