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For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

For each positive integer $n$, consider the threshold set $A^{(n)}=\{a\in A:a\ge 1/n\}$. If it were infinite, one could choose recursively a sequence of distinct elements from it. Every arithmetic average of that sequence would be at least $1/n$, contradicting the assumed limit zero. Hence every $A^{(n)}$ is a finite set.

Every positive real number $a$ exceeds $1/n$ for some $n$, by the Archimedean property. ThereforeA countable union of countable sets is countable, so $\boxed{A\text{ is countable}}$. In fact, under the stipulated infinitude, $A$ is a countably infinite set. Positivity matters: it ensures that these thresholds cover $A$. This is countability from vanishing averages of distinct sequences; ordinary Cesaro convergence of a sequence of one chosen enumeration would not suffice to control the whole set.

A sequence $(a_k)$ converges to $a\in \mathbb{R}$ when for every $\epsilon >0$ there is $N$ such that $|a_k-a|<\epsilon$ for all $k\ge N$. A series ${\sum }_{k=1}^{\infty }{a}_k$ converges to a finite $S$ when its partial sums $s_n={\sum }_{k=1}^n{a}_k$ converge to $S$.

If the series converges, then $a_n=s_n-s_{n-1}\to S-S=0$, so its terms tend to zero. The converse is false: the harmonic series has terms tending to zero but diverges.

For the arithmetic averages, suppose $a_k\to a$. Fix $\epsilon >0$ and choose $K$ so $|a_k-a|<\epsilon /2$ for $k\ge K$. The finite initial contribution $C={\sum }_{k=1}^{K-1}|a_k-a|$ is fixed, andFor sufficiently large $n$, $C/n<\epsilon /2$, provingThis is the Cesaro theorem for convergent sequences. Its hypothesis concerns a convergent sequence; the corresponding series need not be a convergent series.

The dual space here is the algebraic dual space $X^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(X,\mathbb{R})$, consisting of every linear functional; no topology or boundedness is imposed. For a basis $e_1,\dots ,e_n$, define $e_i^{\ast }({\sum }_j{x}_j{e}_j)=x_i$. These are linear functionals satisfying $e_i^{\ast }(e_j)={\delta }_{ij}$. Evaluating ${\sum }_i{a}_i{e}_i^{\ast }=0$ on each $e_j$ gives $a_j=0$, proving linear independence. Conversely, for every $T\in X^{\ast }$,This proves spanning and hence proves, without assuming a dimension theorem, that the dual basis is a basis and $\dim X^{\ast }=n$.

The canonical bidual evaluation map isIts definition makes no choice of basis and it is linear. If $x\ne 0$, one coordinate functional is nonzero on $x$, so $J$ is injective. For any $B\in X^{\ast \ast }$, choose $x={\sum }_{i}B(e_i^{\ast })e_i$. Expanding $T$ in the dual basis gives $T(x)={\sum }_{i}T(e_i)B(e_i^{\ast })=B(T)$, proving surjection explicitly. Thus $J$ is an isomorphism.

If $T_1,\dots ,T_n$ is any basis of $X^{\ast }$, its dual basis $B_1,\dots ,B_n$ in $X^{\ast \ast }$ exists by the result just proved. Put $v_j=J^{-1}(B_j)$. These vectors form a basis of $X$, and $T_i(v_j)=B_j(T_i)={\delta }_{ij}$. Every basis of the finite-dimensional dual is a dual basis, of a uniquely determined basis of $X$.

For a separating subspace of an algebraic dual $W$, choose a basis $T_1,\dots ,T_r$ of $W$. The linear map $x\mapsto (T_1(x),\dots ,T_r(x))$ is injective because $W$ separates nonzero vectors. The rank-nullity theorem implies $n\le r$; but $W\subseteq X^{\ast }$ implies $r\le n$. HenceThe zero-dimensional case also has this conclusion.

For $X=\mathbb{R}[x]$, the monomials $1,x,x^2,\dots$ form an infinite basis, each polynomial using only finitely many of them. The map $T\mapsto (T(1),T(x),T(x^2),\dots )$ identifies $X^{\ast }$ with all real sequences, because any sequence $(a_j)$ defines $T({\sum }_{j=0}^d{c}_j{x}^j)={\sum }_{j=0}^d{c}_j{a}_j$. The linear subspace of finitely supported sequences is proper (it excludes $(1,1,\dots )$) and separating (one coefficient functional detects any nonzero polynomial). Thus

Write $\mathbb{N}=\{1,2,\dots \}$ and let $\chi$ be the given finite coloring of the edges of the complete graph on the positive integers. Construct an increasing sequence $(v_i)$ and nested infinite sets $A_0\supset A_1\supset \cdots$ as follows. Set $A_0=\mathbb{N}$. Having chosen $A_{i-1}$, take $v_i=\min A_{i-1}$, partition $A_{i-1}\setminus \{v_i\}$ by the color of $\{v_i,w\}$, and use the infinite pigeonhole principle to choose an infinite set $A_i$ on which this color is constant, say $r_i$. Every element of $A_i$ exceeds $v_i$.

