Past exam of the options II course of the University of Oxford / 2012 (4304, AO2)

Source paper: Mathematics Part A FOI release, member FOI-20210708-1 - Data/4304 (1).pdf. The cover identifies Options II, AO2, formal code 4304, Thursday 21 June 2012, 9.30am to 12.30pm. All sixteen source pages are readable. The section letters and nineteen printed question labels are retained below, including the unlettered opening tasks in D3 and M5.

Start with $O=(0,0)$ and $E=(1,0)$. A constructible point is a point obtained from these by finitely many ruler-and-compass steps. At each step one may draw the line through two previously obtained distinct points, or the circle with a previously obtained centre and radius equal to the distance between two previously obtained points, and take isolated intersection points of the drawn lines or circles. A constructible line is a line through two distinct constructible points; a constructible circle has constructible centre and a radius obtainable as such a distance. These conventions also permit copying a previously obtained segment with the compass.

Finite construction from a fixed unit segment is essential: a drawing through an arbitrary unspecified real point does not make that point constructible.

Consider an isolated intersection of the two curves. For two distinct lines, solving their linear equations by elimination gives coordinates in $F$ whenever there is an intersection.

For a line and a circle, solve the line for one coordinate with a nonzero coefficient and substitute into the circle equation. The resulting quadratic has coefficients in $F$. A real root lies in $F$ if its discriminant is a square in $F$, and otherwise in $F(\sqrt{D})$ for its positive nonsquare discriminant $D\in F$. The other coordinate is an $F$-linear expression in that root. Thus both coordinates lie in the same real quadratic extension. A tangent intersection has discriminant zero and already lies in $F$.

Write two circle equations in normalized form $x^2+y^2+u_{i}x+v_{i}y+w_i=0$. Subtracting eliminates the quadratic terms. If the circles are distinct and meet, this gives a nontrivial line, reducing the problem to the preceding case. Concentric distinct circles do not intersect. ThereforeThe isolated-intersection qualification is implicit in the construction operation. If coincident curves were allowed and every point on their common curve counted as an intersection point, the printed assertion would be false: a rational line contains points with transcendental coordinates.

Induct through the finitely many construction steps. Initially all available coordinates lie in $\mathbb{Q}$. If coordinates obtained so far lie in a real field $F$, the line through two of its points has coefficients in $F$. A circle with centre in $F^2$ and radius a distance between two such points has equationwhose coefficients lie in $F$, since the squared distance $r^2$ lies in $F$; the radius itself need not lie there.

Part (b) therefore puts every new intersection in $F$ or a real quadratic extension of $F$. If a step produces several points, adjoin their coordinates successively, omitting extensions that add nothing. A finite construction ends in a real field $K$ containing the desired coordinates, with a tower of genuine quadratic extensionsThe tower law for field extensions yieldsA construction involving no nontrivial extension has $n=0$ and degree one.

A constructible regular polygon can be translated, rotated, and rescaled by constructible operations to one inscribed in the unit circle with a vertex at $(1,0)$. Thus the real and imaginary parts of $\zeta =e^{2\pi i/p}$ are constructible. Part (c) puts both in a real field $K$ of degree $2^n$ over $\mathbb{Q}$. Consequently $\zeta \in K(i)$, and $[K(i):\mathbb{Q}]=2^{n+1}$ because $K\subset \mathbb{R}$.

For prime $p$, the supplied irreducibility of $1+x+\cdots +x^{p-1}$ shows that $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. The intermediate-field tower law implies that $p-1$ divides $2^{n+1}$. Every positive divisor of a power of two is a power of two, soThis necessary condition includes $p=2$ with $r=0$. It does not assert sufficiency for arbitrary integers of this form.

The identity element satisfies $e\ast x=ex{e}^{-1}=x$. For $g,h\in G$,Both action axioms hold, so conjugation is a group action on $G$.

By definition, $g$ stabilizes $x$ exactly when $gx{g}^{-1}=x$. Multiplying on the right by $g$ gives the equivalent condition $gx=xg$. HenceThus the stabilizer in this action is the centralizer of the element.

Let $H=Z(x)$. The map $G/H\to C(x)$ sending $gH$ to $gx{g}^{-1}$ is well defined: multiplying $g$ on the right by an element commuting with $x$ leaves its conjugate unchanged. It is onto by the orbit definition. If $gx{g}^{-1}=hx{h}^{-1}$, then $h^{-1}g$ commutes with $x$, so $gH=hH$, proving injectivity.

Every coset has $|H|$ elements, so $|C(x)|=|G|/|H|$. ThereforeThis is the orbit-stabilizer theorem specialized to conjugation.

