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A linear functional $f$ dominated by a finite real convex function $\varphi$ on a vector subspace $W$ extends with the same domination to the whole real vector space. To add $v\notin W$, choose $g(v)$ between the supremum of $(f(w)-\varphi (w-tv))/t$ and the infimum of $(\varphi (w+sv)-f(w))/s$, over $w\in W$ and $s,t>0$. Convexity, applied to $(s(w_1-tv)+t(w_2+sv))/(s+t)$, makes every lower candidate at most every upper candidate. The candidates at $w=0,s=t=1$ bound the interval by finite numbers. Chain unions preserve domination, so Zorn's lemma gives a full extension. No homogeneity of $\varphi$ is assumed.

Every proper ideal of a unital commutative ring is contained in a minimal prime over that ideal. A maximal ideal supplies a prime containing it, and an intersection of a decreasing chain of primes is still prime. Zorn's lemma, with reverse inclusion, therefore supplies a minimal member. No Noetherian assumption is needed.

For an explicit basis extension in ${\mathbb{R}}^2$, start with $A=\{u\}$, where $u=(1,1)$. Its linear span is the line $y=x$. Adding $w=(1,-1)$ preserves linear independence, because $w$ is outside that line, and gives a basis. Every vector has the finite linear combinationThe vector $2u=(2,2)$ cannot be added while preserving linear independence: it is already in the linear span. Many choices outside the line work, so maximality does not select a unique basis.

For a finite-dimensional vector space with a supplied finite spanning set, a greedy algorithm scans the list, retaining each vector precisely when it is outside the linear span of those already retained. Starting with a linearly independent $A$, the same scan adjoins only the needed vectors. Retained vectors remain linearly independent, and every skipped vector is in the final linear span, so the result is a basis containing $A$. Finite reasoning needs no axiom of choice.

For an arbitrary vector space, this finite stopping argument is unavailable. Accepting Zorn's lemma asserts existence of a maximal element once every chain in a partial order has an upper bound; it supplies no general computational recipe. Over ZF, Zorn's lemma, the axiom of choice and the well-ordering theorem are equivalent. Particular infinite vector spaces can still have explicit bases, such as $1,x,x^2,\dots$ for the vector space of univariate polynomials over a specified field; infinity alone does not force a nonconstructive argument.^[ref]

A real linear functional $g$ on a vector subspace, dominated by a sublinear function $p$, can be extended across one new vector $v$ by setting $\tilde{g}(m+tv)=g(m)+tc$. The displayed lower and upper bounds are compatible because $g(m+n)\le p(m-v)+p(n+v)$. A value between them preserves domination for both signs of $t$. Chain unions and Zorn's lemma then yield the full Hahn-Banach theorem.

Here $\varphi$ is a finite real convex function on the whole real vector space; no topology or continuity is required. We prove the convex domination form of the Hahn-Banach theorem by extending across one vector. Suppose a linear functional $f$ is already dominated on a vector subspace $W$ and choose $v\notin W$. For $w\in W$ and positive $t,s$, the proposed extension $g(w+\tau v)=f(w)+\tau c$ must satisfy the boundsTo prove compatibility, take $w_1,w_2\in W$ and $t,s>0$. The convex combination of $w_1-tv$ and $w_2+sv$ with weights $s/(s+t)$ and $t/(s+t)$ lies in $W$. Domination there and convexity giveRearranging proves every lower candidate is at most every upper candidate. Also the candidates $w=0,t=s=1$ showThus both endpoints are finite, and some $c\in [L,U]$ exists. For $\tau >0$ the upper bound gives domination of $f(w)+\tau c$; for $\tau <0$ the lower bound gives it; and for $\tau =0$ domination was assumed. This defines a dominated linear functional on $W+\mathbb{R}v$ extending $f$.

Order all dominated extensions by inclusion of their domains and agreement of values. A chain has the union extension, still linear and dominated. By Zorn's lemma there is a maximal extension. The one-vector construction would enlarge any proper domain, so its domain is $V$. Hence there is a full linear extension with $\boxed{g{|}_W=f,g\le \varphi }$. A convex function here need not be positively homogeneous; this is why the bounds included all positive $s,t$.

Now fix $y\in V$. The translated convex function ${\phi }_y(z)=\varphi (y+z)-\varphi (y)$ vanishes at zero. Apply the just-proved extension theorem to the zero linear functional on $\{0\}$ to obtain a linear functional ${\ell }_y$ with ${\ell }_y(z)\le {\phi }_y(z)$ for every $z$. SetThis is an affine function, satisfies $a_y(x)\le \varphi (x)$ for every $x$, and has $a_y(y)=\varphi (y)$. Let $A=\{a_y:y\in V\}$. Every member is an affine minorant, while the member indexed by $x$ attains $\varphi (x)$ at $x$. Thus the exact representation isThis is the finite convex function as supremum of affine minorants property. The linear functionals are algebraic; without a topology no assertion of continuous affine minorants is intended.

We take $\mathbb{N}=\{1,2,\dots \}$. An ultrafilter $p$ on $\mathbb{N}$ is a proper filter on a set which decides every subset: it contains $\mathbb{N}$ but not $\varnothing$, is closed under finite intersections and passage to supersets, and contains exactly one of $A$ and $\mathbb{N}\setminus A$ for every $A\subseteq \mathbb{N}$. A principal ultrafilter has the form ${\delta }_n=\{A:n\in A\}$; all other ultrafilters are nonprincipal ultrafilters.