The infinite pigeonhole principle applied again to $(r_i)$ gives an infinite set of indices $I$ and a color $r$ such that $r_i=r$ for all $i\in I$. For $i<j$ in $I$, nesting gives $v_j\in A_i$, so $\chi (\{v_i,v_j\})=r$. Thus the infinite set $X=\{v_i:i\in I\}$ is a monochromatic complete graph:This proves the required infinite case of Ramsey's theorem; finiteness of the color set is used at both applications of the infinite pigeonhole principle.

Let $\rho$ be the cyclic logarithmic coloring from part (iii). Define a finite coloring of the positive integers byThe inner logarithm is at least $1$ whenever $n\ge 2$, so this is defined everywhere. Suppose an increasing infinite sequence $(x_i)$ made every product with distinct indices monochromatic. Fix an index $i$ with $x_i\ge 2$ and write $a={\log }_2{x}_i>0$. The sequence is unbounded, so choose $j>i$ with $b={\log }_2{x}_j\ge 28a$. Both ordered products are among the supposedly monochromatic values, but their logarithms satisfyIndeed, the first inequality is equivalent to $b\ge 28a$, and the second uses $a>0$. Part (iii) therefore gives $\chi (x_i{x}_j^2)\ne \chi (x_j{x}_i^2)$, a contradiction. Thus the assertion with all distinct ordered indices is false. The obstruction needs an infinite unbounded sequence; it does not assert that every pair of different positive integers gives different colors.

The dynamical proof of Hindman's theorem gives the following result. The Hindman theorem asserts that every finite coloring of the positive integers admits an infinite strictly increasing sequence $(a_i)$ for which all nonempty finite sums of distinct terms have one color. We will in fact arrange $a_{r+1}>a_1+\cdots +a_r$, so all these sums also have unique representations.

Extend the given coloring arbitrarily to a point $x\in [k{]}^{\mathbb{Z}}$, retaining the prescribed colors at every positive coordinate. Let $X$ be its forward orbit closure under the left shift. The product topology makes $X$ a compact metric space, and the left shift is a continuous map. There is a nonempty minimal subsystem $Y\subseteq X$: order the nonempty closed forward-invariant subsets by reverse inclusion, use compactness and the finite intersection property to intersect any chain, and apply the Zorn lemma. Minimality also implies $\mathcal{L}(Y)=Y$. The result permitted in the question now supplies a minimal point $y\in Y$ proximal to $x$.

We need the joint return lemma for a proximal minimal pair, which we prove here. It suffices to consider an open neighborhood $U$ of $y$ in $X$. Choose an open neighborhood $V$ of $y$ with $\overline{V}\subseteq U$. Every forward orbit in $Y$ meets $V$, and a finite subcover of $\{{\mathcal{L}}^{-j}V:j\ge 0\}$ on $Y$ supplies a bound $J$ on the needed return index. For a compatible metric $d$, choose $\eta >0$ smaller than the distance from $\overline{V}$ to $X\setminus U$ when the latter is nonempty. By uniform continuity of the finitely many maps ${\mathcal{L}}^j$, $0\le j\le J$, some $\delta >0$ ensuresProximality supplies arbitrarily large $t$ with $d({\mathcal{L}}^{t}x,{\mathcal{L}}^{t}y)<\delta$. To see that the times can be large under the definition using an infimum over $t\ge 0$, either $x=y$, in which case this is automatic, or injectivity of the left shift makes every finite collection of distances strictly positive, so a sufficiently smaller proximal distance occurs beyond that collection. Some $j\le J$ has ${\mathcal{L}}^{t+j}y\in V$. Then ${\mathcal{L}}^{t+j}x\in U$ also. Hence arbitrarily large positive $n$ satisfyThis uses the minimal point property for bounded returns and proximality for closeness; closeness alone would not guarantee a return near $y$.

