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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].

On a separable metric space, a sequence uniformly bounded in bounded Lipschitz norm has a pointwise convergent subsequence. Enumerate a countable dense subset and use the Bolzano-Weierstrass theorem and diagonal subsequence argument to obtain convergence there. Uniform control of the Lipschitz constant then makes the subsequence Cauchy at every point. The pointwise closure of the bounded Lipschitz unit ball preserves the original norm bound after rescaling.

Every real sequence has a subsequence converging to an extended real number. A bounded sequence has a real convergent subsequence by the Bolzano-Weierstrass theorem. If it is unbounded above, every tail is unbounded above, so choose increasing indices $n_j$ with $a_{n_j}>j$. These terms tend to positive infinity. If it is unbounded below, use $a_{n_j}<-j$.

A real-valued function that is continuous on a nonempty compact space attains both a global minimum and a global maximum. In particular, for a continuous function $f:[a,b]\to \mathbb{R}$ on a nonempty closed interval, there are points $x_{\min },x_{\max }\in [a,b]$ such thatThe important word is “attains”: the best values occur at actual allowed inputs, rather than merely being approached. The interval version is ^[ref].

For a worked example, take $f(x)=(x-1{)}^2$ on $[-1,2]$. Its global minimum is $f(1)=0$, and its global maximum is $f(-1)=4$. Nonnegativity gives the first answer; $|x-1|\le 2$ on the specified closed interval gives the second. The endpoint $x=-1$ is essential: looking only for a zero derivative would find the minimum and miss the maximum.

For comparison, the continuous function $g(x)=x$ on the open interval $(0,1)$ has infimum $0$ and supremum $1$, but neither is a value of $g$. Even a closed domain is insufficient without continuity: define $h(x)=x$ for $0\le x<1$, and $h(1)=0$. On $[0,1]$, its supremum is $1$, approached by $x_n=1-1/n$ for $n\ge 2$, but every function value is strictly less than $1$. This endpoint change makes $h$ discontinuous, preventing attainment. On the unbounded real line, the continuous function $x\mapsto x$ has neither a global minimum nor a global maximum. These examples show what can fail when a hypothesis is dropped; functions outside the hypotheses may still have extrema.

The extreme value theorem is an existence tool for optimization: first establish that a continuous objective function on a nonempty compact feasible set has a best value, then investigate where it occurs. It neither locates the optimizing point nor promises uniqueness or differentiability. For example, the absolute value function $|x|$ on $[-1,2]$ has its global minimum at zero although no derivative exists there; a constant function has every input as a minimizer. This role in optimization is explained in University of Toronto MAT237 notes, section 1.4: compactness and the extreme value theorem.

A short proof outline identifies the mechanism. The continuous image of a compact space is compact; a nonempty compact subset of the real line contains its infimum and supremum. For a closed interval, one can instead choose inputs whose outputs approach the supremum, extract a convergent subsequence by the Bolzano-Weierstrass theorem, and use continuity to make its limiting output equal that supremum. The closed interval retains the limiting input. Repeating the argument for the infimum supplies the minimum.

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]

Let $f$ be continuous on a closed bounded interval $[a,b]$. First prove that $f$ is a bounded function. If it failed, we could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x_{\ast }$. Because the interval is closed, $x_{\ast }\in [a,b]$. Continuity gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j\to \infty$.

Now let $M=\sup \{f(x):x\in [a,b]\}$, which exists by completeness of the real numbers and the bound just proved. Choose $y_n\in [a,b]$ with $f(y_n)>M-1/n$. A convergent subsequence has limit of a sequence $y_{\ast }\in [a,b]$, and continuity gives $f(y_{\ast })=M$. Applying the same argument to $-f$ gives attainment of the infimum. This proves the extreme value theorem: the maximum and minimum both exist and are attained. The argument also covers a one-point interval, where the conclusions are immediate.

Here is a nested-interval proof of the Bolzano-Weierstrass theorem. Put every term of the bounded real sequence in a closed interval $I_0=[-M,M]$, with $M>0$. Bisect this interval and choose a closed half containing infinitely many terms. Repeating gives nested closed intervals $I_j=[l_j,r_j]$, each containing infinitely many sequence terms, with length $2M/2^j$.

By the least-upper-bound property, $x_{\ast }={\sup }_j{l}_j$ exists. For each $j$, every lower endpoint lies at most $r_j$, while $l_j\le x_{\ast }$; hence $x_{\ast }\in I_j$. Choose increasing indices $n_j$ with $a_{n_j}\in I_j$, possible because each interval contains infinitely many terms. ThenThus every bounded real sequence has a convergent subsequence.

