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Define the radius of convergence of the power series bywith $R=\infty$ when this set is unbounded. It is nonempty because the power series converges at zero.

If $|z|<R$, the definition of supremum supplies a point $w$ of convergence with $|z|<|w|$. The terms $a_n{w}^n$ of a convergent series tend to zero and are bounded, say $|a_n{w}^n|\le M$. ThereforeThe right-hand side has a convergent geometric series, since $|z|/|w|<1$. The comparison test proves absolute convergence at $z$.

If $|z|>R$ and the power series converged at $z$, then $|z|$ would belong to the set whose supremum is $R$, an immediate contradiction. ThusNo assertion about the boundary $|z|=R$ follows from this argument. If $R=0$ the first assertion is vacuous; if $R=\infty$ the second is vacuous.

For $x_0>a$, defineThis nonempty set is bounded above by $f(x_0)$, so $l\le f(x_0)$. Given $\epsilon >0$, the supremum property supplies $s<x_0$ with $l-\epsilon <f(s)\le l$. Whenever $s<t<x_0$, monotonicity givesThis proves the left-hand one-sided limit $f^{-}(x_0)=l$.

For $x_0<b$, defineHere $r\ge f(x_0)$. Choose $u>x_0$ with $r\le f(u)<r+\epsilon$, using the infimum property. For $x_0<t<u$, monotonicity gives $r\le f(t)\le f(u)<r+\epsilon$, proving the right-hand one-sided limit $f^{+}(x_0)=r$. Thuswhere the relevant one-sided limits exist, including the indicated single one-sided limit at each endpoint.

At an interior point, continuity implies both one-sided limits equal $f(x_0)$. Conversely, equality $f^{-}(x_0)=f^{+}(x_0)$ and the displayed inequalities force both to equal $f(x_0)$, giving continuity. Therefore

True. Write $\alpha =\sup A$ and $\beta =\inf B$. Every $a\in A$, $b\in B$ satisfies $a-b\le \alpha -\beta$, giving an upper bound for the set $A-B$ of pairwise differences. For every $\epsilon >0$, the definitions of supremum and infimum give $a\in A$ and $b\in B$ withThen $a-b>\alpha -\beta -\epsilon$. Hence no smaller upper bound works, andHere $A-B$ means pairwise subtraction, rather than set difference.

The extreme value theorem gives finite real numbersSince these values are attained, the set of strict upper bounds in the definition of $M$ is precisely $(\beta ,\infty )$, and the set of strict lower bounds in the definition of $m$ is precisely $(-\infty ,\alpha )$. They are nonempty and bounded on the required sides. By the definitions of infimum and supremum,

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 $\boxed{a_n\to s}$.

Choose $M$ with $|b_n|\le M$. Every tail of a sequence is nonempty and bounded, so its supremum $c_n$ exists and lies in $[-M,M]$. Removing one term cannot increase a supremum, hence $c_{n+1}\le c_n$. The bounded monotone sequence theorem, applied to $-c_n$, provesThus $L$ is the limit superior of $(b_n)$.

For $\epsilon >0$, eventually $c_n<L+\epsilon$, and then $b_n\le c_n<L+\epsilon$. Only finitely many indices can satisfy $b_n\ge L+\epsilon$. If only finitely many satisfied $b_n>L-\epsilon$, some entire tail of a sequence would have $b_n\le L-\epsilon$. Its supremum would also be at most $L-\epsilon$, contradicting $c_n\ge L$.

Inductively choose $n_k>n_{k-1}$ so late that $c_{n_k}<L+1/k$ and $b_{n_k}>L-1/k$. There are infinitely many choices for the latter inequality, so both requirements can be met. Then $|b_{n_k}-L|<1/k$, proving $\boxed{b_{n_k}\to L}$. This constructs a subsequence attaining the limit superior.

