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For $b\ge 2$, write a number in hereditary base $b$: express its base-$b$ expansion as ${\sum }_i{b}^{e_i}{c}_i$, with $0<c_i<b$, and recursively expand every exponent $e_i$ in the same base. Let $B_{b,c}(m)$, for $c>b$, be the number obtained by replacing every occurrence of the base $b$ in this hereditary expression by $c$. Coefficients and the symbols for addition and exponentiation are retained.

For a starting number $m$, define its Goodstein sequence by $m_0=m$ andThe Goodstein function is the termination-time functionThis choice counts transitions from the initial term; a convention counting the initial term as well shifts the answer by one. For example the sequence starting at $3$ is $3,3,3,2,1,0$, so $\mathcal{G}(3)=5$. Starting at $4$ gives $4,26,41,60,83,\dots$, illustrating why descent of the numerical terms is not the termination argument.

Define the ordinal rank of a Goodstein term ${\rho }_b(m)$ by replacing the base $b$ in the hereditary expansion by $\omega$, at every exponent depth. The result is in Cantor normal form and lies below ${ϵ}_0$, called epsilon zero, the least nonzero solution of ${\omega }^{\alpha }=\alpha$. Finite hereditary expressions have finite nesting depth, so their ranks are bounded by a sufficiently high finite tower of powers of $\omega$.

For a fixed base, ${\rho }_b$ is strictly increasing. To see this, compare two ordinary base-$b$ expansions at their largest differing exponent. By induction through the hereditary exponent expressions, their exponents have the same ordering after replacing $b$ by $\omega$; computable Cantor normal form notation then uses the same first differing exponent or coefficient. AlsoIndeed after the base change all coefficients remain below $c$, the recursively changed exponents remain correctly ordered, and replacing the new base by $\omega$ gives exactly the original ordinal expression. These facts can be proved together by structural induction for primitive recursive functions on the hereditary expression.

Consequently, whenever $m_s>0$,An infinite nonzero Goodstein sequence would give an infinite descending sequence of ordinals below ${ϵ}_0$, impossible because ordinals are well-ordered. Thus the Goodstein function is total. Its steps are effective finite operations, so simulating the sequence until zero also proves that it is a total computable function.

The same ordinal notation system gives the requested decidable well-order. Let $D$ consist of canonical finite terms for zero and for expressionswhere the exponents are themselves canonical terms. Membership in $D$ is decidable by recursively checking the syntax, positive coefficients and decreasing exponents. Comparison is decidable by recursive computable Cantor normal form notation, first of exponents, then coefficients, and then the remaining summands. All recursive calls inspect proper subterms, so the algorithm terminates.

Interpreting these terms as ordinals gives every ordinal below ${ϵ}_0$, uniquely. One way to connect this directly to hereditary base representations is to choose, for a given finite ordinal term, a base larger than every coefficient anywhere in that term. Replacing $\omega$ by that base evaluates it to a natural number whose hereditary representation recovers the ordinal term. Thus all bases together supply the complete notation system; a single fixed base is not enough.

Choose an effective injective coding of finite terms by natural numbers, and enumerate the valid codes in increasing numerical order. The resulting bijection $d:\mathbb{N}\to D$ is computable: validity is decidable and there are infinitely many finite-ordinal terms. Its inverse is computable by counting valid codes below a given code. DefineThis is a decidable relation on all of $\mathbb{N}$. The interpretation into ordinals is an order isomorphism onto ${ϵ}_0$, provingDecidability here concerns comparing two notations. Well-foundedness is supplied by their ordinal interpretation, the same mathematical descent principle used to prove Goodstein's theorem.

No such ordinal exists. For $\alpha >1$, ordinal multiplication gives ${\alpha }^2=\alpha \alpha \ge \alpha \cdot 2=\alpha +\alpha >\alpha$. Since ordinal exponentiation is nondecreasing in the exponent for base $\alpha >1$, ${\alpha }^{\omega }\ge {\alpha }^2>\alpha$.

For every countable ordinal $\beta$, ${\omega }^{\beta }$ is countable: successor stages use a countable product, and countable limit stages a countable union. Hence ${\omega }^{{\omega }_1}={\sup }_{\beta <{\omega }_1}{\omega }^{\beta }\le {\omega }_1$. Conversely ${\omega }^{\beta }\ge \beta$ by transfinite induction, so these values are cofinal in ${\omega }_1$. Therefore ${\omega }^{{\omega }_1}={\omega }_1$.

It is not the least fixed point. Set $a_0=1$, $a_{n+1}={\omega }^{a_n}$ and define epsilon zero as ${\epsilon }_0={\sup }_n{a}_n$. This is countable, and continuity of ordinal exponentiation at limits gives ${\omega }^{{\epsilon }_0}={\sup }_n{\omega }^{a_n}={\epsilon }_0$. Any fixed point of $\alpha \mapsto {\omega }^{\alpha }$ dominates every $a_n$ by induction, hence dominates ${\epsilon }_0$. Thus the least fixed point is ${\epsilon }_0<{\omega }_1$.

Zorn lemma states that a nonempty partially ordered set in which every chain in a partial order has an upper bound has a maximal element. Here is a proof from the axiom of choice and Hartogs theorem.

Suppose there is no maximal element. Every chain $C$ has a strict upper bound: choose an upper bound $b$ and then an element strictly greater than $b$. The family of sets of strict upper bounds is a set of nonempty subsets of the underlying set $P$. The axiom of choice is used here to choose, simultaneously for every chain $C$, one strict upper bound $b(C)$.