For a sequence with no convergent subsequence, take $a_n=n$: any subsequence has $n_j\ge j$, so its terms tend to positive infinity rather than to a finite limit of a sequence. For an unbounded sequence with a convergent subsequence, take $a_{2n}=0$ and $a_{2n-1}=n$. Its odd terms are unbounded, while its even subsequence converges to zero.

We prove the equivalence by a cycle of implications. First assume closedness and boundedness. Given a sequence $(x_j)$ in $E$, boundedness puts each coordinate in a fixed closed bounded real interval. Apply the one-dimensional Bolzano-Weierstrass theorem to its first coordinate to extract a convergent subsequence, then apply it to the second coordinate of that subsequence, and continue through the finitely many coordinates. The final subsequence converges coordinatewise to some $x\in {\mathbb{R}}^n$. It also converges in the Euclidean norm, since the sum of the squares of its finitely many coordinate differences tends to zero. Since $E$ is a closed set, this limit belongs to $E$. Thus closedness and boundedness imply the required subsequence property. No higher-dimensional compactness theorem has been assumed.

A real sequence has limit $a\in \mathbb{R}$ if, for every $\epsilon >0$, there is $N$ such that $|a_n-a|<\epsilon$ for all $n\ge N$. It tends to $+\infty$ if, for every real $M$, eventually $a_n>M$; it tends to $-\infty$ if eventually $a_n<M$ for every real $M$.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a real convergent subsequence. This proves the claim for bounded sequences. If the sequence is unbounded above, every tail remains unbounded above because a finite prefix is bounded. Choose indices $n_j$ increasing with $a_{n_j}>j$. This subsequence tends to $+\infty$. If it is unbounded below, choose $a_{n_j}<-j$ to obtain a subsequence tending to $-\infty$. This proves the extended real subsequence theorem.

For the final example, a bounded sequence with vanishing increments need not converge. TakeThe mean value theorem givesNevertheless, along $n_j=\lfloor (\pi /2+2\pi j{)}^2\rfloor$, the square-root rounding error tends to zero, so $a_{n_j}\to 1$. Along $m_j=\lfloor (3\pi /2+2\pi j{)}^2\rfloor$, it tends to $-1$. These two subsequences have different limits, proving nonconvergence.

To prove boundedness on $[a,b]$, suppose instead that $f$ is unbounded and choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence converging to $x_{\ast }\in [a,b]$. Continuity then gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j$. Hence $f$ is bounded.

Let $M={\sup }_{x\in [a,b]}f(x)$ and $m={\inf }_{x\in [a,b]}f(x)$. Choose $d_n$ with $M-1/n<f(d_n)\le M$. Another application of Bolzano-Weierstrass theorem yields a subsequence tending to $d\in [a,b]$, and continuity gives $f(d)=M$. Applying the same argument to $-f$ gives $c\in [a,b]$ with $f(c)=m$. ThusThis proves the extreme value theorem here, including both attainment assertions.

For $g$, its two limits supply $R>0$ such that $|g(x)|<1$ whenever $|x|>R$. On $[-R,R]$, continuity and the result just proved give a finite bound $K$. Consequently $|g(x)|\le \max \{K,1\}$ on the whole real line.

For the final value assertion, if $c=g(a)$, take $x=a$. Otherwise $0<c<g(a)$. The positive-infinity limit gives a point $b>a$ with $|g(b)|<c$, so $g(b)<c<g(a)$. The intermediate value theorem on $[a,b]$ now gives $x\in (a,b)$ withPositivity of $c$ is essential to this tail argument; a function can tend to zero without ever attaining zero.

For a bounded function, a partition of an interval is $P:0=x_0<x_1<\cdots <x_N=1$. Put $m_j={\inf }_{[x_{j-1},x_j]}f$ and $M_j={\sup }_{[x_{j-1},x_j]}f$. Its lower Darboux sum and upper Darboux sum areThe lower Darboux integral is ${\sup }_{P}L(f,P)$ and the upper Darboux integral is ${\inf }_{P}U(f,P)$. Riemann integrability means that these two real numbers coincide, and their common value is the Riemann integral.

Equivalently, the Riemann integrability criterion says that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. To see equivalence, all lower sums are at most all upper sums: refine any two partitions to their common refinement and use Darboux sum refinement monotonicity. Thus a small gap forces the difference of the two integrals below $\epsilon$. Conversely, if both integrals equal $I$, choose a lower sum above $I-\epsilon /2$ and an upper sum below $I+\epsilon /2$; their common refinement has gap below $\epsilon$.

