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Fix $\mathrm{Im}\lambda >-c$. Truncate $C_1$ at $\pm R-ic$ and close it with the upper semicircle centred at $-ic$, oriented counterclockwise. For large $R$ the only pole of the integrand inside is $z=\lambda$. Uniformly on the arc, $|z-\lambda |$ is comparable to $R$ and $f(z)=o(R^{-1})$, so the arc integral is $o(R^{-1})$ and tends to zero. The line integral is absolutely convergent at infinity for the same decay reason. The Cauchy integral formula givesThe right-hand side is holomorphic on this open half-plane by the hypothesis, proving the required holomorphic function. The contour never passes through $\lambda$ because its imaginary part is strictly above the line.

Let $a,b\in {\mathbb{R}}^2\setminus T$. If $a=b$, use the constant path. Otherwise consider the countably many lines through $a$ and a point of $T$, and through $b$ and a point of $T$. Choose a line $L$ whose direction is not parallel to any of these lines; such a direction exists because there are uncountably many directions but only a countable set of forbidden ones. Every forbidden line meets $L$ in at most one point. Choose $c\in L$ outside those countably many intersections and outside $\{a,b\}$.

Neither segment $[a,c]$ nor $[c,b]$ meets $T$: a forbidden point on either segment would put $c$ on the corresponding forbidden line. The piecewise linear pathlies in the complement. This proves that the plane minus a countable set is path connected.

If a homeomorphism $h:{\mathbb{R}}^2\to \mathbb{R}$ existed, its restriction after removing a point $a$ would give a homeomorphism between ${\mathbb{R}}^2\setminus \{a\}$ and $\mathbb{R}\setminus \{h(a)\}$. The first is a path-connected space and therefore connected; the second consists of two separated intervals. Since homeomorphisms preserve connectedness, $\boxed{{\mathbb{R}}^2\ncong \mathbb{R}}$.

If a nonempty subset $S\subset \mathbb{R}$ is not an interval, there are $a,b\in S$ and $c\notin S$ with $a<c<b$. The disjoint nonempty relatively open sets $S\cap (-\infty ,c)$ and $S\cap (c,\infty )$ form a separation, so $S$ is not connected.

Conversely, suppose an interval $I$ had a separation into relatively open nonempty sets $A,B$. Choose $a\in A,b\in B$, interchanging the names if necessary so $a<b$. Put $c=\sup (A\cap [a,b])$. Because $B$ contains a relative neighbourhood of $b$, $c<b$; because $A$ contains a relative neighbourhood of $a$, $c>a$. Also $c\in [a,b]\subset I$. If $c\in A$, relative openness supplies points of $A$ just above $c$, contradicting the supremum. If $c\in B$, a relative neighbourhood of $c$ excludes points of $A$ approaching $c$ from below, again contradicting the supremum. Thus every interval is connected, including a singleton and unbounded or open intervals.

TakeIt is closed, because its only accumulation point is $0$ and that point is included, and bounded. The Heine-Borel theorem makes it compact. The connected components of its complement are preciselyEach is an interval and therefore connected; none can be enlarged inside the complement across one of the intervening points of $K$.

True. Define $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one. The restriction of $f$ to $[-n,n]$ is uniformly continuous by the Heine-Cantor theorem; its modulus of continuity combined with $|c_n(x)-c_n(y)|\le |x-y|$ proves uniform continuity of $f_n$ on all of $\mathbb{R}$. For each fixed $x$, once $n\ge |x|$ we have $f_n(x)=f(x)$. HenceThis pointwise approximation by uniformly continuous functions even gives eventual equality on every fixed bounded interval. It need not give uniform convergence on the whole line, as the bounded counterexample in part (i) already demonstrates.

True. If $|f^{\prime }(x)|\le K$ everywhere, the mean value theorem applied between any two distinct points givesThus $f$ is Lipschitz continuous and hence uniformly continuous. No continuity of $f^{\prime }$ is needed: differentiability of $f$ gives continuity on each finite closed interval and is exactly the hypothesis needed for the mean value theorem there.

A real function on a metric space is uniformly continuous if for every $\epsilon >0$ there is a $\delta >0$ such that, simultaneously for all $x,y$, $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$. For a continuous function on a finite closed interval $[a,b]$, suppose this fails. Then there are ${\epsilon }_0>0$ and pairs $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. By compactness, some subsequence $x_{n_j}$ converges to $x\in [a,b]$; then $y_{n_j}\to x$ too. Continuity contradicts the image separation. This proves the Heine-Cantor theorem in this setting. Boundedness of the interval matters: $x^2$ on $[0,\infty )$ is a counterexample if “closed interval” includes unbounded intervals.

