Past exam of the metric spaces and complex analysis course of the University of Oxford

Solutions to all six questions of the University of Oxford Part A metric space and complex analysis paper, sat on 12 June 2023. Source: released examination PDF, compilation pages 65–68; paper pages 1–4. The solutions retain the paper's labels. In 2(b)(i), the printed zero case $m\ne 0$ is inconsistent with its one case; the solution explicitly uses the intended $m\ne n$. The extra closing parenthesis in the printed definition of the supremum norm metric has no mathematical effect.

A metric space $(X,d)$ is a bounded set if there is a finite constant $M\ge 0$ such thatFor nonempty $X$, this is equivalent to containment in an open ball of finite radius: the triangle inequality converts a radius bound into a diameter bound, and conversely a diameter bound gives a ball of any slightly larger radius about a chosen point.

It is a complete metric space if every Cauchy sequence in $X$ has a limit of a sequence belonging to $X$. Explicitly, the hypothesis on $(x_n)$ isand completeness requires an $x\in X$ with $d(x_n,x)\to 0$. ThusBoundedness controls distances; completeness ensures that these limiting points remain in the space.

A subset $F\subseteq X$ is a closed set in $X$ when its complement is an open set in the metric topology. Equivalently,The open ball is taken inside the ambient metric space $X$; closedness always depends on that ambient space.

First suppose $F$ is a closed set. If $x\notin F$, the definition in part (ii) gives an open ball $B(x,r)$ disjoint from $F$. Thus every $y\in F$ has $d(x,y)\ge r$, and the distance to a set satisfies $\mathrm{dist}(x,F)\ge r>0$. Its contrapositive is the required implication.

Conversely, suppose that zero distance to a set forces membership. For $x\notin F$ we have $\mathrm{dist}(x,F)>0$. Choose $0<r<\mathrm{dist}(x,F)$. Then $B(x,r)\cap F=\varnothing$, proving $X\setminus F$ is an open set. HenceFor $F=\varnothing$, use the extended-real convention $\inf \varnothing =+\infty$: both assertions hold, with the implication vacuous. Part (b) only uses nonempty sets, so all its distances to a set are ordinary finite real numbers.

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.

The intersection $G$ is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.

Fix $n$, $x_n\in F_n$, and $\eta >0$. Recursively choose $x_{k+1}\in F_{k+1}$, for $k\ge n$, using the approximation property of an infimum so thatThis is possible because $x_k\in F_k$ and $\mathrm{dist}(x_k,F_{k+1})\le \rho (F_k,F_{k+1})\le 2^{-k}$. The two bounds form convergent geometric series, so the triangle inequality shows $(x_k{)}_{k\ge n}$ is a Cauchy sequence. By completeness it converges to some $x\in X$.

For each fixed $j$, all sufficiently late $x_k$ lie in $F_j$, by nesting. Closedness of $F_j$ gives $x\in F_j$, and hence $x\in G$. In particular, starting with any $x_1\in F_1$ proves $G\ne \varnothing$. Moreover,For each starting point $x_n$, let $\eta \downarrow 0$ in this bound on its distance to a set $G$. This does not assert that a nearest point exists. It proves $\mathrm{dist}(x_n,G)\le 2^{1-n}$. Since $G\subseteq F_n$, the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.

Let $(F_n)$ be a Cauchy sequence for the Hausdorff distance, and putThe bounded set hypothesis makes ${\delta }_n$ finite, and the Cauchy sequence property gives ${\delta }_n\to 0$. Each $A_n$ is nonempty and a closed set, with $A_{n+1}\subseteq A_n$.

We need estimates that survive taking the closure. For every nonempty $F$, the distance to a set is Lipschitz continuous with constant one: the triangle inequality gives $\mathrm{dist}(x,F)\le d(x,y)+\mathrm{dist}(y,F)$, and interchanging $x,y$ provesFor every $m\ge n$, points of $F_m$ have distance at most ${\delta }_n$ from $F_n$. The same holds throughout their union and its closure, by this continuity. Since $F_n\subseteq A_n$,More generally, if $m\ge n$, every point in ${\bigcup }_{k\ge n}{F}_k$ is within ${\delta }_n$ of $F_m\subseteq A_m$. Passing to its closure again, and using $A_m\subseteq A_n$, givesChoose strictly increasing indices $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. Then $(A_{n_j})$ satisfies part (ii), so it converges in Hausdorff distance to the nonempty closed set $G={\bigcap }_j{A}_{n_j}={\bigcap }_n{A}_n$, with $\rho (A_{n_j},G)\le 2^{1-j}$.

For $n\ge n_j$, the triangle inequality gives the explicit full-sequence boundIt tends to zero as $j\to \infty$. ThereforeThis proves Completeness of the Hausdorff hyperspace using the proposed tail closures, rather than assuming their diameters shrink.

A metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ cover $X$.

It is a sequentially compact space if every sequence $(x_n)$ in $X$ has a convergent subsequence $(x_{n_k})$ whose limit of a sequence belongs to $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

Let $(x_n)$ be any sequence in the compact metric space $X$. There must be an $x\in X$ such that every open ball about $x$ contains $x_n$ for infinitely many indices $n$. Otherwise, for each $x\in X$ choose a ball $B(x,r_x)$ containing only finitely many terms, counted by indices. These balls are an open cover. A finite subcover would imply that only finitely many indices occur anywhere in $X$, a contradiction.