By Hartogs theorem there is an ordinal $\kappa$ admitting no injection into $P$. Transfinite recursion defines $x_{\alpha }$ for $\alpha <\kappa$ byAt each stage the previous elements form a chain, and $x_{\alpha }$ is strictly above every one of them. Hence $\alpha \mapsto x_{\alpha }$ is an injection $\kappa \to P$, a contradiction. Therefore a maximal element exists.

An ordinal has rank equal to itself. Thus the rank of a set here is $\sup \{\omega +2,\omega +3,\omega +4\}=\boxed{\omega +4}$.

For a finite position $s\in A^{<\omega }$, let $[s]$ be the cylinder set of infinite plays extending $s$. Because the factors have the discrete topology, these cylinder sets form a basis for the product topology. PutMembership in $D$ means that I has already secured the outcome, regardless of future moves. Since $G$ is an open set, a play belongs to $G$ exactly when one of its prefixes belongs to $D$.

Build the winning-position attractor in an infinite game by transfinite recursion. Start with $W_0=D$, use unions at limit ordinals, and setThis sequence is increasing: the defining operation is order-preserving, and $W_0\subseteq W_1$. It stabilizes at some ordinal. Indeed, if it strictly increased at every successor stage below the successor cardinal number of $|A^{<\omega }|$, choosing one newly added position per stage would inject that larger cardinal into $A^{<\omega }$. Denote the stable set by $W$. This constructs the needed fixed point directly, without an appeal to an unstated fixed-point theorem.

Every position in $W$ has a least entry ordinal, its rank. Rank zero means membership in $D$. A positive rank is a successor $\alpha +1$: at an I-position some extension belongs to $W_{\alpha }$, while at a II-position every extension does. If the empty word belongs to $W$, I always chooses an extension of smaller rank until $D$ is reached. II's choices also decrease rank before that happens. There is no infinite strictly decreasing sequence of ordinals, so some finite prefix lies in $D$. This is a winning strategy in an infinite game for I.

If the empty word is outside $W$, the fixed point equation says that every extension of an I-position outside $W$ is still outside $W$, and that a II-position outside $W$ has at least one extension outside $W$. II chooses such an extension. Every resulting prefix avoids $D$, hence the whole play avoids $G$ by openness. This is a winning strategy in an infinite game for II.

The choices can be made simultaneously by fixing a well-order of $A$, as allowed by the usual axiom of choice. The argument does not assume that $A$ is countable. Every such open game is determined:

Yes. A cardinal fixed point of the singular cardinal enumeration is obtained by countable iteration. Start with ${\kappa }_0={\aleph }_0$ and putThe singular cardinal enumeration is strictly increasing and satisfies ${\sigma }_{\alpha }\ge \alpha$ for every ordinal $\alpha$; the latter follows by transfinite induction for any strictly increasing ordinal-valued enumeration. If equality occurs at some ${\kappa }_n$, that cardinal number is already a witness. Otherwise the sequence is strictly increasing. Put $\kappa ={\sup }_{n<\omega }{\kappa }_n$. This is an uncountable singular cardinal of cofinality $\omega$.

For every $\alpha <\kappa$, some $n$ has $\alpha <{\kappa }_n$, and thereforeAlso ${\sup }_n{\sigma }_{{\kappa }_n}=\kappa$, so ${\sup }_{\alpha <\kappa }{\sigma }_{\alpha }=\kappa$. At a limit ordinal index, if the supremum of all preceding enumerated cardinals is itself singular, it is exactly the next member: every smaller singular cardinal already has a preceding index. ThusThis argument uses continuity only at a singular supremum; the enumeration need not be continuous at a weakly inaccessible cardinal.

The Freiling axiom of symmetry isThe values of $f$ are countable subsets of the real numbers, including finite ones. One may require $x\ne y$ without changing the axiom: apply the displayed version to $x\mapsto f(x)\cup \{x\}$.

The Generalized continuum hypothesis statesHere ${\kappa }^{+}$ is the successor cardinal; equivalently $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$ for every ordinal $\alpha$.

A uniform example is $\boxed{A=\kappa \setminus \{\omega \}}$. It contains the final segment $[\omega +1,\kappa )$, a club set. Therefore $A$ belongs to the club filter. Every two club sets in a regular uncountable cardinal intersect: alternately choose larger points from each and take the countable supremum, which remains below $\kappa$ and lies in both by closure. Hence every member of the club filter is a stationary set, so $A$ is stationary.

However, every finite ordinal belongs to $A$, and their increasing supremum $\omega$ does not. Thus $A$ is not closed and is not a club set. This is a stationary nonclosed member of the club filter for every regular uncountable $\kappa$.

Use the Lévy hierarchy, with bounded formulas in set theory as the quantifier-free level for this purpose. Let $\mathrm{Ord}(\kappa )$ be the bounded ordinal predicate, and let $\mathrm{Bij}(f,\alpha ,\kappa )$ be the bounded predicate saying that $f$ is a graph of a bijection from $\alpha$ onto $\kappa$. Then cardinalhood is expressed byAll quantifiers in $\mathrm{Ord}$ and $\mathrm{Bij}$ are bounded by the displayed sets and the graph. Ordered-pair membership, graph functionality, injectivity and surjectivity have bounded expansions using the usual set coding of ordered pairs. The only unbounded quantifier shown is universal over $f$. The formula is therefore ${\Pi }_1$ and in particular a Pi-one formula modulo ZF.

In ZF, a cardinal number is an ordinal not in bijection with any smaller ordinal, so this formula has exactly the required meaning. Thus cardinalhood is Pi-one definable; the argument does not use the axiom of choice.