A function is Lipschitz continuous if some finite $K\ge 0$ satisfies $|f(x)-f(y)|\le Kd(x,y)$ for every pair. Choosing $\delta =\epsilon /(K+1)$ proves uniform continuity.

The first assertion is false. Take $f(x)=\sin (x^2)$, which is bounded and continuous, andThen $f(x_n)=1$, $f(y_n)=-1$, while $y_n-x_n=\pi /(y_n+x_n)\to 0$. This violates uniform continuity.

For order $48=2^4\cdot 3$, the Sylow theorems give $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$, hence $n_2=1$ or $3$. If $n_2=1$, the Sylow subgroup of order $16$ is a nontrivial proper normal subgroup. If $G$ were simple in the other case, it would be nonabelian (a finite abelian simple group has prime order), and the preceding result would force $48\mid 3!/2=3$, impossible. Thus every group of order 48 is not simple.

An example with no normal Sylow subgroup for $p=2$ isIn $S_4$, $P=⟨(1234),(13)⟩$ is a dihedral group of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a Sylow subgroup of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other Sylow subgroup for $p=2$ is normal either.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

The Sylow theorems state that if a finite group has order $|G|=p^{a}m$ with $p$ prime and $p\nmid m$, it has a subgroup of order $p^a$, called a Sylow subgroup. Every p-group subgroup is contained in a Sylow subgroup, and all Sylow subgroups for the same prime are conjugate subgroups. Their number satisfiesMore precisely $n_p=[G:N_G(P)]$ for a Sylow subgroup $P$, where $N_G(P)$ is its normalizer.

Let ${\mathbf{p}}_1$ and ${\mathbf{p}}_2$ be the incoming and proton momenta. The other incoming atom is at rest, so its total energy is $m_1{c}^2$. By four-momentum conservation, the tritium atom hasThe specified perpendicular emission gives ${\mathbf{p}}_1\cdot {\mathbf{p}}_2=0$. Therefore the relativistic energy-momentum relation impliesDividing by $c^4$ givesHere $E_1,E_2$ are total energies, not just kinetic energies. This identity is conditional on the collision admitting the stated emission: the calculated energies must satisfy the positive-energy mass-shell conditions, in particular $E_2\ge m_2{c}^2$ and $E_3\ge m_3{c}^2$. No nonrelativistic approximation has been used.

Use momentum conservation and conservation of energy in the laboratory inertial frame. The photon has momentum magnitude $E/c$, while the outgoing electron satisfies the relativistic energy-momentum relation. Its total energy and momentum obeySubstitute these into $E_e^2=m^2{c}^4+c^2|{\mathbf{p}}_e{|}^2$. Cancelling $m^2{c}^4$ and $(E_1-E_2{)}^2$ gives $m{c}^2(E_1-E_2)=E_1{E}_2(1-\cos \theta )$. Both photon energies are positive, so division yieldsThis is Compton scattering. The energies include the electron's rest energy; using only its kinetic energy in the mass-shell equation would give the wrong result. The outgoing photon has unchanged energy at $\theta =0$ and smaller energy for a nonzero scattering angle.

The arithmetic-geometric mean inequality gives $a_k+1/a_k\ge 2$, so $u_n\ge 2n\to +\infty$. Hence $(u_n)$ has no finite limit.

For the example, take $\boxed{a_k=1+2^{-k}}$. Every term is positive and different from $1$, andThe bounding geometric series converges, so the comparison test for series proves that the partial sums $v_n$ converge.

In general, if $v_n$ converges, the necessary condition for a convergent series gives $d_k=a_k-1/a_k=v_k-v_{k-1}\to 0$. Since $a_k>0$,Apply the Cesàro theorem for convergent sequences to these summands:Equivalently, $a_k=(d_k+\sqrt{d_k^2+4})/2\to 1$. The series of the differences may converge while the series of the positive sums still diverges linearly.

A sequence $(a_k)$ converges to $a\in \mathbb{R}$ when for every $\epsilon >0$ there is $N$ such that $|a_k-a|<\epsilon$ for all $k\ge N$. A series ${\sum }_{k=1}^{\infty }{a}_k$ converges to a finite $S$ when its partial sums $s_n={\sum }_{k=1}^n{a}_k$ converge to $S$.

If the series converges, then $a_n=s_n-s_{n-1}\to S-S=0$, so its terms tend to zero. The converse is false: the harmonic series has terms tending to zero but diverges.