Take $f_n(x)=x^n$ on $[0,1]$. Each function is continuous, and the pointwise limit is $f(x)=0$ for $0\le x<1$, while $f(1)=1$. NeverthelessThe limiting function is Riemann integrable, since changing a value at one point does not change its Riemann integral. There is no uniform convergence: for every $n$, ${\sup }_{[0,1]}|f_n-f|=1$, with the supremum approached as $x\uparrow 1$ from below. Equivalently, the uniform limit theorem would require the limit to be continuous.

Write $q=P(B\mid A)$, $s=P(B\mid A^c)$ and $p=P(A)$. The law of total probability gives $P(B)=pq+(1-p)s$, a convex combination of the two fixed conditional values. HenceThese are the closed sharp bounds in the usual mixture-model interpretation, obtained by allowing $p$ to range from zero to one. The joint masses $pq,p(1-q),(1-p)s,(1-p)(1-s)$ realize every such mixture.

There is an endpoint qualification if both conditionals must literally be defined by ratios: then $0<p<1$. If $q\ne s$, the marginal lies strictly between them, so the stated endpoints are an infimum and supremum rather than achieved extrema. If $q=s$, every allowed mixture has exactly that common value. This is the second case of sharp bounds from binary conditional probabilities.

We can arrange that neither the supremum nor the infimum is attained. By part (d), choose a continuous function $f$ on $X$ with unbounded range, and put $u=|f|$. Choose $x_n\in X$ so that $t_n=u(x_n)$ increases strictly to infinity. Construct a continuous function $h:[0,\infty )\to \mathbb{R}$ withkeep it constant up to $t_1$ and interpolate linearly on each $[t_n,t_{n+1}]$. Since $t_n\to \infty$, the joining points have no finite accumulation, so this defines a continuous function everywhere on its domain.

Now setBy function composition, $g$ is continuous, and $-1<g(x)<1$ for every $x$. Along the even and odd chosen points,Thus ${\sup }_{X}g=1$ and ${\inf }_{X}g=-1$, and neither is a value of $g$. This proves the bounded continuous witness with neither extremum attained construction, including for subsets that are bounded but not closed.

For $0\le t\le 1$, the triangle inequality givesTaking the supremum proves $\boxed{d(p_n,p)\le \rho (p_n,p)}$. Hence convergence for $\rho$ implies convergence for $d$.

The converse is false. For $n\ge 1$, take $q_n(t)=(1-t{)}^n/n$ and $p=0$. Its supremum norm is $d(q_n,0)=1/n\to 0$, whereas the binomial theorem givesThe alternating coefficients cancel on $[0,1]$; the polynomial coefficient sum norm measures their absolute values before any cancellation.

Assume condition (1). For fixed $a$, the supremumis nondecreasing as $r$ increases. Therefore its limit as $r\downarrow 0$ exists in the extended real numbers and equals ${\inf }_{r>0}{M}_a(r)$. Because $a\in B(a,r)$, it is at least $f(a)$. Condition (1) says precisely that this limit is $f(a)$.

It follows that for every $\epsilon >0$ there is $r>0$ with $M_a(r)<f(a)+\epsilon$. Fix $C\in \mathbb{R}$ and $a$ with $f(a)<C$. Taking $\epsilon =C-f(a)$ produces an open ball on which $f(x)<C$. Thus $\{x:f(x)<C\}$ is an open set, and its complementThis proves (1) implies (2). The property being characterized is upper semicontinuity.

Assume condition (2). Fix $a\in X$ and $\epsilon >0$. The superlevel setis a closed set and does not contain $a$. Its complement is therefore an open neighbourhood of $a$, so some open ball $B(a,r)$ avoids $F$. Consequently $f(x)<f(a)+\epsilon$ throughout that ball, givingThe same bound holds for every smaller radius. Taking $r\downarrow 0$ and then $\epsilon \downarrow 0$ yieldsTogether with the previous implication this proves the equivalence. A supremum may equal $f(a)+\epsilon$ even when every individual value is strictly smaller; the non-strict estimate here handles that possibility.