An element $x$ is central exactly when $gx=xg$ for every $g\in G$. By part (b), this is exactly the assertion that every $g$ stabilizes $x$. That means every conjugate of $x$ is $x$; conversely, a singleton conjugacy class has this property for every $g$.In particular, the singleton classes are in one-to-one correspondence with the elements of the centre.

Conjugacy is an equivalence relation: the identity element gives reflexivity, conjugation by $g^{-1}$ reverses conjugation by $g$, and composing conjugations gives transitivity. Its equivalence classes therefore partition the finite set $G$. Counting this disjoint union givesSeparating singleton classes from the others gives the class equation $|G|=|Z(G)|+{\sum }_{|C_i|>1}|C_i|$.

Write $|G|=p^n$, where $n\ge 1$. By part (c), each conjugacy-class size divides $p^n$ and hence is a power of $p$. Every nonsingleton class consequently has size divisible by $p$. The class equation impliesThe identity element lies in the centre, so the centre is nonempty. Its size is a positive multiple of $p$, and thereforeThis proves that a nontrivial finite p-group has a nontrivial centre.

For an integer $a$ and an odd prime $p$, the Legendre symbol isThe value one refers to a nonzero quadratic residue; divisibility by $p$ is recorded separately by zero.

Euler's criterion states thatFor $p\nmid a$ this says that the power is $1$ for a quadratic residue and $-1$ for a quadratic nonresidue. When $p\mid a$, both sides are zero modulo $p$.

Euler's criterion givesIf either factor is divisible by $p$, both asserted values are zero. Otherwise the two values lie in $\{1,-1\}$, whose elements are distinct modulo an odd prime. The modular congruence is therefore equality:Finally, $a\equiv b(\mathrm{mod}p)$ means they give the same residue class and thus the same set of square-congruence solutions. Divisibility by $p$ is also identical. Hence $\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$.

The quadratic reciprocity states that for distinct odd primes $p,q$,With $q=3$, it gives $\left(\frac{3}{p}\right)=(-1{)}^{(p-1)/2}\left(\frac{p}{3}\right)$. The nonzero square modulo three is one, so $\left(\frac{p}{3}\right)=1$ when $p\equiv 1(\mathrm{mod}3)$, and $-1$ when $p\equiv -1(\mathrm{mod}3)$. An odd prime other than three is congruent to $1,5,7$, or $11$ modulo twelve. Evaluating the two signs givesThus

Reducing modulo $23$ gives $189\equiv 5$. Reciprocity with five has positive sign, sosince the nonzero squares modulo five are one and four. Modulo $59$, $189\equiv 12=4\cdot 3$, and $59\equiv -1(\mathrm{mod}12)$. Thus part (b) givesAs $1357=23\cdot 59$, a solution modulo $1357$ would in particular solve the modular congruence modulo $23$, contradicting the first symbol. Therefore there is no integer solution modulo $1357$.

The monotone convergence theorem states that if measurable functions $0\le f_n\uparrow f$ pointwise on a measure space, then $\int f_n\uparrow \int f$, allowing the limit to be infinite.

Extend each restriction to $I_n$ by zero on $I\setminus I_n$. The functions $f{1}_{I_n}$ are nonnegative and increase pointwise to $f$ because the intervals are increasing and exhaust $I$. HenceFor nonnegative $f$, integrability means finiteness of this integral, soNo boundedness or closedness of the intervals is required.

The function is nonnegative, measurable, and even. Direct integration givesThus it is integrable, with integral $\boxed{2}$.

On finite intervals the function is continuous, butConsequently its nonnegative integral on the half-line is infinite: it is not integrable. The divergence occurs at infinity, not at the origin.

For every real $x$, the numerator is at most $2e^{|x|}$, while the denominator is at least $e^{2|x|}$. ThereforeThe upper bound is integrable by part (a), so the comparison theorem shows that the function is integrable on $\mathbb{R}$.

We prove absolute integrability. On $(0,1]$, $|\sin (1/x)|\le 1$ and the denominator is at least one, so $|f(x)|\le 1$. On $[1,\infty )$, the standard inequality $|\sin t|\le |t|$ givesSubstitution $t=\log x$ yieldsThus ${\int }_0^{\infty }|f(x)|dx\le 1+\pi /2<\infty$, and the function is integrable despite its oscillation near zero.

For a continuously differentiable diffeomorphism $\varphi :J\to I$ between intervals, the integration by substitution statesIt holds as an identity element of extended integrals for nonnegative measurable $h$. For signed or complex-valued measurable $h$, integrability on one side is equivalent to integrability after substitution, and the identity element then holds. The absolute derivative accounts for a decreasing substitution; on oriented finite intervals the equivalent formula uses ${\varphi }^{\prime }$ and oriented endpoints.