This also agrees with the tagged Riemann sum definition: there is $I$ such that every tagged sum on a partition of sufficiently small mesh is within any prescribed tolerance of $I$. Each tagged sum lies between the two Darboux sums. From a partition with small gap, an arbitrary small-mesh partition differs from its refinement at only the cells crossing the finitely many original division points; boundedness makes their total contribution tend to zero. Conversely, on a fine partition tags approximating each cell supremum or infimum force both Darboux sums close to the same tagged-sum limit.

Now suppose $f$ is continuous on $[0,1]$. It is bounded by the result of Question 11. It is also uniformly continuous: otherwise there are $x_n,y_n\in [0,1]$ with $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0>0$. The Bolzano-Weierstrass theorem gives a convergent subsequence of $x_n$; the corresponding $y_n$ have the same limit, contradicting continuity. Given $\epsilon >0$, this uniform continuity gives $\delta >0$ such that points less than $\delta$ apart have function values less than $\epsilon /2$ apart. A partition of mesh below $\delta$ then satisfiesHence every continuous function on the interval is Riemann integrable, with the required proof supplied by its oscillation estimates.

The Baire category theorem states that, in a nonempty complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty open set cannot be covered by countably many nowhere dense sets.

For the Uniform boundedness principle, let $X$ be a Banach space, $Y$ a normed vector space, and $(T_{\alpha })$ a family of bounded linear operators for which ${\sup }_{\alpha }\Vert T_{\alpha }x\Vert <\infty$ for every $x\in X$. PutEach $F_n$ is closed, being an intersection of inverse images of closed balls, and $X={\bigcup }_n{F}_n$. By the Baire category theorem, some $F_n$ has interior. Choose a closed ball $\overline{B}(x_0,r)\subseteq F_n$ with $r>0$. For $\Vert h\Vert \le r$, both $x_0$ and $x_0+h$ belong to $F_n$, and linearity gives $\Vert T_{\alpha }h\Vert \le 2n$. Scaling yieldsCompleteness is needed for the domain, not the codomain.

We next prove the Silverman-Toeplitz theorem. Write $c$ for the real convergent sequence space with the supremum norm. It is a Banach space: the l-infinity sequence space is complete by coordinatewise passage to limits of norm-Cauchy sequences, and $c$ is closed in it. Indeed, if $x^{(m)}\to x$ uniformly and each $x^{(m)}$ converges, approximate $x$ within $\epsilon /3$ by one $x^{(m)}$ and make that sequence's tail oscillation smaller than $\epsilon /3$. Then the tail of $x$ is Cauchy and therefore converges in $\mathbb{R}$.

Suppose $A$ is a regular summability matrix. Fix a row $i$ and consider the finite bounded linear functionalsFor every $x\in c$ their values converge, by the definition of regularity, and are therefore bounded. Their operator norms are exactly ${\sum }_{j\le N}|a_{ij}|$: the upper bound is the triangle inequality, and the finite sign sequence $x_j=\mathrm{sgn}(a_{ij})$ for $j\le N$, zero afterwards, attains it. Applying the Uniform boundedness principle to this row shows ${\sum }_j|a_{ij}|<\infty$.

Consequently the full row functional $A_i:c\to \mathbb{R}$ is continuous and has norm ${\sum }_j|a_{ij}|$, using the same finite sign tests and then letting $N\to \infty$. Regularity makes $(A_i(x){)}_i$ convergent, hence bounded, for every $x\in c$. A second application of the Uniform boundedness principle gives a common row bound $M$. Testing regularity on a coordinate vector $e_j$ and on the constant-one sequence gives, respectively, convergence to zero down each column and convergence of row sums to one. Thus the necessary conditions are

Conversely assume these conditions and let $x_j\to L$. The sequence is bounded, so every row sum defining $A_i(x)$ converges absolutely. WriteGiven $\epsilon >0$, choose $N$ such that $|x_j-L|<\epsilon /(M+1)$ for $j>N$. The tail of the first sum has modulus at most $M\epsilon /(M+1)<\epsilon$, uniformly in $i$. Its finite head tends to zero by columnwise convergence, and $L(s_i-1)\to 0$. Since $\epsilon$ is arbitrary, $A_i(x)\to L$. The three conditions are necessary and sufficient for regularity.

