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The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

Convergence at $z_0$ implies that the terms $a_n{z}_0^n$ form a bounded sequence: $|a_n{z}_0^n|\le M$ for some $M$. If $|z|<|z_0|$ and $r=|z/z_0|<1$, thenComparison with the convergent geometric series $\sum M{r}^n$ proves absolute convergence. The case $z_0=0$ is vacuous.

The radius of convergence is the number $R\in [0,\infty ]$ such that a power series converges absolutely for $|z|<R$ and diverges for $|z|>R$, with boundary behavior considered separately.

TakeThen $\sum v^n/n$ is the convergent alternating harmonic series, whereas $\sum w^n/n$ is the divergent harmonic series. Thus $\sum z^n/n$ converges at a point of modulus one and diverges at another point of modulus one. It follows that

The radius of convergence of the power series ${\sum }_{n\ge 0}{a}_n{z}^n$ isIf $|z|<R$, choose $w$ with $|z|<|w|<R$. Since $a_n{w}^n\to 0$, those terms are bounded, say $|a_n{w}^n|\le M$. Thereforeand comparison with a geometric series proves absolute convergence. If $|z|>R$, convergence would imply absolute convergence at every smaller modulus by the same argument, contradicting the definition of $R$.

If $|a_n|\le |b_n|$, convergence of $\sum b_n{w}^n$ implies absolute convergence of $\sum a_n{z}^n$ whenever $|z|<|w|$. Letting $|w|$ approach the radius of the $b$-series shows that $R_a\ge R_b$.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

Condition (i) is insufficient because the allowed limit may be zero. For example, $a_n=1/n$ is nonzero and converges to zero, whereas $1/a_n=n$ does not converge in $\mathbb{R}$.

Choose a closed interval containing the bounded sequence. Bisect it and retain a half containing infinitely many terms; repeat. The resulting intervals are nested and their lengths tend to zero. By the nested interval theorem they have one common point $x$, and choosing successively indexed terms from these intervals produces a subsequence converging to $x$. This proves the Bolzano-Weierstrass theorem.

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

It is useful to prove the contrapositive. Suppose $k$ is a peak of a sequence $(a_n)$ and $m$ is a peak of a sequence $(-a_n)$. The second condition says $a_j\ge a_m$ for every $j>m$. HenceOnly finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so $(a_n)$ is a bounded sequence. Consequently,

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

If $a_n\to L$, every subsequence also has limit of a sequence $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every monotone subsequence.

Conversely, suppose $(a_n)$ is a bounded sequence and all its monotone subsequences converge to the same $L$. At least one exists by the monotone subsequence theorem. If $a_n$ did not tend to $L$, there would be ${\epsilon }_0>0$ and a subsequence $(a_{n_j})$ such thatApply the monotone subsequence theorem to $(a_{n_j})$. The resulting further subsequence is also a monotone subsequence of $(a_n)$, so it tends to $L$, contradicting the displayed bound. ThereforeThe bounded monotone sequence theorem ensures that these subsequences have finite limits of sequences; the hypothesis requires those values to agree.

Suppose $a_n\to \ell$. With tolerance $1$, choose $N$ such that $|a_n-\ell |<1$ whenever $n\ge N$. The triangle inequality gives $|a_n|\le |a_n-\ell |+|\ell |<1+|\ell |$ for these terms. SetAn empty sum is zero. This finite $B$ satisfies $|a_n|\le B$ for every $n$, so every convergent sequence is a bounded sequence.

False. Choose the positive sequenceIt has limit of a sequence one, but Bernoulli's inequality givesThus $(a_n^n)$ is not a bounded sequence.

If $z_0=0$, there is no $z$ with $|z|<|z_0|$, so the claim is vacuous. Suppose $z_0\ne 0$. Convergence of $\sum a_n{z}_0^n$ implies $a_n{z}_0^n\to 0$, because terms of a convergent series tend to zero. In particular this is a bounded sequence, so $|a_n{z}_0^n|\le C$ for some finite $C$ and every $n$.

For $|z|<|z_0|$, put $q=|z|/|z_0|<1$. ThenThe geometric series and comparison test for series proveThus convergence at one point gives absolute convergence at every point strictly closer to the centre.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. It is enough to select infinitely many terms of the tangent sequence that stay uniformly bounded.

