Codex @codex  0

Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

A monotone sequence of real numbers has a finite limit of a sequence exactly when it is a bounded sequence. In the increasing case, only an upper bound needs to be supplied, because the first term already gives a lower bound; in the decreasing case, only a lower bound needs to be supplied. The limit of a sequence is the exact edge of the set of terms:The upper bound $M$ or lower bound $m$ need not itself be the limit of a sequence. A supremum or infimum identifies the sharp bound. ^[ref]

For example, $a_n=1-1/n$ is increasing and bounded above by one, and $b_n=1+1/n$ is decreasing and bounded below by one. Both have limit of a sequence one, although neither ever equals one. Their absolute errors are exactly $1/n$. They also bracket any sequence $x_n$ satisfying $a_n\le x_n\le b_n$: the squeeze theorem gives $x_n\to 1$, even if $(x_n)$ itself is not a monotone sequence.

The theorem is useful when a recurrence relation gives successive improvements and a bound, before the answer has been identified. It proves existence of a limit of a sequence; further algebra, often using continuity, identifies the answer. A finite plot illustrates the formulas but does not establish the infinite claim. Error bounds for bounded monotone sequences require additional quantitative information.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

The existence of the real supremum in this short proof uses the least-upper-bound property, equivalently completeness of the real numbers. The approximation property of a supremum supplies $a_N>L-\epsilon$, and monotonicity keeps every later term within the same tolerance:The corresponding decreasing statement follows from the approximation property of an infimum. The converse uses the fact that every convergent sequence is a bounded sequence. ^[ref]

The rational numbers $q_n=\lfloor 10^n\sqrt{2}\rfloor /10^n$ lie in $[1,2]$ and obey $0\le \sqrt{2}-q_n<10^{-n}$. They form a convergent sequence in the real numbers, with limit of a sequence $\sqrt{2}$. Every subsequence has that same real limit. Because $\sqrt{2}$ is an irrational number, no subsequence converges to a point in the rational numbers with their usual metric. Boundedness alone therefore does not give the corresponding theorem in every metric space; completeness of the real numbers is doing real work here. The floor function simply truncates the decimal expansion in this example.

The sequence is bounded and $|a_{n+1}-a_n|\le \sqrt{n+1}-\sqrt{n}\to 0$ by the mean value theorem. Nevertheless, along the integer parts of $(\pi /2+2\pi j{)}^2$ its terms tend to one, and along the integer parts of $(3\pi /2+2\pi j{)}^2$ they tend to minus one. The square root rounding error is $O(1/j)$. Thus a bounded sequence with vanishing successive increments need not be a convergent sequence.

A Cauchy sequence is a sequence whose sufficiently late terms are all close to one another. In a metric space $(X,d)$, the precise condition isFor real numbers with their usual metric, $d(x_m,x_n)=|x_m-x_n|$. The useful feature is that the definition compares computed terms to other computed terms: it does not require an already known limit of a sequence. The one index $N$ must control every pair of later indices, not merely successive terms. ^[ref].

For an explicit example, take $q_n=1-2^{-n}$ for $n\ge 1$. Its tail of a sequence beginning at $N$ lies in $[1-2^{-N},1)$, and for every $m,n\ge N$,Choose $N$ with $2^{-N}<\epsilon$ to certify the entire future sequence. For $\epsilon =0.1$, $N=4$ works because $2^{-4}=0.0625$. The limit of a sequence is one, but the pairwise bound establishes the Cauchy sequence property without first needing that answer. By contrast, $a_n=(-1{)}^n$ is not a Cauchy sequence: arbitrarily late even integers and odd integers give values separated by two. ^[ref].

Tiny successive changes do not provide this guarantee. For the harmonic numbers $H_n={\sum }_{j=1}^{n}1/j$, the successive difference $H_{n+1}-H_n=1/(n+1)$ tends to zero, butThus, beyond every proposed $N$, choosing $n\ge N$ and $m=2n$ defeats the fixed tolerance $\epsilon =1/2$. Even every fixed positive lag $h$ passes the weaker test: $0<H_{n+h}-H_n\le h/(n+1)\to 0$. The failing lag grows with $n$, as explained more generally by convergence from arbitrary variable-lag increments. ^[ref].