For the averaging matrix $B$, every absolute row sum and every row sum equals $1$. A fixed column eventually has entry $1/r\to 0$. Thus $B$ is a regular summability matrix, and its transformed values are the Cesàro means. To exhibit a bounded sequence without a Cesaro limit, put $N_0=0$, $N_m=2^{2^m}$ for $m\ge 1$, and setAt the end of the $m$th block,The averages at even block endpoints approach $1$, while those at odd endpoints approach $-1$. Hence $w$ is bounded and has no $B$-limit.

For an arbitrary regular summability matrix $A$, use a gliding hump argument; this also handles negative entries. Put $N_0=0$ and $i_0=0$. Inductively choose $i_m>i_{m-1}$ so large thatThis is possible by the finite-column convergence and convergence of the row sums to one. Since the selected row is absolutely summable, choose $N_m>N_{m-1}$ withDefine a single bounded sequence on the successive disjoint blocks byusing $\mathrm{sgn}(0)=0$. The row's absolute mass on its own block is greater than $3/4-1/8=5/8$, while all coordinates outside that block contribute a term of modulus less than $1/8$, irrespective of the signs subsequently assigned. ThereforeEven and odd selected rows cannot tend to the same limit. Every row series does converge absolutely because $\Vert y{\Vert }_{\infty }\le 1$. This proves bounded divergence for every regular summability matrix.

The row bound makes $T_A:{\ell }^{\infty }\to {\ell }^{\infty }$, $T_{A}x=(A_i(x){)}_i$, a bounded linear operator with $\Vert T_A\Vert \le M$. The summability domain of a regular matrix isSince $c$ is closed, this is a closed linear subspace. It is proper by the constructed $y$. A proper vector subspace of a normed vector space has empty interior: a ball in the subspace can be translated to a ball about zero, and scaling that ball would put every vector in the subspace. ThusFor a countable family of regular matrices, the Baire category theorem on the Banach space ${\ell }^{\infty }$ givesAny vector in this complement has none of the indicated matrix limits. The matrices in this clause must be regular, as in the preceding discussion; without that implicit qualification even a single zero matrix sums every bounded sequence to zero.

Finally take any bounded sequence $x$. By the Bolzano-Weierstrass theorem it has a subsequence $x_{n_i}\to L$ with $n_i$ strictly increasing. Define the selector matrixIts absolute row sums and row sums are $1$, and each fixed column is eventually zero. The Silverman-Toeplitz theorem makes it regular, while $A_i(x)=x_{n_i}\to L$. Thus every bounded sequence is summable by some regular matrix depending on that sequence. This is subsequence selection as regular matrix summability; it reverses the quantifier order of the impossibility result for a fixed matrix.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms:

Set $h=f-g$, an entire function. By the Bolzano-Weierstrass theorem, a subsequence of the distinct zeros $z_i$ converges to a point $a$ in the closed unit disc. An analytic function has a convergent Taylor series in a neighbourhood of each point. If the first nonzero coefficient of $h$ at $a$ had index $m$, thenwith $k$ analytic and nonzero in a sufficiently small disc. The zero $a$ would then be isolated, contradicting the accumulating distinct zeros. All Taylor coefficients at $a$ vanish, so $h$ vanishes on a disc.

To justify continuation across the whole plane, let $Z$ be the set of points at which $h$ vanishes on a neighbourhood. This set is nonempty and open. It is also closed: if points of $Z$ tend to $b$, continuity of each derivative gives $h^{(m)}(b)=0$ for every $m$, and the Taylor series at $b$ gives a zero neighbourhood. Since $\mathbb{C}$ is a connected space, $Z=\mathbb{C}$. This proves the identity theorem here and gives $\boxed{f\equiv g}$.

Without boundedness, the distinct zeros need not have an accumulation point in the plane. For example,are distinct entire functions agreeing at all the indicated positive integers.

Write $L(f)={\sup }_{x\ne y}|f(x)-f(y)|/d(x,y)$ and $B(f)={\sup }_z|f(z)|$. The quantity in the question is the bounded Lipschitz norm $\Vert f{\Vert }_{\mathrm{BL}}=L(f)+B(f)$. Both terms are nonnegative and absolutely homogeneous, and each satisfies the triangle inequality. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a vector space. If the sum is zero, $B(f)=0$ forces $f=0$. Thus it is a norm.

For the sequence bounded by one, both $L(f_i)\le 1$ and $B(f_i)\le 1$. Given real $x$, choose a nearby rational $q$. ThenFirst make $|x-q|$ small, then use convergence at $q$ to make the last term small. The values at every real $x$ form a Cauchy sequence, so $f_i$ converges pointwise to a real function $f$.