Put $q=y(0)$. Inductively, let $F_r=\{0\}\cup \mathrm{FS}(a_1,\dots ,a_r)$, where $\mathrm{FS}$ denotes the finite-sums set, and maintainThe initial condition at $r=0$ is just $y(0)=q$; no condition on $x(0)$ is required. The cylinder set $U_r=\{z:z(s)=q\text{ for every }s\in F_r\}$ is a neighborhood of $y$. Use the proved joint return lemma to choose $a_{r+1}>{\sum }_{i=1}^r{a}_i$ with both ${\mathcal{L}}^{a_{r+1}}x$ and ${\mathcal{L}}^{a_{r+1}}y$ in $U_r$. All new sums belong to $a_{r+1}+F_r$, and the old sums remain in $F_r$, so both inductive conditions persist. Every nonzero sum is positive, where $x$ agrees with the original coloring. ThereforeThis proves the Hindman theorem using only the permitted proximal-minimal existence result and the compactness and return arguments supplied above.

Let $\mathrm{Inj}(<\omega ,X)$ denote the finite repetition-free sequences, including the empty sequence. The map $x\mapsto (x)$ is injective, so this set is infinite when $X$ is infinite. Suppose, for a contradiction, that it contains a countably infinite subset. Fix its given injective enumeration $s_0,s_1,\dots$; this is part of the supposition, not a choice from a family of sets.

Enumerate all pairs $(n,j)$ with $j<\mathrm{length}(s_n)$ by their natural-number pairing codes, recording the corresponding entries $s_n(j)$. If the union of entries were infinite, recursively selecting the first new entry would give an injection $\mathbb{N}\to X$, contradicting Dedekind-finiteness. Therefore the union is a finite set $F$, say of size $m$. A repetition-free sequence over $F$ has length at most $m$, and there are exactlysuch sequences, with the empty product equal to one. All $s_n$ would belong to this finite set, contradicting their distinctness. Finite counting and the least-code construction use no axiom of choice. This proves finite repetition-free sequences preserve Dedekind-finiteness and the required infinitude:

Work in ZF with the law of excluded middle, without the axiom of choice. The paper's terminology is an infinite Dedekind-finite set. The useful common principle is countable union of explicitly ordered finite lists without choice: given an actual sequence of finite lists, enumerate their entries by list number and position using a Cantor pairing function. If the union of entries is infinite, repeatedly take the entry with the least code not already selected. This produces an injection from $\mathbb{N}$ without choosing any enumerations of unordered sets.

Consequently, if a family of objects has an explicitly ordered finite list of labels for each object, and only finitely many objects can be made from any given finite pool of labels, then a countably infinite list of distinct objects forces a countably infinite subset of the label set. For a Dedekind-finite label set this is impossible. If the label set embeds into the object family as well, infinitude is preserved. ThusThe distinction between ordered lists supplied by the data and arbitrary finite subsets is essential: a blanket countable-union theorem for unordered finite sets would introduce a choice principle.

One form of the Ehrenfeucht-Mostowski theorem says that, for every infinite first-order structure $M$ and every total order $I$, an elementary extension contains distinct elements $(a_i{)}_{i\in I}$ forming an order-indiscernible sequence. After a suitable Skolem expansion, their Skolem hull is a first-order model of $\mathrm{Th}(M)$, and every order automorphism of $I$ extends to a structure automorphism of that hull. If $M$ has a definable infinite linearly ordered subset $P$, the generators may lie in $P$ and their order may agree with that definable order. Uniqueness of the extended structure automorphism is asserted in the chosen Skolem expansion, rather than for every structure automorphism of its reduct.

Here is an ultraproduct proof of the Ehrenfeucht-Mostowski theorem. First expand $M$ to $M^{+}$ with Skolem functions for all existential first-order formulas, iterating through the enlarged languages if necessary. This is done in the ordinary classical metatheory; the choice restriction in question 1 is not a restriction on this question. Fix a nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$.

We can obtain an infinite sequence of distinct elements $b_0,b_1,\dots$ in an elementary extension $M_0$ of $M^{+}$ by one preliminary ultrapower. For each $n\ge 1$ choose a finite list of $n$ distinct elements of $M^{+}$. Represent $b_j$ by the function taking the $j$th entry when $n>j$, with an arbitrary default otherwise. Distinctness holds on a cofinite set, so follows from Łoś theorem. In the definably ordered variant, first name any parameters defining $P$, and choose each finite list increasingly inside $P$ instead; then $M_0\models P(b_j)$ and $b_j<b_k$ for $j<k$. Infinitude of an ordered set supplies every finite increasing chain. This preparatory step uses no Ramsey theorem.