Write $h(A,B)={\sup }_{a\in A}\mathrm{dist}(a,B)$, so that $\rho (A,B)=\max \{h(A,B),h(B,A)\}$ is the Hausdorff distance. Since $X$ is a bounded set, there is a finite $M$ with $d(x,y)\le M$. Nonemptiness of $A,B$ therefore gives $0\le h(A,B)\le M$, so $\rho$ is finite. Its symmetry and nonnegativity are immediate, as is $\rho (A,A)=0$.

If $\rho (A,B)=0$, then every $a\in A$ has zero distance to a set $B$. Since $B$ is a closed set, part (a)(iii) gives $A\subseteq B$. Interchanging the sets gives $B\subseteq A$. Thus $\rho (A,B)=0$ implies $A=B$. Closedness is essential here: without it a set and its closure have Hausdorff distance zero.

For the triangle inequality, fix $a\in A$ and $ϵ>0$. The approximation property of an infimum supplies $b\in B$ with $d(a,b)<\mathrm{dist}(a,B)+ϵ$, and then $c\in C$ with $d(b,c)<\mathrm{dist}(b,C)+ϵ$. Consequently,Let $ϵ\downarrow 0$ and take the supremum over $a$ to obtain $h(A,C)\le h(A,B)+h(B,C)$. Applying this inequality in the reverse direction as well yieldsAll metric axioms follow. Boundedness was used to ensure finiteness, and closedness to distinguish different sets; neither was needed for the triangle inequality.

Conservation laws. Use natural units $c=1$ and write $p_i^{\mu }=(E_i,{\mathbf{p}}_i)$ with the $+---$ metric. In the parent's rest frame, which is the final-state centre-of-momentum frame, $P^{\mu }=(M,0)$. Four-momentum conservation and the relativistic energy-momentum relation giveA real massless daughter has strictly positive energy. The kinematic threshold is $M\ge {\sum }_i{m}_i$; below, no decay satisfying these conditions exists. The endpoint discussion assumes open phase space, $M>{\sum }_i{m}_i$, to avoid collapsed threshold configurations.

Two-body energies. Momentum conservation requires ${\mathbf{p}}_2=-{\mathbf{p}}_1$, so both daughters have the same momentum magnitude $p$. Their on-shell relations implyand subtraction eliminates $p$:Since $E_1+E_2=M$, solving the sum and difference equations gives the relativistic two-body decay resultAs a check, their common momentum magnitude iswhere $\kappa$ is the Källén function. It vanishes at the massive two-body threshold and is real on the allowed side. If the symbols $M,m_i$ denote masses in kilograms rather than rest energies, multiply the displayed energy formulas by $c^2$ and the displayed momentum formula by $c$.

Minimum energy in three-body decay. The on-shell inequality already gives $E_1\ge m_1$. For $m_1>0$, the bound is attainable by putting daughter 1 at rest and letting daughters 2 and 3 recoil back to back:The last equation has a solution because $M-m_1>m_2+m_3$; its left side increases continuously from $m_2+m_3$ to infinity. ThusIf $m_1=0$, the same formula describes the lower boundary $E_1\to 0^{+}$. It is an infimum, not an attained physical zero-energy massless particle: this is a soft-particle endpoint.

Maximum energy from the recoil mass. Group daughters 2 and 3 into a recoil system with four-momentum $Q=p_2+p_3=P-p_1$ and squared invariant mass $s_{23}=Q^2$. Directly in the parent rest frame,Maximizing $E_1$ therefore means minimizing the recoil invariant mass. This approach does not assume a recoil rest frame when that system is null.

If both recoil daughters have positive masses, use their pair rest frame. Their total energy there is at least $m_2+m_3$, with equality when both are at rest relative to one another. Equivalently, their future-directed four-velocities satisfy $p_2\cdot p_3\ge m_2{m}_3$, givingEquality means the two recoil daughters have a common velocity in the parent frame: their momenta are parallel and ${\mathbf{p}}_2/m_2={\mathbf{p}}_3/m_3$. Daughter 1 recoils in the opposite direction. HenceAt this endpoint their total energy and its sharing areThese explicitly exhibit a permitted configuration attaining the bound when both recoil masses are positive.