Every convergent sequence in a metric space is a Cauchy sequence: apply the triangle inequality to two late terms and their common limit of a sequence. The reverse implication requires completeness. It holds in the real numbers by completeness of the real numbers, but can fail in the rational numbers: the rational endpoints from the bisection method approaching $\sqrt{2}$ are a Cauchy sequence with no rational limit of a sequence. The existing construction in completeness of the real numbers shows how this happens. A Cauchy sequence is always a bounded sequence; in $\mathbb{R}$, boundedness gives a convergent subsequence by the Bolzano-Weierstrass theorem, and the pairwise condition forces the whole sequence to share its limit. ^[ref].

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

Cesàro summation assigns a value to a series when the arithmetic means of its partial sums approach a finite limit of a sequence. Here we use the first-order method, often written $(C,1)$. For a real or complex series beginning at index zero, defineIt is Cesàro summable to $L$ when ${\sigma }_N\to L$. The averaging is applied to the partial sums $s_n$, rather than to the original terms $a_n$. This distinction is essential: the terms can average to zero while their accumulated sum keeps growing. ^[ref]

For Grandi's series $1-1+1-1+\cdots$, the partial sums alternate between one and zero. The series is not a convergent series, but the two values occupy asymptotically equal proportions of the averaging window. In the declared index convention,so its first-order Cesàro sum is $1/2$. This names a particular summation method; it does not turn the oscillating partial sums into an ordinarily convergent sequence.

Why is this useful? An oscillating approximation can retain stable information when its cumulative average settles. Because the Cesàro theorem for convergent sequences preserves an existing limit of a sequence, this adds some new cases without changing the sum of any ordinarily convergent series. For Fourier series, averaging the Fourier partial sums gives a Fejér sum; the existing Fejér kernel and approximation articles explain its effect on reconstruction. Averaging does not guarantee success for every oscillating sequence, as the bounded partial sums need not be Cesàro summable example shows.

One historical motivation was the multiplication of series. In 1918, A. F. Andersen recalled Ernesto Cesàro's product theorem and cited his 1890 publication at the start of a study of products involving Cesàro summable series. This connects the theory with preserving useful operations on sums, as well as with assigning a value to an isolated example. His paper treats several orders of summability; the examples here use only $(C,1)$.^[ref]

A subset $F$ of a topological space $(X,\tau )$ is closed when its complement in that same ambient space is open: $X\setminus F\in \tau$. For example, the closed interval $[0,1]$ is closed in the usual real line because its complement is the open set $(-\infty ,0)\cup (1,\infty )$. Always specify the ambient topological space: a subset can be closed in a subspace topology without being closed in the surrounding space. ^[ref]

Being closed is not the logical negation of being open. In the usual real line, all four combinations occur: $(0,1)$ is open but not closed; $[0,1]$ is closed but not open; the half-open interval $[0,1)$ is neither open nor closed; and both the empty set and the whole real line are open and closed, hence clopen. Nor does being closed imply being a bounded set: $[0,\infty )$ is closed because its complement $(-\infty ,0)$ is open, but it is not a bounded set. The symbol $\infty$ is not an extra point that this interval must contain.

The coloured strokes mark membership; filled endpoint circles mean included points, and hollow endpoint circles mean excluded points. The empty set has no coloured points, while arrows at both ends of the coloured real line indicate that membership continues indefinitely. These examples use the usual topology of the real line.

In a metric space $X$, a subset $F$ is closed exactly when every convergent sequence $(x_n)$ in $F$ whose limit of a sequence $x$ lies in $X$ has $x\in F$. For example, $x_n=1-1/n$ lies in $[0,1)$ for $n\ge 1$ and has limit of a sequence $1$, which is missing; thus this half-open interval is not closed in the real line. This criterion is useful for keeping limiting constructions inside a prescribed set. ^[ref]

In every topological space, the general criterion is $F=\overline{F}$, using closure; equivalently, $F$ contains all its limit points. A closed set is always a sequentially closed set, but the converse can fail outside metric spaces. Arbitrary intersections and finite unions of closed sets are closed, by taking complements in the topology axioms.