For a finite subset $F=\{i_1<\cdots <i_r\}$ of $I$, define the ultrafilter ${\mathcal{U}}_F$ on ${\mathbb{N}}^F$ by nested membership, in this order:where $(\mathcal{U}n)Q(n)$ means $\{n:Q(n)\}\in \mathcal{U}$. This is the ordered Fubini product of ultrafilters. Closure under intersections and the decision between a set and its complement hold at each nested level, proving it is an ultrafilter. For $F=\varnothing$ take the principal ultrafilter on the one-point product. PutIf $F\subseteq G$, pull a function back along the projection ${\mathbb{N}}^G\to {\mathbb{N}}^F$. A test independent of an omitted coordinate is unchanged by its ultrafilter quantifier, so projection pushes ${\mathcal{U}}_G$ to ${\mathcal{U}}_F$. The induced map $N_F\to N_G$ is therefore an elementary embedding by Łoś theorem. These maps are coherent, giving a directed system indexed by finite subsets of $I$.

Take its directed limit of elementary embeddings $N$. One can verify elementarity directly: representatives of a finite tuple occur at a common stage; functions and atomic relations are interpreted there. In the existential step, a witness in the limit occurs with the parameters at some later common stage, and elementarity pulls the existence statement back. Thus every $N_F$ embeds elementarily into $N$, which contains an elementary copy of $M^{+}$.

For $i\in F$, let $a_i$ be the class of the coordinate function $\mathbf{n}\mapsto b_{n_i}$. Projection coherence makes this independent of $F$. If $i<j$, then for each fixed $n_i$ the set $\{n_j:n_j\ne n_i\}$ is cofinite. The nested test therefore gives $a_i\ne a_j$. In the ordered variant, the same argument with $n_j>n_i$ gives $P(a_i)$ and $a_i<a_j$.

For every first-order formula $\phi$ of the expanded language and every increasing tuple $i_1<\cdots <i_r$, Łoś theorem givesThe right-hand side depends on the first-order formula and tuple length, not on the indices. This proves order indiscernibility in the expanded language. Let $H$ be the Skolem hull of the generators in $N$. The Tarski-Vaught test gives $H\prec N$, and its reduct is a first-order model of $\mathrm{Th}(M)$.

An order automorphism $\pi$ of $I$ acts byIndiscernibility makes this well defined: combine the finite supports of two term expressions into one increasing tuple, and apply indiscernibility to their logical equality. Applying it to relation first-order formulas proves preservation of all relations; the inverse is induced by ${\pi }^{-1}$. Every element of the hull is such a term, so the extension is unique among structure automorphisms preserving the Skolem expansion. MoreoverThis proves the theorem, including its ordered version and structure automorphism conclusion, using ultrapowers and a directed limit throughout.

Use untyped lambda calculus and normal-order beta reduction. Write $c_n=\lambda fx.f^{n}x$ for the Church numeral of $n$. We will construct a closed term $F$ such thatwhen $f(\mathbf{n})$ is defined, and with no head normal form otherwise. The latter condition implies that the result is not in beta equivalence with any Church numeral, so both definedness and the output are represented.

First represent all total primitive recursive functions. The zero function is $\lambda \mathbf{x}.c_0$, the successor function isand the $i$th projection function is $\lambda x_1\cdots x_k.x_i$. Function composition in recursion theory is represented by $\lambda \mathbf{x}.G(H_1\mathbf{x})\cdots (H_r\mathbf{x})$. This composition claim is used here for total functions, whose inner terms all reduce to numerals.

For primitive recursion use Church pairs and iteration. PutSuppose $G,H$ represent the total constituents of $f(\mathbf{x},0)=g(\mathbf{x})$ and $f(\mathbf{x},n+1)=h(\mathbf{x},n,f(\mathbf{x},n))$. DefineAfter $j$ iterations the pair reduces to $\mathsf{Pair}{c}_j{c}_{f(\mathbf{x},j)}$, by mathematical induction on $j$. Its second projection is the required result. This lambda definition of primitive recursion by pair iteration and structural induction on the primitive recursive construction give terms for all primitive recursive functions.

For an arbitrary partial computable function use the Kleene normal form theorem with its program index fixed. There are total primitive recursive functions $C(\mathbf{x},s)$ and $U(s)$ such that $C$ is zero exactly when $s$ codes a valid halting computation history on $\mathbf{x}$, and $U$ extracts that history's output. Checking finite coded configurations and each transition is primitive recursive. Determinism ensures that every accepted history on an input has the same output. HenceIf the input computation never halts, no history is accepted. Let $\hat{C},\hat{U}$ be the total numeral-representing terms already obtained. The Church numeral zero test is $\mathsf{Zero}=\lambda n.n(\lambda z.\mathsf{F})\mathsf{T}$, returning $\mathsf{T}$ exactly on zero. A Church Boolean applied to two arguments selects the appropriate branch without evaluating the other.