Put $C={\int }_0^{\infty }(1+x^4{)}^{-1}dx=\pi /(2\sqrt{2})$. The map $x=y^{-1/4}$ is a decreasing diffeomorphism of the positive half-line onto itself andApplying the nonnegative version before assuming finiteness proves both existence and value:Tonelli's theorem permits either iterated integral of a nonnegative measurable function on a product of sigma-finite measure spaces and identifies both with its extended product integral. Fubini's theorem states that for an absolutely integrable measurable function on such a product, almost every section is integrable, the section-integral functions are integrable, and the two iterated integrals equal the product integral.

The given two-variable function is continuous and positive on the positive quadrant. For fixed $y>0$, substituting $t=x{y}^{1/4}$ givesTonelli now givesThis proves absolute integrability, so Fubini is applicable.

For $x\ne 1$, integrate the displayed partial fraction identity element in $y$ up to $R$:At $x=1$ the section integral is ${\int }_0^{\infty }(1+y{)}^{-2}dy=1$, agreeing with the continuous limit of this expression. Therefore Fubini identifies the remaining integral and givesThe apparent singularity at one is removable, with limiting integrand value $1/4$. Positivity and the finite product integral already establish convergence at both ends.

The closure of $A$ is the intersection of every closed subset containing it:The collection is nonempty because it contains $X$. Arbitrary intersections of closed sets are closed, so this is the smallest closed set containing $A$.

Every subset is contained in its closure, giving $\overline{A}\subseteq \overline{\overline{A}}$. On the other hand, $\overline{A}$ is already closed by its definition as an intersection of closed sets. It is therefore one of the closed supersets used when taking its own closure, so $\overline{\overline{A}}\subseteq \overline{A}$. HenceThis is the idempotence of the closure operation.

If $x\notin \overline{A}$, some closed superset $F$ of $A$ misses $x$. Its complement is an open set containing $x$ and disjoint from $A$. Conversely, if an open set $U$ contains $x$ and misses $A$, the closed set $X\setminus U$ contains $A$ and misses $x$, so $x\notin \overline{A}$.

Negating this equivalence gives the neighbourhood characterization

An accumulation point of $A$ is a point $x\in X$ such that every open neighbourhood of $x$ contains a point of $A$ different from $x$:The point itself need not belong to $A$. Requiring a different point distinguishes accumulation from mere membership in the closure.

In a metric space,The sequence may equivalently be required to have pairwise distinct terms. The metric hypothesis makes a countable sequence sufficient; arbitrary topological spaces need not have this sequential characterization.

Take $A_1={\mathbb{Q}}^2$. Every open disk contains infinitely many points with both coordinates rational, so every point of the plane is an accumulation point of $A_1$. Every disk also contains infinitely many points with an irrational first coordinate, and these lie in the complement. ThusThe infinitely-many-points observation ensures that one can avoid the centre of the disk itself in both cases.

TakeIts only accumulation point is $(0,0)$, which is absent from $A_2$. Every listed point is isolated, so none belongs to $A_2^{\prime }$. ConsequentlyFor completeness, away from the origin a sufficiently small disk meets only finitely many of the listed points and can be shrunk to exclude all but its centre, proving there are no further accumulation points.

Use three nested sequences:For fixed $n,m$, letting $k\to \infty$ produces the point $(1/n+1/(n^{2}m),0)$. Letting $m\to \infty$ produces $(1/n,0)$, and letting $n\to \infty$ produces $(0,0)$. Accordingly, putThe successive derived sets areTo see that no limits are missing, take any convergent sequence of distinct points from $A_3$. If $n$ is unbounded, a subsequence has $n\to \infty$ and converges to the origin. Otherwise fix $n$ on a subsequence. Unbounded $m$ then forces limit $(1/n,0)$; with $n,m$ fixed, distinctness forces unbounded $k$, giving the corresponding point of $B$. The same argument, with one index fewer at each stage, proves the next two equalities. Distinct $n$-clusters are separated away from the origin because their first coordinates lie in the disjoint intervals $[1/n,1/n+1/n^2]$.

The original set has positive second coordinates, the first derived set has isolated $B$-points on the axis, the second has infinitely many $C$-points, and the third is a singleton. All three required successive inequalities follow. This is a three-stage accumulation tower.

The quotient topology is defined byPreimages commute with arbitrary unions and finite intersections and take the empty and whole sets to the corresponding sets, so this definition indeed gives a topology.