To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple $x\ne y,z$:Pass to the pointwise limit and take the independent suprema over $(x,y)$ and $z$. This gives $\boxed{\Vert f{\Vert }_{\mathrm{BL}}\le 1}$, the pointwise closure of the bounded Lipschitz unit ball.

To obtain diagonal compactness for bounded Lipschitz functions, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in $[-1,1]$, so the Bolzano-Weierstrass theorem supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The diagonal subsequence argument gives one subsequence converging at every rational. Applying the preceding argument proves there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one.

The Bolzano-Weierstrass theorem states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence. More generally, a bounded sequence in ${\mathbb{R}}^d$ has a subsequence converging to a point of ${\mathbb{R}}^d$.

Let $E_{\ast }={\inf }_{\mathrm{deg}q\le n}\Vert f-q{\Vert }_{\infty }$, and choose a minimizing sequence $p_m$ with $\Vert f-p_m{\Vert }_{\infty }\to E_{\ast }$. Since $E_{\ast }\le \Vert f{\Vert }_{\infty }$, discarding finitely many terms ensures $\Vert p_m{\Vert }_{\infty }\le 2\Vert f{\Vert }_{\infty }+1$.

Fix $n+1$ distinct nodes $a_0,\dots ,a_n$. The Lagrange interpolation formula expressesThe vector of node values is bounded in ${\mathbb{R}}^{n+1}$, so the Bolzano-Weierstrass theorem supplies a subsequence converging to $(v_0,\dots ,v_n)$. Put $p=\sum v_j{\ell }_j$, of degree at most $n$. Because the ${\ell }_j$ are fixed bounded functions,along that subsequence. The reverse triangle inequality implies $\Vert f-p{\Vert }_{\infty }=E_{\ast }$. Thus a best uniform approximation exists, with no appeal to an unproved abstract existence result.

Here ${\mathcal{P}}_n$ is the vector space of real polynomials of degree at most $n$.

Two Lipschitz equivalent norms $N_1,N_2$ satisfy $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for all $v$, for constants $0<c\le C<\infty$. These inequalities give the same convergent sequences in both norms. A subset of a metric space is closed exactly when it contains the limits of all convergent sequences from that subset. Therefore closedness is the same in the two norms.

For finite dimension, fix a basis $e_1,\dots ,e_d$ and let $x={\sum }_j{x}_j{e}_j$. For any norm $N$,The reverse triangle inequality then makes $N$ continuous in the Euclidean coordinate norm. If no positive lower bound existed on the Euclidean unit sphere, choose $x_m$ there with $N(x_m)\to 0$. By the Bolzano-Weierstrass theorem, a subsequence converges to $x$ on the unit sphere. Continuity would give $N(x)=0$, although $x\ne 0$, a contradiction. Thus $N$ is bounded below by a positive multiple of the Euclidean norm. Comparing any two norms through that coordinate norm proves all norms in finite dimension are Lipschitz equivalent.

On $C[0,1]$, nonnegativity, absolute homogeneity and the triangle inequality for $\Vert f{\Vert }_1={\int }_0^1|f|$ follow from the corresponding real-number properties and integration. If a continuous $f$ is nonzero at a point, it has absolute value bounded below by a positive constant on a nontrivial one-sided or two-sided interval around that point. Its integral is then positive. Hence $\Vert f{\Vert }_1=0$ implies $f=0$, completing the norm axioms.

For $n\ge 2$, put $g_n(x)=\max \{1-n|x-1/2|,0\}$ and $f_n=1-g_n$. Then $f_n(1/2)=0$, butThe limit function $1$ is not in the indicated point evaluation kernel. Consequently that set is not closed in the integral norm. This is the point-evaluation kernel is not closed in the integral norm phenomenon.

Finally $\Vert g_n{\Vert }_{\infty }=1$ while $\Vert g_n{\Vert }_1=1/n$. Thus no constant can bound the uniform norm by a multiple of the integral norm. Although $\Vert f{\Vert }_1\le \Vert f{\Vert }_{\infty }$ on this unit interval, the converse comparison fails: the two norms are not Lipschitz equivalent. The finite-dimensional theorem cannot be applied to this infinite-dimensional function space.

True. Boundedness follows from $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$. To prove closedness of the Minkowski sum, let $z_j=x_j+y_j\in X+Y$ converge to $z$. By the Bolzano-Weierstrass theorem, the bounded sequence $x_j$ has a subsequence converging to $x$. Closedness gives $x\in X$. Along that subsequence, $y_j=z_j-x_j\to z-x$, which belongs to $Y$ by closedness. Thus $z=x+(z-x)\in X+Y$. This proves that the sum is bounded and closed.

We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.