Take the fixed-point combinator $Y=\lambda h.(\lambda z.h(zz))(\lambda z.h(zz))$. For the input tuple defineExpanding the displayed abbreviations gives a genuine finite combinator. At a numeral code $c_j$, the total predicate computation terminates. If it returns a positive numeral, the false Church Boolean selects the next search code. If it returns zero, the true Church Boolean selects $\hat{U}{c}_j$, which reduces to the output numeral. If the least accepted code is $s$, a finite sequence of these tests reaches it and then produces $c_{U(s)}$.

If there is no accepted code, every test is false and head reduction moves through codes $0,1,2,\dots$ forever. At no finite point can a head variable or an outer numeral abstraction be produced: the applied search term must first unfold the fixed point and evaluate the next total test, and the decoder branch is always discarded. Thus it has no head normal form. The standard head-normalization property, or the Church-Rosser theorem together with normalization of normal-order reduction, rules out beta equivalence to a numeral. In particular, we have avoided a lazy-composition pitfall: an outer function that discards an argument need not force a divergent inner computation. The decoder is reached only after the guarded search succeeds. This proves lambda representation of partial computable functions:

The strips are nested. Every connection permitted inside $T_k$ is also permitted inside $T_{k+1}$, so $q_k(n)\le q_{k+1}(n)$ for every fixed separation. ConsequentlyThe preceding nonnegative bound makes this a decreasing sequence bounded below. By monotone convergence of real sequences,There is no assertion that this limit of a sequence is strictly positive: positivity at each fixed width need not survive an increasing-width limit of a sequence.

One can also identify the limit of a sequence. Let $q(n)$ be the percolation two-point connection probability in the whole square lattice. Every finite connecting graph path has a bounded vertical extent, so $q_k(n)\uparrow q(n)$. Commuting infima gives the strip approximation to the planar connection decay rate:The same positive-association argument identifies the final infimum with the whole-plane normalized logarithmic limit of a sequence. This is an interchange of infima justified by monotonicity at fixed $n$. For example, when $p>1/2$, the threshold proved in question 2 and uniqueness give an infinite percolation cluster with root probability $\theta (p)>0$. The Harris-FKG inequality makes the probability that both endpoints belong to it at least $\theta (p{)}^2$, so $q(n)\ge \theta (p{)}^2$. The whole-plane rate, and hence ${\lim }_k{f}_k(p)$, is then zero, despite the strict positivity of every finite-strip rate.

First take $x\ne y$. Sample a uniform spanning tree using Wilson's algorithm rooted at $y$, with $x$ processed first. The first walk runs from $x$ until its first hit of $y$, so its loop erasure is exactly the unique $x$-to-$y$ graph path of the resulting tree. Later attachments cannot alter that graph path. This is the uniform-spanning-tree path representation of loop-erased random walk.

Instead root Wilson's algorithm at $x$ and process $y$ first. The resulting tree still has the same uniform law, while its first attachment has the law of loop-erased random walk from $y$ to $x$. A tree's graph path between two fixed graph vertices has the same unoriented edge set in both directions. Pushing the common tree law through this graph path map therefore givesHere $\mathcal{E}$ means the set of unoriented edges. In fact, reversing the complete graph vertex sequence from the first loop-erased walk has the law of the second, giving reversibility of loop-erased random walk. If $x=y$ and first hitting allows time zero, both graph paths are empty and the equality is immediate.

This is a distributional assertion, not a pathwise identity between chronological loop erasure and reversal. For example, the valid walk trajectory $x,a,b,a,c,b,y$ in a graph containing the triangle on $a,b,c$ and the edges $xa,by$ erases forward to $x,a,c,b,y$. Erasing the reversed trajectory and then reversing the result gives $x,a,b,y$, a different graph path. Wilson's root-independent tree law is what proves the stated equality of probability distributions.

For the diagonal operator on sequence space,Equality holds at the first coordinate vector, so $\Vert K\Vert =1$ and $K$ is a bounded linear operator. Every diagonal entry is nonzero, giving the null space $\mathcal{N}(K)=\{0\}$. A datum has a preimage exactly when its weighted coordinates belong to the l2 sequence space. HenceFinite sequences lie in the range and are dense in ${\ell }^2$, so $\mathcal{R}(K{)}^{\perp }=\{0\}$ andThe sequence $f^{(n)}=e_n/n$ converges to zero in the data norm, but $K^{†}{f}^{(n)}=e_n$ has norm one. Thus $K^{†}$ is unbounded and discontinuous. In particular the dense range is not closed; the datum $f_j=1/j$ also proves that it is a proper subset of ${\ell }^2$.