To prove existence, start with the cofinite filter and order its proper filter extensions by inclusion. The union of a chain is again a proper filter on a set: the finitely many sets involved in an intersection all belong to one member of the chain, and no member contains the empty set. Thus Zorn's lemma supplies a maximal proper filter $p$. If $A\notin p$, adjoining $A$ must make the generated filter improper; equivalently, some $B\in p$ has $A\cap B=\varnothing$. Then $B\subseteq \mathbb{N}\setminus A$, so $\mathbb{N}\setminus A\in p$. This proves that $p$ is an ultrafilter. It contains the complement of every finite set and hence no finite set. In particular, it contains no singleton and is nonprincipal.

The Stone-Čech compactification of the natural numbers $\beta \mathbb{N}$ is the set of all ultrafilters with basic open setsWe have $\hat{A}\cap \hat{B}=\widehat{A\cap B}$ and $\beta \mathbb{N}\setminus \hat{A}=\widehat{\mathbb{N}\setminus A}$, so these basic sets define a topology and are clopen sets. If $p\ne q$, choose $A\in p\setminus q$. The disjoint basic neighbourhoods $\hat{A}$ and $\widehat{\mathbb{N}\setminus A}$ separate them. Thus the space is Hausdorff.

For compactness, it suffices to show every basic open cover has a finite subcover, since an arbitrary open cover can be refined by basic open sets. Suppose $\{\widehat{A_i}:i\in I\}$ has no finite subcover. The complements $\mathbb{N}\setminus A_i$ have the finite intersection property: if a finite intersection were empty, the corresponding finite family of $\widehat{A_i}$ would cover every ultrafilter. Their generated proper filter on a set extends to an ultrafilter, by the same Zorn's lemma argument. This ultrafilter contains every $\mathbb{N}\setminus A_i$, and therefore belongs to none of the alleged cover members, a contradiction. We have proved that $\beta \mathbb{N}$ is a compact Hausdorff space. The map $n\mapsto {\delta }_n$ embeds the discrete natural numbers densely: any nonempty $\hat{A}$ contains ${\delta }_n$ for some $n\in A$.

For addition on the Stone-Čech compactification of the natural numbers, defineand putThe map $D_q$ preserves intersections and complements, because translation does and $q$ is an ultrafilter. It also sends $\mathbb{N}$ to $\mathbb{N}$ and the empty set to itself. Consequently the displayed rule defines an ultrafilter $p+q$. On principal points it satisfies ${\delta }_n+{\delta }_m={\delta }_{n+m}$, so it extends ordinary addition.

Write $({\forall }^{p}n)\phi (n)$ to mean that $\{n:\phi (n)\}\in p$. The ultrafilter addition rule is then an iterated quantifier, with its order fixed. Expansion givesOrdinary addition is an associative operation, and the quantifiers are in the same order on both lines. Thus $\boxed{(p+q)+r=p+(q+r)}$. This argument does not interchange ultrafilter quantifiers and does not assert that this extended addition is commutative.

For fixed $q$, the translation $R_q:p\mapsto p+q$ hasa basic clopen set. Therefore addition is continuous in its left variable. This is the left-continuity convention used here: right translations are continuous, giving a left-topological semigroup. Continuity in both variables jointly is not part of the assertion.

Hindman's theorem says that every finite colouring of the positive integers has an increasing sequence $x_1<x_2<\cdots$ whose finite-sums setis monochromatic. Assume, as permitted, that the addition just defined has an idempotent ultrafilter $p$, so $p+p=p$. This ultrafilter is nonprincipal: a principal ${\delta }_n$ would satisfy ${\delta }_n+{\delta }_n={\delta }_{2n}\ne {\delta }_n$ for positive $n$. Hence every member of $p$ is infinite and every cofinite set belongs to $p$.

Exactly one cell $A$ of the finite colour partition belongs to $p$. Define its star set bySince $A\in p+p$, we have $D_p(A)\in p$ and therefore $A^{\ast }\in p$. Moreover, for each $n\in A^{\ast }$, the set $A-n$ belongs to $p$. Idempotence gives $D_p(A-n)\in p$, and translation of the defining formula showsThis proves the idempotent-ultrafilter star-set lemma rather than merely invoking it.

Choose $x_1\in A^{\ast }$. Suppose $x_1<\cdots <x_k$ have been chosen with every nonempty finite sum in $A^{\ast }$. Choose $x_{k+1}$ fromThis is a finite intersection of members of the ultrafilter $p$, and so it is nonempty. The new sums are $x_{k+1}$ and $s+x_{k+1}$; all lie in $A^{\ast }$ by the choice. Mathematical induction yields an increasing infinite sequence withThus the assumed idempotent gives the full Hindman theorem, with distinct increasing generators and all their nonempty finite sums of one colour.

Put $e=1$ and write $G_{\ge e}$ for the nonnegative cone, including the identity. Every element of this cone is its own absolute value in a lattice-ordered group, so the desired condition is equivalent toWe prove this equivalence with normal-valued lattice-ordered group without assuming any characterization of normal values. Membership of $g$ in a convex lattice subgroup is equivalent to membership of $|g|$, so it suffices to discuss values of elements $t>e$. A value in a lattice-ordered group of $t>e$ means a convex lattice subgroup maximal among those omitting $t$. Its cover of a value in a lattice-ordered group will be denoted $V^{\ast }$: normal-valued means $V◃V^{\ast }$, not $V◃G$.