If exactly one recoil daughter is massless, say $m_3=0$ and $m_2>0$, work in the rest frame of daughter 2. For any positive photon-like daughter energy there,The lower recoil-mass bound is approached only as $E_3^{(2)}\to 0^{+}$. Thus the same general formula with $m_3=0$ is the supremum of $E_1$ for a genuine three-body event, rather than an attained maximum. Interchanging 2 and 3 covers the other mixed case. The distinction is important because the paper's category “at least one has mass” includes this soft-particle endpoint as well as the attainable two-massive-particle case.

If both recoil daughters are massless, instead computeIt can be exactly zero at nonzero $E_2,E_3$ when their momenta are parallel. Then $|\mathbf{Q}|=E_2+E_3$, and momentum conservation makes $|{\mathbf{p}}_1|=|\mathbf{Q}|$. Solving $E_1+|{\mathbf{p}}_1|=M$ and $E_1^2-|{\mathbf{p}}_1{|}^2=m_1^2$ givesBoth recoil daughters can have positive energies: set $E_2=x|{\mathbf{p}}_1|$ and $E_3=(1-x)|{\mathbf{p}}_1|$ for any $0<x<1$. This endpoint is attained without a soft daughter, even though their combined four-momentum is null and has no rest frame.

Combining the three-body decay energy endpoints, the closed kinematic range isFor masses in ordinary units, multiply both endpoint expressions by $c^2$. Open endpoints occur precisely in the soft limits described above. These bounds specify kinematic possibilities; the decay amplitude can still vanish at a boundary, so a possible endpoint need not have a nonzero event density.

For a launch angle $\theta$, the projectile motion equations give a flight time $2u\sin \theta /g$, a projectile range $R=u^2\sin (2\theta )/g$, and an apex height $u^2{\sin }^2\theta /(2g)$. Thus the ceiling requires ${\sin }^2\theta \le 2gH/u^2$.

Set $z={\sin }^2\theta$. Then $R^2=4u^{4}z(1-z)/g^2$, which increases up to $z=1/2$. The optimum is therefore $z=\min (1/2,2gH/u^2)$, giving the projectile range under a ceilingThe printed expression is valid in the first regime. A higher ceiling leaves the usual $45^{\circ }$ launch admissible, so it cannot reduce the maximum range. The angle in the ceiling-limited regime is $\arcsin \sqrt{2gH/u^2}$. Equality at the roof is allowed in this ideal geometrical model; if touching the roof is forbidden, the same value is a supremum approached from below.

Take downward distance $s$ and downward speed $v=ds/dt\ge 0$, with constant gravitational acceleration $g>0$ and quadratic drag coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent drag force is ${\mathbf{F}}_d=-mk|\mathbf{v}|\mathbf{v}$. For this downward fall, Newton's second law becomesThe terminal velocity is obtained by setting the net force to zero:From rest, the speed increases monotonically towards this limiting value. In fact $v(t)=v_{\infty }\tanh (\sqrt{gk}t)$; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule $dv/dt=vdv/ds$. Writing $w=v^2$ gives a linear ordinary differential equation:Consequently,Here $k$ has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

Orient the $x$ axis along the displacement from the earlier event to the later one, so $\Delta x=d=1000\mathrm{m}$ and $\Delta t=10^{-6}\mathrm{s}$. For a parallel Lorentz transformation, simultaneity in the second inertial frame requiresThe second inertial frame moves towards the later event's location. The positive value $d^2-c^2\Delta t^2$ establishes spacelike separation, which permits this change of simultaneity. For a general velocity $\mathbf{V}$, the condition is $\mathbf{V}\cdot \Delta \mathbf{r}=c^2\Delta t$: only its component along the separation is fixed at the displayed value, and transverse components are allowed provided $|\mathbf{V}|<c$. Thus the parallel velocity has the smallest possible speed.