A map is continuous exactly when the preimage of every open set is open. By the definition in part (i), every open $U\subseteq X/\mathcal{R}$ has open preimage $p^{-1}(U)$. Thus $p$ is continuous, directly from the defining property of the quotient topology.

If $g$ is continuous, so is the composition $g\circ p$. Conversely, assume the composition is continuous. For every open set $V\subseteq Y$,is open in $X$. The quotient definition then says that $g^{-1}(V)$ is open in $X/\mathcal{R}$. Hence $g$ is continuous andThis is the universal property of the quotient topology.

Let two distinct quotient points correspond to two equivalence classes. By hypothesis those classes lie in disjoint open saturated sets $U,V\subseteq X$. Saturation gives $p^{-1}(p(U))=U$, so the quotient definition makes $p(U)$ open, and likewise $p(V)$.

The image sets contain the two quotient points and are disjoint. Indeed, a class meeting both $U$ and $V$ would be contained in each because both sets are saturated, contradicting their disjointness. Thus the quotient is Hausdorff.

Define $\tilde{f}:X/\mathcal{F}\to Y$ by $\tilde{f}([x])=f(x)$. Its definition is independent of the representative because the classes are precisely the fibres of $f$. It is onto since $f$ is onto, and injective because two points with equal $f$-values lie in the same fibre.

The relation $\tilde{f}\circ p=f$ and part (iii) imply continuity of $\tilde{f}$. Since $X$ is compact and $p$ is continuous, $X/\mathcal{F}$ is compact. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Therefore

Every iterate $g^n(x,y)=(x+n,(-1{)}^{n}y)$ is an isometry of the strip with its Euclidean subspace metric. Let $a,b$ lie in distinct orbits. Since the first-coordinate difference in $a-g^{n}b$ grows without bound as $|n|\to \infty$, the positive distances $d(a,g^{n}b)$ have a positive minimum $\delta$. Only finitely many indices can give a distance below any fixed bound, and none gives zero.

Choose $0<\epsilon <\delta /3$. Saturate the relative open balls by settingThese are open saturated neighbourhoods of the two orbits. Any two centres $g^{n}a,g^{m}b$ are at least $\delta$ apart, since their distance equals $d(a,g^{m-n}b)$. Their radius-$\epsilon$ balls therefore cannot intersect. Hence $U\cap V=\varnothing$, and part (a)(iv), an instance of positive orbit separation implies a Hausdorff quotient, proves

Let $D=[0,1]\times [-1,1]$ and $q=p\circ i:D\to S/{\mathcal{R}}^{\prime }$. This is continuous. It is onto: given $(x,y)\in S$, choose an integer shift bringing $x$ into $[0,1]$, with the corresponding possible sign change in $y$; the resulting representative lies in $D$.

Two representatives in $D$ are in the same orbit only if their first coordinates differ by an integer. An interior point can therefore be equivalent only to itself. Distinct boundary representatives have first coordinates zero and one, and the iterate $g$ identifies exactly $(0,y)$ with $(1,-y)$. Thus the fibres of $q$ are precisely the equivalence classes defining the Möbius band $M$.

The rectangle $D$ is compact, and the target is Hausdorff by part (i). Applying part (a)(v) to $q$ therefore yieldsThe compactness argument supplies continuity of the inverse without requiring a discontinuous choice of one representative per orbit.

A $C^1$ diffeomorphism between open subsets of ${\mathbb{R}}^n$ is a bijection $f:U\to V$ for which both $f$ and $f^{-1}$ are continuously differentiable. More generally, a $C^k$ diffeomorphism requires that regularity for both maps; a smooth diffeomorphism has derivatives of every order. In the $C^1$ setting used here,

The inverse function theorem states that if $f:U\to {\mathbb{R}}^n$ is $C^1$ on an open set and $Df(a)$ is invertible, then there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $f(a)$ such that $f:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse satisfiesIf $f$ is $C^k$ or smooth, the inverse has the same regularity.

The planar implicit function theorem states that if $F$ is $C^1$ near $(a,b)$, $F(a,b)=0$, and $F_y(a,b)\ne 0$, then in some product neighbourhood the equation $F(x,y)=0$ is equivalent to $y=h(x)$ for a unique $C^1$ function with $h(a)=b$. Moreover,If instead $F_x(a,b)\ne 0$, exchange the coordinates to solve locally for $x$ as a function of $y$.

The function is smooth, $f(0,0)=0$, andAt the origin both partial derivatives equal one. In particular, $f_y(0,0)=1\ne 0$, so the implicit function theorem supplies a neighbourhood in whichIn fact $h$ is smooth because $f$ is smooth.