We first record facts about values and the coset order. Zorn's lemma supplies a value omitting any $t>e$, and supplies one containing any specified convex lattice subgroup that omits $t$. If $V$ is a value of $t$, every convex lattice subgroup strictly containing $V$ contains $t$. Therefore $V^{\ast }=⟨V,t{⟩}_c$ is the least such subgroup and there are no intermediate convex lattice subgroups. Moreover, if $C\cap D=V$ and both strictly contain $V$, then both contain $t$, a contradiction. The criterion proved in part (i) thus makes $V$ a prime convex lattice subgroup: its right cosets are totally ordered. Equivalently, for $u,v\ge e$ with $u\wedge v=e$, at least one belongs to $V$. The equivalence follows because the coset projection preserves finite meets; in proving this, inequalities with two elements of $V$ can be replaced by their join.

The right-coset order gives the useful testIndeed, $u\le pv$ gives $(u{v}^{-1})\vee e\le p\vee e\in P$, and convexity gives membership. Conversely the displayed positive part itself supplies such a $p$. We will also use the basic lattice-ordered group identity that $(r\vee e)\wedge (r^{-1}\vee e)=e$: the positive and negative parts of one element are disjoint.

Here is a squared bound from comparison at values. Suppose $p,q,x\ge e$ and $Vpq\ge Vx$ for every value $V$ of $x$. Then $p^2{q}^2\ge x$. To prove it, putand suppose $d>e$. Choose a value $P$ of $d$. If $x\in P$, then $e\le d\le x$, so $d\in P$, a contradiction. Otherwise extend $P$ to a value $V$ of $x$.

If $p\notin V$, set $c=(x{q}^{-2}{p}^{-1})\vee e$. Since $p{q}^2\ge pq$ and $Vpq\ge Vx$, test (2) gives $c\in V$. The element $(p{c}^{-1})\vee e$ is outside $V$: membership would imply $e\le p\le ((p{c}^{-1})\vee e)c\in V$, contradicting $p\notin V$. It is consequently outside $P$. Its inverse positive part isbecause $p^{-1}\le e$. These two positive parts have meet $e$, so primeness of $P$ forces $d\in P$, again a contradiction.

If $p\in V$, then $Vpq=Vq$, and $c=(x{q}^{-1})\vee e\in V$. Necessarily $q\notin V$, since otherwise $Vx\le V$ would put $x$ in $V$. Hence $t=p^{2}q\notin V$. As before, $(t{c}^{-1})\vee e$ is outside $V$ and outside $P$, while its inverse positive part ishere $t^{-1}\le e$. Primeness again yields the contradiction $d\in P$. Thus $d=e$, proving the bound. This auxiliary argument uses no normality assumption.

Suppose first that all values are normal in their covers. Given $a,b\ge e$, put $x=ab$. If $x=e$, both factors are $e$ and (1) holds. Otherwise choose any value $V$ of $x$. Since $e\le a,b\le x$, both factors belong to $V^{\ast }$. The quotient group $V^{\ast }/V$ is a linearly ordered group, because $V$ is prime and is normal in $V^{\ast }$. It has no nontrivial proper convex subgroup of an ordered group, since such a subgroup would lift to an intermediate convex lattice subgroup between $V$ and $V^{\ast }$.

This quotient is an Archimedean ordered group. If $u>e$ and all powers $u^n$ were bounded by some $v$, the principal convex lattice subgroup generated by $u$ would be a nontrivial proper subgroup: it cannot contain $v$, since $v\le u^m$ would contradict $u^{m+1}\le v$. That contradicts the preceding paragraph.

For completeness, Archimedean linearly ordered groups are Abelian by the following elementary argument. If there is a least positive element $s$, its powers are cofinal and every element lies between consecutive powers; absence of any element between $e$ and $s$ makes the group cyclic. Otherwise, for every $c>e$ there is $d>e$ with $d^2<c$: choose $e<s<r<c$ and put $d_0=r{s}^{-1}$. If $d_0^2<r$, use $d=d_0$; if not, multiplication of $d_0^2\ge r$ gives $s^2\le r<c$, so use $d=s$. If positive $g,h$ did not commute, interchange them to have $gh<hg$ and put $c=(gh{)}^{-1}hg>e$. Choose $d^2<c$ and, by the Archimedean property, integers $m,n\ge 0$ with $d^m\le g<d^{m+1}$ and $d^n\le h<d^{n+1}$. Two-sided order invariance givesa contradiction. Arbitrary nonidentity elements can be replaced by their positive inverses without changing whether they commute. This proves commutativity without assuming the characterization being proved.

It follows that $Vba=Vab=Vx$. Apply the squared-bound lemma with $p=b$, $q=a$ and $x=ab$, for all values of this $x$. We obtainTaking $a=|f|$ and $b=|g|$ gives the required inequality.