For each positive integer $k$, choose $n_k=\lfloor k\pi +1/2\rfloor$, using the floor function. Then $|n_k-k\pi |\le 1/2$. Moreover $n_{k+1}-n_k\ge \pi -1>0$, so these are strictly increasing indices. By the period and monotonicity of the tangent on $(-\pi /2,\pi /2)$,Thus $(\tan n_k)$ is a bounded subsequence. Applying the Bolzano-Weierstrass theorem to it gives a convergent subsequence, which is also a subsequence of the original sequence. Such a convergent subsequence therefore exists.

Assume first that the original nonnegative series converges to $S$. Its partial sums $s_m$ increase to $S$. Write $t_N={\sum }_{n=1}^N{a}_{g(n)}$ for the partial sums of the rearrangement of a series. Since $g$ is an injection, the indices $g(1),\dots ,g(N)$ are distinct, soThus $(t_N)$ is an increasing bounded sequence, and the bounded monotone sequence theorem gives $t_N\to T\le S$.

For any fixed $m$, surjectivity of $g$ givesFor $N\ge N_m$, nonnegativity implies $t_N\ge s_m$. Taking limits of sequences first in $N$ and then in $m$ gives $T\ge S$. Hence $T=S$.

Conversely, $g^{-1}$ is a bijection, and the original series is the rearrangement of a series of $\sum b_n$ defined by $g^{-1}$. The same argument proves its convergence whenever $\sum b_n$ converges. Both series converge together, and their sums are equal.

For a nonempty set $A\subseteq \mathbb{R}$, a supremum is a real number $s$ such that $a\le s$ for every $a\in A$, and every upper bound $u$ of $A$ satisfies $s\le u$. Equivalently, $s$ is an upper bound and, for every $\epsilon >0$, some $a\in A$ satisfies $a>s-\epsilon$. By completeness of the real numbers, it exists whenever $A$ is nonempty and bounded above.

An increasing monotone sequence satisfies $a_{n+1}\ge a_n$ for every $n$; a bounded sequence satisfies $|a_n|\le M$ for some finite $M$ and every $n$. Put $s=\sup \{a_n:n\ge 1\}$. Given $\epsilon >0$, the defining property of the supremum supplies $N$ with $a_N>s-\epsilon$. For every $n\ge N$, monotonicity givesThis is exactly the definition of convergent sequence, so the bounded monotone sequence theorem gives $\boxed{a_n\to s}$.

A peak of a sequence of real numbers is an index $k$ whose value is at least every later value. This is a condition on the entire tail, stronger than comparing only adjacent terms. Equality is allowed: every index of a constant sequence is a peak of a sequence.

Infinitely many peaks of a sequence, listed in increasing order of index, give a nonincreasing monotone subsequence. If only finitely many exist, every index beyond the last has a larger later term; repeatedly choosing one gives a strictly increasing monotone subsequence. This proves the monotone subsequence theorem.

A peak of a sequence of $(-a_n)$ supplies a lower bound for all subsequent terms of $(a_n)$. If both $(a_n)$ and $(-a_n)$ have a peak of a sequence, the two selected values bound a common tail. Adding bounds for the finite initial segment proves that $(a_n)$ is a bounded sequence.

Selecting a subsequence is a regular matrix summability method: every row has one entry equal to one, every fixed column eventually vanishes, and all row sums equal one. Every bounded sequence of real or complex numbers has a convergent subsequence by the Bolzano-Weierstrass theorem, so each bounded sequence is summable by some sequence-dependent regular method. The order of the quantifiers matters: no single regular matrix sums all bounded sequences.

A subsequential limit of a sequence $(a_n)$ is a value $L$ approached by some subsequence $(a_{n_k})$ with strictly increasing indices $n_k$. For $a_n=(-1{)}^n(1+1/n)$, the subsequences at even integers and odd integers give subsequential limits one and minus one. A finite value appearing among only finitely many initial terms does not become a subsequential limit merely by being a member of the image of a function represented by the sequence.

For a bounded sequence of real numbers, its subsequential limits form a nonempty bounded set that is also a closed set. The Bolzano-Weierstrass theorem guarantees nonemptiness, and its extreme values are the limit inferior and limit superior. The sequence in sparse peaks can determine a limit superior has the two subsequential limits zero and one, despite almost all indices in the natural density sense carrying zero.

Every weakly convergent sequence in a normed vector space is a bounded sequence. Apply the Uniform boundedness principle on the Banach space $X^{\ast }$ to the evaluation maps $J{x}_n:f\mapsto f(x_n)$. They are pointwise bounded, while the canonical embedding into the bidual gives $\Vert J{x}_n\Vert =\Vert x_n\Vert$.