Use transfinite induction on nonzero $\alpha$. Part (b) gives $\alpha ={\omega }^{\delta }n+\zeta$, with $n$ a positive natural number and $\zeta <{\omega }^{\delta }\le \alpha$.

If $\zeta =0$, this is already a one-term Cantor normal form. Otherwise $0<\zeta <\alpha$, so the induction hypothesis supplies a finite Cantor normal form for $\zeta$. Its leading exponent $\eta$ satisfies ${\omega }^{\eta }\le \zeta <{\omega }^{\delta }$, hence $\eta <\delta$. All later exponents are smaller still. Prefixing ${\omega }^{\delta }n$ therefore givesThe number of terms is finite because the inductive representation of the strictly smaller remainder is finite. Equivalently, repeated extraction cannot produce an infinite strictly descending sequence of ordinals. Since $\alpha \ne 0$, the number of terms is positive.

For nonzero $\alpha$, consider the exponents $\eta$ with ${\omega }^{\eta }\le \alpha$. The set is nonempty because it contains zero, and bounded because ${\omega }^{\eta }\ge \eta$ by transfinite induction. Let $\delta$ be its supremum. If this supremum is a successor it is attained; if it is a limit, continuity of ordinal exponentiation gives ${\omega }^{\delta }\le \alpha$. The zero case is immediate. Thus $\delta$ itself belongs to the set and is its maximum. Maximality givesPut $\rho ={\omega }^{\delta }>0$. Since ${\omega }^{\delta +1}=\rho \cdot \omega ={\sup }_{N<\omega }\rho N$, there is a least positive natural number $N$ with $\alpha <\rho N$. We have $N\ge 2$ because $\rho \le \alpha$. Set $n=N-1$. ThenFor ordinals $\xi \le \alpha$, the final segment of $\alpha$ after its initial segment $\xi$ has an order type $\zeta$ satisfying $\alpha =\xi +\zeta$. Apply this with $\xi =\rho n$. If $\zeta \ge \rho$, monotonicity would give $\alpha \ge \rho n+\rho =\rho (n+1)$, a contradiction. ThereforeThese intervals also make the leading exponent and coefficient unique. The remainder is unique because addition by a fixed left prefix is strictly increasing in its right operand.

The recursive definitions of ordinal addition and ordinal multiplication arewhere the limit clauses apply to nonzero limit ordinals. Ordinal exponentiation has ${\alpha }^0=1$, ${\alpha }^{\beta +1}={\alpha }^{\beta }\cdot \alpha$, and for $\alpha >0$, ${\alpha }^{\lambda }={\sup }_{\beta <\lambda }{\alpha }^{\beta }$. Complete the zero-base convention by $0^{\beta }=0$ for $\beta >0$, with $0^0=1$. Stating the zero-base exception avoids including the value $0^0$ in a naive supremum at a positive limit.

For finite $n$, induction gives $1+n=n+1$. Now prove commutativity for finite $m,n$ by induction on $m$. The case $m=0$ is immediate. Associativity, the finite identity just obtained, and the induction hypothesis giveThus $\boxed{m+n=n+m\text{ for finite ordinals}}$.

It fails for general ordinals: $\boxed{1+\omega =\omega <\omega +1}$. The equality follows from the limit clause, since the finite sums $1+n$ are cofinal in $\omega$.

True in ZF. Let $A$ be a set that is not finite. For each natural number $n$, formEach $E_n$ is a subset of $\mathcal{P}(A)$ and is nonempty: a nonfinite set has an $n$-element subset for every finite $n$, by finite induction. The sets $E_n$ are pairwise disjoint, hence pairwise distinct. Consequentlyis an injection from $\mathbb{N}$ into $\mathcal{P}(\mathcal{P}(A))$. This is a double-power-set countability bound without choice.

Part (d), applied to the set $\mathcal{P}(A)$, says that its power set is not countably infinite. Combining that result with the displayed injection givesThere is no hidden countable choice: the proof uses the entire definable family of $n$-element subsets, rather than choosing one subset for each $n$.

False. Take $\kappa ={\aleph }_0$, $\lambda =1$, and ${\lambda }^{\prime }=2$. ThenFor example the Cantor pairing function $⟨m,n⟩=(m+n)(m+n+1)/2+n$ bijects ${\mathbb{N}}^2$ with $\mathbb{N}$. Thus the exponent increases strictly while the cardinal power remains the same. This counterexample is completely choice-free.

False. Take $\kappa =2^{{\aleph }_0}$ and $\lambda ={\aleph }_0$, both infinite. The explicit pairing bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ and the currying law for cardinal exponentiation giveThe power-set cardinal need not be assumed well-orderable. Every bijection used in this calculation is explicitly available in ZF, so the counterexample does not use choice.

True in ZF. Suppose $\mathcal{P}(A)$ were countably infinite. The singleton injection would inject $A$ into the natural numbers. The set $A$ cannot be finite, since its power set would then be finite. Its image is therefore an infinite subset of $\mathbb{N}$, which has a canonical increasing enumeration. This gives a bijection $A\cong \mathbb{N}$ without choice.

But then $\mathcal{P}(A)\cong \mathcal{P}(\mathbb{N})$, contradicting Cantor's theorem that the latter is strictly larger than $\mathbb{N}$. HenceThe proof does not assume that an arbitrary infinite set contains a countably infinite subset; here the given injection into the natural numbers supplies its well-ordering.

True in ZF. Let $i:A\hookrightarrow A^{\prime }$ and $j:B\hookrightarrow B^{\prime }$ witness the two cardinal inequalities. Because $A$ is nonempty, fix one element $a_0\in A$. Send $f:B\to A$ to $\hat{f}:B^{\prime }\to A^{\prime }$ defined byThe first case is well defined by injectivity of $j$. If $f\ne g$, choose a point where their values differ; the images differ at its $j$-image because $i$ is injective. Thus $f\mapsto \hat{f}$ is injective, proving the desired inequality of function-space cardinalities.

Fixing one element of a single known nonempty set does not invoke the axiom of choice. The nonzero-base hypothesis is what permits this constant extension outside the smaller domain.

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.

True in ZF. For a set $A$, the singleton map injects $A$ into its power set. No map $f:A\to \mathcal{P}(A)$ is onto: the diagonal setdiffers from $f(a)$ at the element $a$ for every $a$. Thus there is no bijection between $A$ and its power set. The injection together with this failure proves $\boxed{\kappa <2^{\kappa }}$.

If strict comparison is formulated as injection in one direction but not the other, the Cantor-Schröder-Bernstein theorem, which requires no choice, excludes an injection of the power set back into $A$. This is Cantor's theorem in a choice-free form.