Conversely, assume (1). We need one-sided bounds for adjoining a positive element. If $H$ is any convex lattice subgroup and $a\ge e$, thenFor the first equality, denote the indicated set by $S$. The inequality (1) allows a positive power of $a$ to move past a positive element of $H$, at the cost of squaring:Also $xy\le |x||y|$ and $(xy{)}^{-1}\le |y||x|$. Applying the bound in both orders, and taking the join of the two resulting elements of $H$ and the larger exponent, bounds $|xy|$ in the same form. Inversion does not change $|g|$. A common bound obtained from $h\vee k$ and the larger exponent proves closure under join, meet and order intervals. Thus $S$ is a convex lattice subgroup containing $H$ and $a$. Conversely every such subgroup contains each $h{a}^n$ and everything whose absolute value it bounds, proving equality. The same proof in the opposite group proves the second equality, since (1) with the variables interchanged reads $ba\le a^2{b}^2$.

Let $V$ now be any value and $W=V^{\ast }$. For each $a\in W_{\ge e}\setminus V$, the subgroup $⟨V,a{⟩}_c$ strictly contains $V$ and is contained in $W$, so the cover property makes it equal to $W$. Formula (3) therefore gives power cofinality in the cover of a value: for every $b\in W_{\ge e}$ there are powers withHere the left cosets carry the order obtained by inversion from the right-coset order; equivalently $bV\le a^{m}V$ means $b\le a^{m}v$ for some $v\in V$.

Take $v\in V_{\ge e}$ and $a\in W_{\ge e}\setminus V$. If $u=a^{-1}va\notin V$, then $u\in W_{\ge e}\setminus V$ andThe inequality uses $a\ge e$. But (4), applied to $u$ and the target $a^2$, gives $V{a}^2\le V{u}^n$ for some $n$. This contradicts $Va<V{a}^2$; equality of those two cosets would itself imply $a\in V$. Hence $a^{-1}va\in V$.

To obtain the other inclusion, suppose $w=av{a}^{-1}\notin V$. ThenLeft-coset cofinality in (4) gives $a^{2}V\le w^{n}V$, contradicting $aV<a^{2}V$. Thus $av{a}^{-1}\in V$ too. Every element of $V$ is a quotient of its two nonnegative lattice parts, so both conjugation inclusions hold for all $v\in V$. Positive elements of $W$ lying in $V$ already normalize $V$, and every element of $W$ is a quotient of two nonnegative elements. ConsequentlyAll values have this property, so $G$ is normal-valued. At no stage was normality in the whole group assumed or required.

Finally, in a linearly ordered group, nonnegative $a,b$ are comparable. If $a\le b$, then $ab\le b^2\le b^2{a}^2$; if $b\le a$, then $ab\le a^2\le b^2{a}^2$. Thus (1) holds in every o-group. The proved converse gives every o-group is normal-valued.

For each $g\ne 1$, choose a value in a lattice-ordered group $P_g$, using Zorn's lemma. Such a value is a prime convex lattice subgroup. Otherwise the last construction in Question 1(i) would give two larger convex lattice subgroups whose intersection is $P_g$; maximality among subgroups omitting $g$ would put $g$ in both, hence in their intersection. Thus the right coset set ${\Omega }_g=P_g\backslash G$ is a totally ordered set.

Right multiplication defines an action ${\rho }_g(a):P_{g}h\mapsto P_{g}ha$. It is well-defined, preserves the total order, and has inverse ${\rho }_g(a^{-1})$. The coset formulas from Question 1(i) show pointwise preservation of both join and meet, becauseand likewise for the meet.

Take the disjoint union of these chains, arranged in any fixed order sum of the indices $g\ne 1$. Let every $a\in G$ act on each component by ${\rho }_g(a)$. This gives a lattice-ordered group homomorphism into the order automorphisms of one chain. It is injective: a nonidentity $a$ moves the identity coset in ${\Omega }_a$, since $P_{a}a\ne P_a$. Its image is closed under pointwise join and meet, giving exactly the required lattice-ordered permutation group. This is the Holland representation theorem.

First pass from the preorder $R$ to a partial order. Define the indifference relation $y\sim z$ by $yRz$ and $zRy$. Reflexivity and transitivity make this an equivalence relation. On the quotient set $C={\mathbb{R}}^L/\sim$, putTransitivity makes this independent of the representatives, and the quotient relation is antisymmetric, hence a partial order.

We need its order-extension property, including freedom to order incomparable classes. Here is the argument. Order all partial order extensions of $⊵$ by inclusion. The union of a chain of extensions is again a partial order: any finite list of comparisons needed to check an axiom occurs together in one member of the chain. Zorn's lemma therefore supplies a maximal extension. If two classes $A,B$ remain incomparable, adjoin $A⊵B$ and take the transitive closure of a relation. The only possible new comparisons have the form $X⊵A⊵B⊵Y$. A cycle would require a previously existing path $B⊵A$, contradicting incomparability. The enlarged relation is still a partial order, contradicting maximality. Consequently the maximal extension is a total order. This also proves the Szpilrajn extension theorem used here.

Lift such a total order to the original bundles by comparing their equivalence classes. The lifted preference relation $R^{\prime }$ is complete and transitive, contains $R$, and preserves its strict preference: if $yPz$, their distinct classes satisfy $[y]⊵[z]$, and antisymmetry of the extended order prevents the reverse comparison. In particular, the class $\mathcal{P}$ is nonempty and $R\subseteq {\bigcap }_{R^{\prime }\in \mathcal{P}}{R}^{\prime }$.