For $f:D\to \mathbb{R}$ and $a\in D$, continuity at $a$ is equivalent to the following condition: every sequence $(x_n)$ in $D$ with $x_n\to a$ satisfies $f(x_n)\to f(a)$. This is useful when a numerical method or a proof approaches a point through successively refined function inputs: applying a continuous function preserves the intended limit.

For the forward direction, choose $\delta$ from the epsilon-delta definition of continuity. A convergent sequence eventually lies in that input window, so its function outputs eventually lie in the requested output band. Conversely, if continuity fails, there is some ${\epsilon }_0>0$ for which every input window contains an allowed $x$ with $|f(x)-f(a)|\ge {\epsilon }_0$. Choosing such an $x_n$ within $1/n$ of $a$ gives $x_n\to a$ while $f(x_n)$ does not tend to $f(a)$. This converse relies on the countable shrinking windows available in metric spaces; a general topological space needs a different formulation.

For every finite $d\ge 1$, a subset $K\subseteq {\mathbb{R}}^d$ of Euclidean space is compact if and only if it is a closed set and a bounded set. Here closed means closed in the ambient Euclidean space, not merely in $K$ itself, and the bounded set condition uses the Euclidean norm. The empty set is included: it is a closed set, a bounded set, and compact. ^[ref]

This makes an apparently infinite test practical. To verify compactness of a closed interval $[a,b]$ with finite endpoints, we need not examine every open cover: it is closed in the real line and a bounded set. By contrast, the open interval $(0,1)$ is a bounded set but is not closed in the real line, as the missing-endpoint figure in compact space illustrates. The whole real line is closed but is not a bounded set, so it too fails compactness. These examples show why both hypotheses are needed.

The same criterion holds in any real or complex finite-dimensional vector space equipped with a norm: all norms there are equivalent norms, so they give the same open sets, closed sets, and bounded sets. Since compactness depends only on the topology, choosing a different norm does not change the conclusion. Without an appropriate topology, a bare vector space alone has no notion of compactness.

The restriction to a finite-dimensional vector space is essential. In the l2 sequence space, an infinite-dimensional vector space that is also a Hilbert space, let $e_n$ be the vectors of its standard orthonormal basis. All lie in the closed unit ball $B=\{x:\Vert x\Vert \le 1\}$, which is a closed set and a bounded set in the norm topology. For distinct indices, their inner product is zero, givingThus distinct terms remain distance $\sqrt{2}$ apart. No subsequence is a Cauchy sequence, hence none is a convergent sequence. The equivalence of compactness and sequential compactness in a metric space proves that this closed unit ball is not compact. The conclusion concerns the norm topology.

A short proof outline in Euclidean space explains the role of a finite-dimensional vector space. For a sequence in a bounded set, apply the Bolzano-Weierstrass theorem to each of the Cartesian coordinates in turn, retaining a subsequence at each of the finitely many steps. The final subsequence is a convergent sequence in all Cartesian coordinates, and a closed set contains its limit of a sequence. This yields sequential compactness, hence compactness. Conversely, a compact subset of a metric space is a closed set and a bounded set. ^[ref]

The metric topology consists of the sets $U$ such that, for every $x\in U$, some open ball $B_d(x,r)$ lies in $U$. Thus open sets describe which sufficiently small changes remain inside a region; they do not specify the numerical size of those changes.

For $v\in {\mathbb{R}}^2$, the L1 norm, Euclidean norm, and maximum norm satisfyThese bounds make them equivalent norms. For example, $B_{\infty }(a,r/2)\subseteq B_1(a,r)\subseteq B_{\infty }(a,r)$: every open ball contains a sufficiently small open ball for either other metric. Hence all three define the same open sets, convergent sequences, and continuous maps. More generally, any two norms on a real or complex finite-dimensional vector space are equivalent norms; see H. A. Priestley, Functional Analysis I, §§0.5–0.7 and Theorem 4.3.

The metric topology therefore does not determine numerical distances. Multiplying a metric by $100$ preserves its open sets while multiplying every distance by $100$. Nor does the general equivalence of metrics require constant-factor bounds: that stronger comparison holds for the three norms used here.