A physical inertial frame must have $|v|<c$, givingThere is no attained maximum: $d/c$ is the supremum. For the concrete boundary case $\Delta t=1000\mathrm{m}/c$, the events have null separation and the formula would require $v=c$; no inertial frame can make them simultaneous.

For steady currents in vacuum, Ampère's law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enclosed}}$. A rectangular loop with one long side inside an ideal long solenoid and the other outside gives $B\ell ={\mu }_0(n\ell )I$, because the exterior field is negligible and the perpendicular sides contribute zero. Thus $B={\mu }_{0}nI$, approximately uniform away from the ends.

Write the electron electric charge as $q_e=-e$, with $e>0$. The Lorentz force is $\mathbf{F}=-e\mathbf{v}\times \mathbf{B}$; it does no work. For perpendicular motion, $m{v}^2/r_L=evB$ givesHere ${\omega }_c$ is the positive angular speed; the rotation sense is determined by the negative electron electric charge.

The parallel velocity is unaffected, while the perpendicular component rotates. Hence the path is a helix with $v_{\parallel }=v\cos \theta$ and $r_L=mv\sin \theta /(eB)$. For on-axis entry into a solenoidal tube, the transverse circle passes through that axis: its guiding centre is displaced by $r_L$, and its maximum distance from the tube axis is $2r_L$, not $r_L$.

Using the nonrelativistic approximation appropriate to $1\mathrm{keV}$,The cyclotron period is $2\pi /{\omega }_c=0.893\mathrm{ns}$, so the flight lasts $4.783$ revolutions. The electron therefore crosses the axis again four times before leaving. At $3^{\circ }$, $2r_L=0.279\mathrm{mm}<a$. Wall clearance requires $2mv\sin \theta /(eB)\le a$, givingThe transit is longer than a full orbit, so the maximum transverse excursion really is sampled before exit.

For the solenoidal time-of-flight energy window, the prescribed time fixes $v_{\parallel }=l/t_0=v_0=\sqrt{2E_0/m}$, while $0\le v_{\perp }\le eBa/(2m)$. ThereforeThese are total kinetic energies. The upper value is the ideal grazing limit; if touching the wall counts as rejection, it is a supremum approached from below.

A normal mode is a motion in which all generalized coordinates oscillate at one frequency with fixed amplitude ratios. Small oscillations about stable equilibrium in a time-independent conservative system give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$, with positive mass matrix and symmetric stiffness matrix, allowing a generalized normal-mode eigenproblem.

For the three-to-one mass double-pendulum modes, the angles are absolute displacements from vertical, as printed. To quadratic order,For $\mathbf{\theta }=\mathbf{A}\cos \omega t$ and $q={\omega }^2\ell /g$, the determinant isThusThe low mode has the strings leaning together, with the lower angle twice as large; the high mode has opposite angular signs. Both reverse together after half their respective periods.

For the specified release, decompose $(\alpha ,\alpha )=(3\alpha /4)(1,2)+(\alpha /4)(1,-2)$. Thereforeand the lower mass's horizontal displacement isThis is the supremum of two incommensurate harmonic modes, conventionally described as the maximum-amplitude estimate. Strictly, ${\omega }_{+}/{\omega }_{-}=\sqrt{3}$ is irrational, so the opposite cosine extrema are approached arbitrarily closely but never coincide at a finite time. The quoted value is a supremum of the ideal linearized motion.

For a signed measure define its total variation measure byThe value at the empty set is zero. For disjoint measurable sets $A_j$ with union $A$, any finite partition $E_r$ of $A$ givesTaking the supremum proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus {\bigcup }_{j\le k}{A}_j$ givesIf one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\epsilon \downarrow 0$ and $k\to \infty$ proves countable additivity, including infinite values. Thus $|\mu |$ is a positive measure. This uses only countable additivity of $\mu$, without assuming a Hahn decomposition.