For the reverse inclusion, take a pair with $\neg (yRz)$. If $zRy$, then $zPy$, so every member of $\mathcal{P}$ strictly ranks $z$ above $y$ and excludes $y{R}^{\prime }z$. If neither comparison holds, their classes are incomparable. First adjoin $[z]⊵[y]$, which creates no cycle by the preceding argument, and then extend to a total order. Its lift belongs to $\mathcal{P}$ and again excludes $y{R}^{\prime }z$. Thus every pair outside $R$ is outside at least one member of $\mathcal{P}$, provingThe intersection proof is an order-theoretic fact about any preorder. The congruence axiom additionally ensures exact rationalization of the observed demand correspondence: every chosen bundle weakly outranks every affordable bundle, while an affordable unchosen bundle is strictly worse by part (b). Hence each completion's budget-maximizing set is exactly the observed chosen set.

The Tychonoff theorem states that an arbitrary product of compact spaces, with the product topology, is compact; no Hausdorff hypothesis is necessary. We prove it using Zorn's lemma and the axiom of choice.

First, every proper filter on a set extends to an ultrafilter. Order its proper extensions by inclusion. The union of a chain is again a proper filter: any finite collection of its members occurs in one member of the chain, and the empty set occurs in none. Zorn's lemma gives a maximal proper filter $\mathcal{U}$. If $A\notin \mathcal{U}$, adjoining $A$ must destroy properness, so some $F\in \mathcal{U}$ has $F\cap A=\varnothing$, and hence $X\setminus A\in \mathcal{U}$. This proves the decision property of an ultrafilter.

Next prove the ultrafilter characterization of compactness. In a compact space the closed sets $\overline{F}$, for $F\in \mathcal{U}$, have the finite intersection property, so their intersection contains a point $x$. Every open neighbourhood $O$ of $x$ belongs to $\mathcal{U}$: otherwise its closed complement belongs to $\mathcal{U}$ and contains $x$, a contradiction. Thus the ultrafilter converges to $x$. Conversely, if an open cover has no finite subcover, the closed complements have the finite intersection property and generate a proper filter. Extend it to an ultrafilter. A limit point lies in some member $O$ of the cover, so both $O$ and its complement would belong to the ultrafilter. This is impossible. These arguments also prove the closed-set form of compactness directly by taking complements.

Now let $X={\prod }_{j\in J}{X}_j$. If a factor is empty, the product is empty and compact. Otherwise the axiom of choice makes $X$ nonempty. For an ultrafilter $\mathcal{U}$ on $X$, its coordinate imageis an ultrafilter. Compactness of $X_j$ supplies at least one limit $x_j$; use the axiom of choice to choose one for every $j$. A basic open neighbourhood of $x=(x_j{)}_j$ restricts finitely many coordinates. Each corresponding inverse image belongs to $\mathcal{U}$, so their finite intersection also does. Hence $\mathcal{U}$ converges to $x$ in the product topology. The ultrafilter criterion proves the theorem. The explicit choices and the maximal-filter argument are exactly the places where choice enters.

The Banach-Alaoglu theorem states that the closed unit ball of the continuous dual space $V^{\ast }$ of a real or complex normed vector space $V$ is compact in the weak-star topology $\sigma (V^{\ast },V)$. Completeness of $V$ is not required.

For each $v\in V$, let $D_v=\{z\in \mathbb{K}:|z|\le \Vert v\Vert \}$, a compact scalar disk or interval. By the Tychonoff theorem, $P={\prod }_{v\in V}{D}_v$ is compact. Inside $P$, impose the equationsEach equation defines a closed subset, since it concerns continuous coordinate projections into a Hausdorff scalar field. Their intersection $K$ is therefore compact. A point in $K$ defines a linear functional $\ell (v)=z_v$, and the coordinate bounds give $|\ell (v)|\le \Vert v\Vert$. Conversely, every $\ell \in V^{\ast }$ with $\Vert \ell \Vert \le 1$ gives such a point. Thus $K$ is exactly the image of the dual unit ball under $\ell \mapsto (\ell (v){)}_{v\in V}$. The induced product topology is precisely pointwise convergence on $V$, which is the weak-star topology. This proves

Let $p$ be the Minkowski functional of $E$:The contained ball makes $E$ absorbing, so $p$ is finite, nonnegative and satisfies $p(v)\le \Vert v\Vert /ϵ$. It is positively homogeneous. To check subadditivity, if $v\in sE$ and $w\in tE$, the defining property of a convex set gives $(v+w)/(s+t)\in E$, whence $p(v+w)\le s+t$; let $s,t$ decrease to the respective infima. Thus $p$ is a sublinear functional. We have $p(e)\le 1$ for every $e\in E$. Also $p(x)\ge 1$: otherwise $x\in tE$ for some $0<t<1$, which would put $x$ in $E$ because $E$ is a convex set and $0\in E$.

We prove the dominated form of the Hahn-Banach theorem needed here. Suppose $g$ is linear on a subspace $M$ and $g\le p$ there. To extend it to $M+\mathbb{R}v$ with $v\notin M$, choose the value $c$ at $v$ betweenEvery lower candidate is at most every upper candidate, becauseTaking $m=n=0$ among the candidates shows the two endpoints are finite: the lower supremum is at least $-p(-v)$ and at most $p(v)$, and the upper infimum lies between them and $p(v)$. Define $\tilde{g}(m+tv)=g(m)+tc$. For $t>0$, the upper inequality applied to $m/t$ gives domination by $p(m+tv)$. For $t<0$, apply the lower inequality to $m/(-t)$ and multiply by $-t$. At $t=0$ domination already holds. Hence this is a dominated one-dimensional extension.