For the arithmetic averages, suppose $a_k\to a$. Fix $\epsilon >0$ and choose $K$ so $|a_k-a|<\epsilon /2$ for $k\ge K$. The finite initial contribution $C={\sum }_{k=1}^{K-1}|a_k-a|$ is fixed, andFor sufficiently large $n$, $C/n<\epsilon /2$, provingThis is the Cesàro theorem for convergent sequences. Its hypothesis concerns a convergent sequence; the corresponding series need not be a convergent series.

For the constant monomial $f_0=1$, the binomial theorem gives $B_n(f_0)(x)=(x+1-x{)}^n=1$. For $f_1$, the binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ givesThus $\boxed{B_n(f_0)=1,B_n(f_1)=x}$, including $x=0,1$ by evaluation of the polynomials.

For $m\ge 1$ and $n\ge 2$, substitute the preceding coefficient identity into the Bernstein polynomial sum and write $j=k-1$. This yields the Bernstein monomial recurrenceThe coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n{)}^{m-1}\to 1$ and the others tend to zero. The weights in any Bernstein polynomial sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le 1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x{x}^{m-1}=x^m$: the $\ell =0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the supremum norm proves uniform convergence on $[0,1]$.

As an independent quantitative bound, take a binomial distribution variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb{E}[(K/n{)}^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the Cauchy-Schwarz inequality and the variance of $K/n$ giveFor $m=0$ the error is exactly zero. Finally the requested sum equals $2^m{B}_{2n}(f_m)(1/2)$, so

For $n\ge 2$, use the binomial coefficient identity $(n/k)\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. It reduces the left side to $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)(k/n{)}^{m-1}$. Put $t=(k-1)/(n-1)$. Then $k/n=((n-1)t+1)/n$, and the binomial theorem givesThus the required coefficients areFor $m=1$ the sum has the single coefficient $a_{n,1,0}=1$. For $k=1$, the power $t^0$ denotes the constant monomial $1$.

The original PDF allows $n=1$, but its displayed ratio $(k-1)/(n-1)$ is then $0/0$. Literally that expression is undefined. The meaningful endpoint identity is obtained before normalization:This polynomial identity is valid at $n=k=1$, giving $1$ on both sides. The normalized formula and the Bernstein monomial recurrence below are used only for $n\ge 2$.

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.

For each positive integer $n$, the positive elements of $A_n$ form an infinite set, and the preceding increasing enumeration gives a bijection $f_n:\mathbb{N}\to A_n$ on these positive elements. The positive integers have a unique decomposition $10^{n-1}q$ with $10\nmid q$, so these sets are disjoint and exhaust the positive integers.

The printed natural numbers include $0$. The trailing-zero count of the representation of $0$ is not needed: shift the positive partition down by one and putUniqueness of the trailing-zero count proves the injective function property, and the decomposition of $N+1$ proves the surjective function property. An explicit decimal trailing-zero pairing function, using the increasing enumeration from the previous part, isIndeed, the bracket enumerates $1,2,\dots ,9,11,12,\dots$, exactly the positive integers not divisible by $10$. This handles both the positive index $n$ and the element $0$ without silently changing the PDF's convention.

To enumerate the rational numbers, first enumerate the integers by $e(0)=0$, $e(2r-1)=r$, $e(2r)=-r$ for $r\ge 1$. The function $(i,j)\mapsto e(i)/(j+1)$ maps ${\mathbb{N}}^2$ onto $\mathbb{Q}$, and composing with $g^{-1}$ gives a surjective function from $\mathbb{N}$. Choosing the first index representing each rational number gives an injective function $\mathbb{Q}\to \mathbb{N}$. Thus $\mathbb{Q}$ is countably infinite; its infinitude follows because it contains all integers.

Use the well-order of the natural numbers to list $A$ in increasing order. Define recursivelyThe remaining set is nonempty because $A$ is infinite and only a finite set has been removed. Each chosen value is larger than the preceding one, so $f$ is an injective function.

To prove it is a surjective function, fix $a\in A$. If $a$ were never selected, every selected minimum would be at most $a$. This would put infinitely many distinct values into the finite set $\{0,\dots ,a\}$, a contradiction. HenceThe increasing enumeration of an infinite subset of natural numbers uses no arbitrary choice: each next element is the uniquely determined minimum.

A set $A$ is a countable set if there exists an injective function $A\to \mathbb{N}$, with $\mathbb{N}=\{0,1,2,\dots \}$. Equivalently, it is a finite set or has a bijection with $\mathbb{N}$. An infinite countable set is called a countably infinite set.

The empty set is included: the empty function into $\mathbb{N}$ is injective. This convention distinguishes “countable” from “countably infinite” and will also cover the finite threshold sets in the last part.