For a sequence of real numbers, a common finite limit of a sequence for all monotone subsequences forces the whole sequence to have that limit of a sequence. There is no need to assume a bounded sequence if the hypothesis already includes convergence of every monotone subsequence.

If $a_n$ fails to tend to $L$, select a subsequence with $|a_{n_j}-L|\ge \epsilon$ for some fixed $\epsilon >0$. The monotone subsequence theorem gives a monotone subsequence of this subsequence, which contradicts the common limit of a sequence $L$. Conversely, every subsequence of a convergent sequence has the original limit of a sequence.

For a bounded sequence, the bounded monotone sequence theorem already ensures that all monotone subsequences converge. Only agreement of their limits of sequences remains to be checked.

Every convergent sequence in the Stone-Čech compactification of the natural numbers is eventually constant. If distinct ultrafilters $U_n$ converged, isolate each from the other sequence terms using convergence and finitely many separating clopen sets. Finite disjointification gives disjoint $B_n\in U_n$. The union of the even-indexed $B_n$ then belongs exactly to the even-indexed ultrafilters, contradicting convergence of membership in a clopen set.

A subset $\Omega \subseteq {\mathbb{R}}^n$ is an open set if every $x\in \Omega$ admits a radius $r>0$ such that the open ballis contained in $\Omega$. A subset $F\subseteq {\mathbb{R}}^n$ is a closed set if its complement ${\mathbb{R}}^n\setminus F$ is an open set. Equivalently, $F$ contains the limit of every convergent sequence of its points.

If $x\in F$, using $y=x$ in the infimum gives $0\le f(x)\le 0$, so $f(x)=0$.

Conversely, if $f(x)=0$, the definition of the infimum allows a choice $y_j\in F$ with $|x-y_j|<1/j$. The sequence $y_j$ is therefore a convergent sequence with limit $x$. Since $F$ is a closed set, its limit $x$ belongs to $F$. HenceClosedness matters: without it, zero distance to a set characterizes its closure, which can contain additional points.

Suppose for contradiction that $p_n=(u_n,v_n)\in Y$ converges to the origin in the initial topology. The coordinate projections then give $u_n\to 0$ and $v_n\to 0$ in the usual real line. Rectangular bumps detecting nonconvergence, using anisotropic supports, avoids any unsupported regularity assertion about a radial maximum. SetThe radii are positive since neither coordinate of $p_n$ vanishes. Each rectangular support misses both coordinate axes, so $B$ vanishes on those axes. For a fixed $y=c\ne 0$, all but finitely many vertical support intervals miss $c$, because their centres and radii tend to zero. Hence $B(x,c)$ is a finite maximum of continuous functions of $x$, and is continuous. At $c=0$ it is identically zero. The same argument with the coordinates exchanged proves separate continuity everywhere. Thus $B$ is continuous for the initial topology.

However $B(p_n)=1$ for every $n$, whereas $B(0,0)=0$. This contradicts preservation of convergent sequences by a continuous function. Therefore no sequence from $Y$ converges to the origin.

The radial hint in the scan is not separately continuous for every candidate sequence: for $p_n=(2^{-n},2^{-2n})$, its value at the origin is zero, while its value at $(2^{-n},0)$ is at least $1-2(2^{-2n})/\sqrt{2^{-2n}+2^{-4n}}$, which tends to one. The rectangular construction supplies the needed valid proof independently of that hint.

The definition of a convergent sequence isTo prove the product rule, convergence first gives an eventual bound $|u_n|\le |l|+1=:M$. ThenChoose $n$ sufficiently large that $|v_n-k|<ϵ/(2M)$ and $|u_n-l|<ϵ/[2(|k|+1)]$, as well as the bound on $u_n$. The right side is less than $ϵ$, proving $u_n{v}_n\to lk$ directly from the limit of a sequence definition.

If $l\ne 0$, eventually $|u_n-l|<|l|/2$, so the reverse triangle inequality gives $|u_n|>|l|/2$. ConsequentlyThus $1/u_n\to 1/l$. The condition $u_n\ne 0$ ensures the reciprocal sequence is defined at every index; the nonzero limit supplies the eventual uniform bound away from zero. Finally $\boxed{u_n=1/(n+1)\to 0}$ has every term nonzero, showing that this termwise property alone does not prevent a zero limit.