Order all dominated extensions by inclusion of their domains. Chain unions preserve linearity and domination, so Zorn's lemma gives a maximal extension. The one-dimensional argument shows its domain is the whole vector space. This proves the requisite dominated Hahn-Banach theorem, rather than invoking an unproved separation form.

On the line $\mathbb{R}x$, start with $g(tx)=tp(x)$. For $t\ge 0$ this equals $p(tx)$; for $t<0$ it is nonpositive and therefore at most $p(tx)$. Extend it to $T\le p$ on $V$. ThenApplying domination to $v$ and $-v$ also givesThus $T$ is a continuous linear functional, andThis proves separation from an absorbing convex set with no closedness assumption on $E$. The proof also gives the norm-preserving real extension theorem used earlier: take $p(v)=C\Vert v\Vert$ when the original functional has norm at most $C$.

We prove the extension by Zorn's lemma, retaining the full domination condition. Consider all pairs $(Y,S)$ extending $(Y_0,T_0)$, with $Y$ a subspace and $S$ linear, satisfying $S(y)\le p(x+y)-q(x)$ for every $x\in X,y\in Y$. Order these pairs by extension. A chain has an upper bound: take the union of its subspaces and maps. The union is a subspace, and its map is well defined and linear because any finitely many vectors lie together in one member of the chain. Its domination is inherited there. Thus a maximal extension exists.

Suppose its domain $Y$ is proper, and choose $z\notin Y$. A linear extension to $Y+\mathbb{R}z$ has the form $S_a(y+tz)=S(y)+ta$. Define the admissible values for a sandwich extension bywith $x,x^{\prime }\in X$ and $y,y^{\prime }\in Y$. Part (i) shows every quantity in the supremum is at most every quantity in the infimum, hence $A\le B$. Both bounds are real and finite: choosing zero vectors provides $A\ge -p(-z)$ and $B\le p(z)$, and the pairwise comparison then bounds them from the other sides. Choose $a\in [A,B]$.

For $t>0$, positive homogeneity and the upper bound on $a$, used with $x/t,y/t$, giveFor $t=-s<0$, the lower bound on $a$ applied to $x/s,y/s$ giveswhich rearranges to $S(y)-sa\le p(x+y-sz)-q(x)$. For $t=0$ this is the existing domination. Thus $S_a$ is a legitimate larger extension, contradicting maximality. The maximal domain is therefore all of $X$; call its map $T$.

Setting the first vector in the domination to zero yields $T(v)\le p(v)$. Setting it to $v$ and the domain vector to $-v$ yields $-T(v)\le -q(v)$. ConsequentlyThe one-dimensional extension has checked both signs of its coefficient; positive homogeneity alone does not permit treating negative coefficients as positive ones.

Let $C=\Vert T\Vert$. We construct the extension by a deterministic recursion over the given dense list, using the one-dimensional dominated extension of a real linear functional. Neither Zorn's lemma nor any selection of a family of extensions is needed.

Suppose $f:W\to \mathbb{R}$ is already a linear functional satisfying $|f(w)|\le C\Vert w\Vert$, and $v\notin W$. An extension to $W+\mathbb{R}v$ must have the formDefineFor any $w,z\in W$, the bound on $f$ and the triangle inequality giveThus every quantity in the supremum defining $L_v$ is at most every quantity in the infimum defining $U_v$. Taking $w=0$ and $z=0$ gives finite bounds, specificallyThe completeness of the real numbers defines these endpoints uniquely. Set $a=L_v$; this is an explicit rule, not an arbitrary choice from a nonempty interval.

For $t>0$, the upper-endpoint inequality with $w/t$ yields $f(w)+ta\le C\Vert w+tv\Vert$. For $t<0$, write $t=-s$ with $s>0$ and use the lower-endpoint inequality with $w/s$ to obtain the same bound. For $t=0$ it is the original bound. Applying the upper bound also to the negative of the vector givesThe representation $w+tv$ is unique because $v\notin W$, so $f_v$ is a well-defined bounded linear functional extending $f$.

Start with $W_0=U$ and $f_0=T$. At step $n$, if $y_n\in W_{n-1}$ leave the functional unchanged; otherwise use exactly the supremum-endpoint rule with $v=y_n$. Put $W_n=U+\mathrm{span}\{y_1,\dots ,y_n\}$. This defines compatible functionals $f_n$ on $W_n$ with the common bound $C$. Their union $f_{\infty }$ on $W_{\infty }={\bigcup }_n{W}_n$ is linear and bounded by $C$. The subspace $W_{\infty }$ is dense in $V$ because it contains every $y_n$.

