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An ordered group is Archimedean when $a\ge 1$ and $a^n\le b$ for all positive integers force $a=1$. In an additive Abelian group this excludes a positive element whose every integer multiple stays below one fixed bound. This generalizes the Archimedean property of the real numbers.

Dropping either condition destroys the implication. The increasing sequence $a_n=n$ has a lower bound one but no upper bound, and no finite limit of a sequence. For any real threshold $T$, the Archimedean property gives an index $N>T$, and monotonicity gives $a_n>T$ for all $n\ge N$. Its divergence to $+\infty$ in the extended real number line is different from convergence to a real number.

Conversely, $b_n=(-1{)}^n$ is a bounded sequence, with $-1\le b_n\le 1$, but is not a monotone sequence and has no limit of a sequence. Its subsequence at even integers is constantly one, and its subsequence at odd integers is constantly minus one. These two subsequential limits cannot both equal a single limit of a sequence. ^[ref]

The ambient real numbers also matter. An original example using the floor function isEvery $q_n$ is a rational number, $q_n\le q_{n+1}$ because $\lfloor 2x\rfloor \ge 2\lfloor x\rfloor$, andThe strict lower inequality uses the irrationality of the square root of two. Hence this bounded sequence converges to $\sqrt{2}$ in the real numbers, but has no limit of a sequence in the rational numbers. Convergence to a rational candidate would also be convergence in the real numbers, contradicting uniqueness of the limit of a sequence. The formulas give $q_1=1$, $q_2=5/4$, $q_3=11/8$, $q_4=11/8$: this counterexample even has a plateau. This illustrates why completeness of the real numbers matters: the rational numbers lack the least-upper-bound property needed to contain the required supremum. ^[ref]

If every sequence of distinct elements of a positive-real set $A$ has arithmetic averages tending to zero, then $A$ is a countable set. For each positive integer $n$, the threshold set $\{a\in A:a\ge 1/n\}$ must be finite: an infinite sequence inside it would have averages at least $1/n$. The Archimedean property makes $A$ their union, so the countable union of countable sets theorem applies. The hypothesis concerns every distinct-element sequence, not merely one enumeration.

For a nonempty set $A\subseteq \mathbb{R}$, a real number $\ell$ is its infimum precisely when it is a lower bound and every other lower bound is at most $\ell$. It is the greatest lower bound, not necessarily the minimum.

For $C=\{1/n:n\in {\mathbb{N}}_{>0}\}$, zero is a lower bound. Every positive candidate $\epsilon$ fails: the Archimedean property gives an integer $n>1/\epsilon$, hence $1/n<\epsilon$. Thus $\inf C=0$, although zero is never a member of $C$. This is the lower-edge counterpart of approximating an unattained supremum.

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.

To prove the Archimedean property, suppose the positive integers were bounded above in $\mathbb{R}$. By the least upper bound axiom they would have a finite supremum $s$. Since $s-1$ is not an upper bound, some positive integer $k$ satisfies $k>s-1$. Then the positive integer $k+1>s$, contradicting the definition of $s$. Thus the positive integers are unbounded. Equivalently, for any positive real $a,b$, some positive integer $n$ satisfies $na>b$, by applying unboundedness to $b/a$.

For fixed $m$, put $r_m={\cos }^2(m!\pi x)$. The standard properties needed are $0\le {\cos }^{2}t\le 1$, and ${\cos }^{2}t=1$ exactly when $t\in \pi \mathbb{Z}$. If $0\le r<1$, then $r^n\to 0$; for $0<r<1$ this also follows from the Bernoulli inequality, since $r^{-n}\ge 1+n(r^{-1}-1)\to \infty$. Therefore the inner limit isIf $x=p/q$ is a rational number, with $q\ge 1$, then $q$ divides the factorial $m!$ whenever $m\ge q$. Hence $m!x$ is an integer for all sufficiently large $m$, and the outer limit is one. This includes the endpoints $x=0,1$. If $x$ is an irrational number, $m!x$ cannot be an integer for any $m$, since division by $m!$ would make $x$ rational. Then every inner limit is zero. The factorial cosine iterated limit detects rationality result is consequentlyEach inner limit was taken with $m$ fixed; the argument does not interchange the two limits or assert a joint limit as both indices grow.

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.

No. The parameter-free partial typeis finitely satisfiable in the ordered real field: choose a number larger than the finitely many bounds. By compactness theorem it extends to a complete type over the empty set. But no real number realizes it, by the Archimedean property. Therefore the ordered real field is not an omega-saturated model.

The named unit and addition make every integer bound expressible, even without extra parameters. This is why the obstruction is present in the ordered-field language and absent in the two reducts above.

Write each member as $3-2/(n+1)$. Every member is strictly below $3$, so $3$ is an upper bound. If $u<3$, the Archimedean property supplies $n\in \mathbb{N}$ with $n+1>2/(3-u)$; then $3-2/(n+1)>u$. Thus no smaller number is an upper bound, proving $\sup S=3$. Since $3\notin S$, there is no greatest element. Equivalently, each member is followed by a larger member because $2/(n+2)<2/(n+1)$.