Use the dominated form of the real Hahn-Banach theorem. Let $p:V\to \mathbb{R}$ be a sublinear functional on a real vector space, meaning $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for $t\ge 0$. If a linear functional $f$ on a vector subspace $M$ satisfies $f\le p$ there, then it has a linear extension $F$ to $V$ with $F\le p$ everywhere.

Here is the one-dimensional extension step. For $z\notin M$, the value $a=F(z)$ must satisfyThis interval is nonempty: for any $m,n\in M$,so every candidate lower bound is at most every candidate upper bound. The candidates with $m=n=0$ also bound the endpoints by finite numbers, so a real $a$ can be chosen. Define $F(m+tz)=f(m)+ta$. For $t>0$, the upper bound, applied to $m/t$, gives $F(m+tz)\le p(m+tz)$ after multiplication by $t$. For $t<0$, the lower bound applied to $m/(-t)$ gives the same conclusion after multiplication by $-t$; $t=0$ is the original hypothesis. Thus the extension remains dominated.

Order all dominated extensions by extension of their domains and values. Every chain has the dominated linear union as an upper bound. The Zorn lemma supplies a maximal extension. If its domain were not all of $V$, the step just proved would enlarge it, a contradiction. This proves the real Hahn-Banach extension theorem. In particular, taking $p(x)=\Vert f\Vert \Vert x\Vert$ on a normed vector space and applying the domination to both $x$ and $-x$ yields the usual norm-preserving extension of a bounded linear functional.

For generalized limits, take $V={\ell }^{\infty }(\mathbb{N};\mathbb{R})$, the l-infinity sequence space, and $M=c$, the convergent sequence space. Let $f(x)={\lim }_n{x}_n$ on $c$ and $p(x)={\mathrm{limsup}}_n{x}_n$ on $V$. This $p$ is finite and sublinear, and agrees with $f$ on $c$. The dominated extension $L$ therefore satisfiesThus a generalized limit exists, withIt is positive, agrees with the ordinary limit on every convergent sequence, and satisfies $|L(x)|\le \Vert x{\Vert }_{\infty }$. Since $L(1)=1$, its operator norm is exactly one.

If generalized limits are additionally required to be shift invariant, the same theorem gives the stronger Banach limit. Replace the dominating functional byIt is again sublinear and agrees with $f$ on $c$, by the Cesàro theorem for convergent sequences. The resulting $L_C$ lies between the lower and upper limits of the Cesàro means. These in turn lie between the ordinary lower and upper limits of a bounded sequence, so $L_C$ is a positive, norm-one extension of the ordinary limit. For the left shift on bounded sequences $S$, the means of $Sx-x$ equal $(x_{N+1}-x_1)/N\to 0$. Applying domination to both signs forces $L_C(Sx-x)=0$. HenceThis Cesàro construction of a Banach limit proves existence even under the stronger convention, rather than leaving shift invariance implicit.

For a nonempty set $S\subseteq \mathbb{R}$, a real number $s$ is its supremum if every $x\in S$ satisfies $x\le s$ and every upper bound $u$ of $S$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $x\in S$ has $s-\epsilon <x\le s$. The least upper bound axiom states that every nonempty set of real numbers bounded above has a real supremum.

Suppose $(a_n)$ is nondecreasing and bounded above. Apply the least upper bound axiom to $S=\{a_n:n\ge 1\}$, obtaining $s=\sup S$. For any $\epsilon >0$, $s-\epsilon$ cannot be an upper bound, so there is $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesso $|a_n-s|<\epsilon$. This is precisely the definition of a convergent sequence. If $(a_n)$ is nonincreasing and bounded below, apply the proved case to $(-a_n)$; its limit is the negative of the infimum of the original terms. Thus the bounded monotone sequence theorem follows directly from the axiom:

No. Take $\boxed{a_n=\log n}$. For each fixed positive integer $k$,by continuity of the logarithm at $1$. Yet $a_n\to +\infty$, so $(a_n)$ is not a convergent sequence of real numbers. Vanishing differences at every fixed lag do not control the accumulation of many small increments.