To make the final passage to the closure equally free of the axiom of choice, for $v\in V$ let $j_n(v)$ be the least positive integer with $\Vert y_{j_n(v)}-v\Vert <2^{-n}$. Density ensures that this integer exists, and leastness specifies it uniquely. DefineThe limit exists becauseAny other approximating sequence in $W_{\infty }$ gives the same limit, by the same norm bound. Applying this independence to sums and scalar multiples of approximating sequences proves linearity of $\tilde{T}$. It also shows $\tilde{T}=f_{\infty }$ on $W_{\infty }$ and $|\tilde{T}(v)|\le C\Vert v\Vert$. Restriction to $U$ therefore givesThe upper bound follows from construction and the lower bound from restriction to $U$. This also covers $C=0$, when all extensions are zero. No completeness of $V$ or closedness of $U$ was assumed. The argument establishes the choice-free Hahn-Banach extension in a separable space by uniquely specified operations at every stage.

The Gelfand-Mazur theorem says that a nonzero complex unital Banach algebra in which every nonzero element is invertible is the complex field itself: $a\mapsto \lambda$ where $a=\lambda 1$. This is an isometric isomorphism under the usual normalization $\Vert 1\Vert =1$; without that normalization its norm is simply $\Vert \lambda 1\Vert =|\lambda |\Vert 1\Vert$. We first prove the required spectral and resolvent facts rather than assume them.

For any element $a$ of a complex unital Banach algebra $A$, define its spectrum of an element byIf $\Vert b\Vert <1$, the partial sums of ${\sum }_{n\ge 0}{b}^n$ converge in the complete algebra. Multiplying a partial sum by $1-b$ on either side gives $1-b^{N+1}\to 1$, proving the Neumann series inverse. For $|\lambda |>\Vert a\Vert$ this givesThus the spectrum is bounded. Under $\Vert 1\Vert =1$, $\Vert R(\lambda ,a)\Vert \le (|\lambda |-\Vert a\Vert {)}^{-1}$; for an unnormalized identity the same estimate with numerator $\Vert 1\Vert$ suffices. Also $\lambda R(\lambda ,a)\to 1$ as $|\lambda |\to \infty$.

If ${\lambda }_0$ is outside the spectrum and $R_0=R({\lambda }_0,a)$, then for $|h|\Vert R_0\Vert <1$,This follows by factoring $({\lambda }_0+h)1-a=({\lambda }_{0}1-a)(1+h{R}_0)$ and applying the just-proved series. The series converges locally in norm, proving that the resolvent set is open, the spectrum is closed, and the resolvent of an element is holomorphic. Subtracting inverse equations also gives the resolvent identityconsistent with the derivative $R^{\prime }(\lambda ,a)=-R(\lambda ,a{)}^2$ from the local series.

The spectrum is nonempty. If it were empty, the resolvent would be an entire algebra-valued function tending to zero at infinity. Choose a bounded linear functional $\ell :A\to \mathbb{C}$ with $\ell (1)=1$, extending the functional on $\mathbb{C}1$ by the complex Hahn-Banach theorem. Then $\ell (R(\lambda ,a))$ is an entire function, bounded outside a disk by the Neumann estimate and bounded inside by continuity and compactness. The Liouville theorem makes it constant, and its decay makes that constant zero. But $\lambda \ell (R(\lambda ,a))\to \ell (1)=1$, a contradiction. We have proved that every spectrum is a nonempty compact subset of the disk $|\lambda |\le \Vert a\Vert$.

Now assume every nonzero element of $A$ is invertible. For any $a$, choose $\lambda \in {\sigma }_A(a)$. The noninvertible element $a-\lambda 1$ must be zero. Thus every element is a scalar multiple of the identity, uniquely because $1\ne 0$, andNo commutativity assumption was needed for this part; it is forced by the conclusion. Complex scalars are essential to the spectral argument.

For the maximal ideal space and its correspondence with characters of an algebra, now take $A$ to be a commutative complex unital Banach algebra. Every proper ideal lies in an algebraic maximal ideal by Zorn's lemma: a chain's union is an ideal not containing $1$. Every proper ideal also has proper closure. Otherwise it contains an element $m$ with $\Vert 1-m\Vert <1$, which is invertible by the Neumann series, forcing the ideal to contain $1$. A maximal ideal $I$ is therefore closed, since its closure is a proper ideal containing it.

The quotient $A/I$, with its quotient norm, is a complex Banach algebra. Completeness follows from closedness of $I$, and submultiplicativity follows by taking infima over representatives. It is a division algebra: if $a\notin I$, maximality gives $I+Aa=A$, so $ba+i=1$ for some $b\in A$, $i\in I$. Hence $a+I$ is invertible. By the proved Gelfand-Mazur theorem, $A/I\cong \mathbb{C}$, with the identity mapping to $1$. Composing this isomorphism with the quotient map produces a nonzero multiplicative complex linear functional ${\varphi }_I$ with kernel $I$.

Conversely let $\varphi :A\to \mathbb{C}$ be a nonzero multiplicative complex linear functional, without assuming continuity. Nonzeroness and multiplicativity give $\varphi (1)=1$, hence $\varphi (\lambda 1)=\lambda$ and surjectivity onto $\mathbb{C}$. Its kernel is a maximal ideal. If $a-\varphi (a)1$ were invertible, applying $\varphi$ to its product with an inverse would give $0=1$. Therefore $\varphi (a)\in {\sigma }_A(a)$, soThis proves automatic continuity of characters; with normalized identity their operator norm is one. Finally a kernel determines its character uniquely: for every $a$, its class in $A/I=\mathbb{C}1$ is a unique scalar multiple of the unit, and every character with kernel $I$ must take that scalar value. We have obtained the bijectionThe commutativity in this correspondence cannot be dropped: ${\mathrm{Mat}}_2(\mathbb{C})$ has the maximal two-sided ideal $0$, but has no scalar-valued character, since such a character would have zero kernel and give an impossible injective linear map into $\mathbb{C}$.

For use in the next question, the spectrum equals character values in a commutative Banach algebra. We have proved one inclusion. For the other, if $a-\lambda 1$ is noninvertible, its generated ideal is proper, hence lies in a maximal ideal. The corresponding character takes $a$ to $\lambda$. ThusGive ${\mathcal{M}}_A$ the Gelfand topology, namely the weak-star topology of pointwise convergence on $A$ through these characters. It is Hausdorff and compact: the characters lie in the dual unit ball and form a weak-star closed subset, described by the closed equations $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Apply the Banach-Alaoglu theorem. The Gelfand transform $a\mapsto \hat{a}$, where $\hat{a}(\varphi )=\varphi (a)$, is consequently a contractive unital algebra homomorphism into $C({\mathcal{M}}_A)$; it separates distinct characters. This argument proves compactness and continuity, but does not assert injectivity or isometry for an arbitrary Banach algebra.

For Hartogs theorem, collect all well-order relations on all subsets of a set $x$. They form a set, since the relations belong to the power set of $x\times x$. Replacement collects their ordinal order types in a set $T$. DefineIf there were an injection $\alpha \to x$, transfer the well-order of $\alpha$ to its image. Its type would put $\alpha$ in $T$, giving $\alpha +1\le \alpha$, impossible. Thus some ordinal does not inject into $x$, without using choice.

Assume the axiom of choice and let a nonempty partially ordered set $P$ have an upper bound for every chain. Suppose it has no maximal element. Choose, from every subset of $P$ with a nonempty set of strict upper bounds, one such strict upper bound. Every chain has a strict upper bound: first take an upper bound and then a strictly larger element, using the absence of maximal elements. Take a Hartogs ordinal $\alpha$ for $P$. Transfinite recursion chooses $p_{\beta }$, for each $\beta <\alpha$, strictly above all previously chosen elements. Their predecessors form a chain at every stage, so this recursion is defined and strictly increasing. It gives an injection $\alpha \to P$, contradicting the choice of $\alpha$. Hence $P$ has a maximal element, proving Zorn's lemma.

For the well-ordering theorem, take a choice function on the nonempty subsets of $x$ and recursively choose an element of $x$ not previously selected. If this process had not exhausted $x$ at any stage below its Hartogs ordinal $\alpha$, it would define an injection $\alpha \to x$, again impossible. Therefore it exhausts $x$ at some ordinal stage $\beta <\alpha$. Order each element by its unique selection index; this transfers the ordinal well-order to all of $x$. The empty set is already well-ordered. These proofs use transfinite recursion, not an unstated fixed-point theorem.

Write $f$ as the ring homomorphism $A\to B$. By the permitted assumption, $B$ is a finitely generated $A$-algebra. We will prove that every element of $B$ is integral over $A$, including any nilpotent contribution. If $B=0$, there is nothing to prove.

We use the integral closure as an intersection of valuation rings in its general form: if an integral domain $R$ lies in a field $K$, then its integral closure inside $K$ is the intersection of all valuation rings of $K$ containing $R$. Here $K$ need not be the fraction field of $R$. One direction follows from valuation rings are integrally closed. For the other, if $0\ne x\in K$ is not integral, the ideal $x^{-1}R[x^{-1}]$ is proper: an expression for $1$ in that ideal, multiplied by a sufficiently large power of $x$, would be a monic equation for $x$. Choose a maximal ideal containing this ideal, and apply the valuation domination lemma to the corresponding local ring inside $K$. The resulting valuation ring $V$ contains $R$ and has $x^{-1}$ in its maximal ideal, so $x\notin V$. This proves the claimed intersection criterion. The valuation domination lemma is obtained by choosing, using Zorn's lemma, a maximal local overring dominating the given local ring; the usual alternative between adjoining $z$ and adjoining $z^{-1}$ forces this maximal overring to be a valuation ring.

The Noetherian ring $B$ has finitely many minimal prime ideals ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_r$. SetTake any valuation ring $V$ of $K_i$ containing $R_i$. The maps $B\to D_i\to K_i$ and $A\to R_i\to V$ form a square for the valuative criterion for properness. Its lift gives $B\to V$, whose composite with $V\hookrightarrow K_i$ is the original map. Consequently the image $D_i$ is contained in $V$. The intersection criterion proves that every element of $D_i$ is integral over $R_i$.

Fix $b\in B$. For each $i$, choose a monic polynomial $P_i(T)\in A[T]$ whose value at the image of $b$ in $D_i$ is zero; coefficients over $R_i$ can be lifted to $A$. The product $P(T)={\prod }_i{P}_i(T)$ is monic and satisfiesThe nilradical of a Noetherian ring is nilpotent. Explicitly, choose finitely many generators $n_j$ with $n_j^{e_j}=0$; every product of $1+{\sum }_j(e_j-1)$ generators then contains a vanishing power. Thus $(\sqrt{(0)}{)}^N=0$ for some $N$, and $P(T{)}^N$ is a monic polynomial annihilating $b$. Hence $B$ is integral over $A$.