Now choose successively $n_k>n_{k-1}$ with $x_{n_k}\in B(x,1/k)$; infinitely many available indices make this possible. Then $d(x_{n_k},x)<1/k\to 0$, so $(x_{n_k})$ is a convergent subsequence. ThusThe proof counts indices rather than distinct points, so it also handles a value repeated infinitely often.

The source prints $m\ne 0$ in the zero case. Since the coordinates are indexed by positive integers, that overlaps its $m=n$ case and does not define a sequence. We use the intended coordinate vectors, with $e_m^{(n)}=1$ for $m=n$ and $e_m^{(n)}=0$ for $m\ne n$.

Every $e^{(n)}$ belongs to $X$, because its entries are bounded by one and its coordinate sequence eventually vanishes. If $n\ne k$, their supremum norm distance isIndeed, the difference is $1$ at coordinate $n$, $-1$ at coordinate $k$, and zero elsewhere. No subsequence is a Cauchy sequence, since arbitrarily late distinct terms remain distance one apart. Every convergent sequence is a Cauchy sequence, so no convergent subsequence exists. Therefore $\boxed{X\text{ is not sequentially compact}}$.

Yes. Let $(x^{(k)}{)}_{k\ge 1}$ be any sequence in $A$, where $x^{(k)}=(x_m^{(k)}{)}_{m\ge 1}$. At coordinate $m$, the values lie in the compact interval $[-1/m,1/m]$. By part (a)(ii) we can successively extract a convergent subsequence for coordinate one, a further one for coordinate two, and so on. Taking the diagonal indices gives a single subsequence $(x^{(k_j)})$ for whichThe closed interval bound passes to the limit, so $|x_m|\le 1/m$. Hence $x_m\to 0$ and $x=(x_m)\in A$.

Coordinate convergence alone would not imply supremum norm convergence. Here the common tail bound supplies the missing estimate. Given $ϵ>0$, choose $M$ with $2/(M+1)<ϵ$. For $m>M$,For the finitely many $m\le M$, coordinate convergence makes all differences less than $ϵ$ once $j$ is large. Therefore $d(x^{(k_j)},x)<ϵ$, andThis is an instance of compactness from uniformly vanishing sequence tails. The common $1/m$ bound is exactly what the coordinate vectors in part (i) lack.

By the permitted sequential compactness of the product metric space $X\times X$, the sequence of pairs $({\varphi }^{(k)}x,{\varphi }^{(k)}y)$ has a convergent subsequence, and hence a Cauchy subsequence. Choose two of its indices $r<s$ sufficiently late thatThe distance-increasing assumption iterates: $d({\varphi }^{(r)}u,{\varphi }^{(r)}v)\ge d(u,v)$. With $n=s-r\ge 1$, use $u=x,v={\varphi }^{(n)}x$, and then $u=y,v={\varphi }^{(n)}y$, to obtainBoth summands are nonnegative, soThe use of paired iterates ensures that the same positive return time works for both points.

Choose $n\ge 1$ as in part (i). Since distances between iterates are nondecreasing, the triangle inequality givesLetting $ϵ\downarrow 0$ and combining with the given reverse bound proves that $\varphi$ is an isometry:For surjectivity, fix $x\in X$ and use part (i) with $y=x$ and $ϵ=1/k$ to choose $n_k\ge 1$ satisfying $d(x,{\varphi }^{(n_k)}x)<1/k$. Put $z_k={\varphi }^{(n_k-1)}x$, interpreting ${\varphi }^{(0)}$ as the identity. By sequential compactness, a convergent subsequence satisfies $z_{k_j}\to z\in X$. The assumed continuity yieldsThus every $x$ has a preimage, and $\boxed{\varphi \text{ is surjective}}$. If $X$ is empty, both conclusions hold vacuously. This proves the distance-increasing self-map theorem for sequentially compact metric spaces.

At every $z=x+iy\in U$, the Cauchy-Riemann equations areTo derive them from complex differentiability, approach $z$ along real increments $h$. The complex derivative givesApproaching along imaginary increments $ih$ instead givesThese limits exist because $f$ is holomorphic. Equating their real and imaginary parts proves the equations, together with $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$.

A sufficient condition is that $U$ is a connected space. Suppose, towards a contradiction, that $|f|$ is the same constant $R$ throughout $U$. If $R=0$, then $f=0$, contradicting $f(a)\ne f(b)$.

If $R>0$, differentiate $u^2+v^2=R^2$ in $x$ and $y$, and use the Cauchy-Riemann equations from part (a). They giveThe coefficient matrix for $(u_x,v_x)$ has determinant $-(u^2+v^2)=-R^2\ne 0$. Hence $u_x=v_x=0$, so $f^{\prime }=0$ everywhere. A holomorphic function with zero complex derivative is locally constant: within a small disc, differentiate along a straight segment. Local constancy and connectedness imply constancy on $U$, again a contradiction.

Consequently,This is the constant-modulus holomorphic function principle. The extra hypothesis matters: on two disjoint open discs, define $f=1$ on one and $f=-1$ on the other. It is holomorphic and nonconstant, yet $|f|=1$ everywhere.

The function $h(z)=1/(z+1)$ is holomorphic on $U$ and hasIts imaginary part is exactly the prescribed $v$. Thus $f-h$ is a real-valued holomorphic function. The Cauchy-Riemann equations force both partial derivatives of its real part to vanish. Since $\mathbb{C}\setminus \{-1\}$ is a connected space, $f-h$ is a single real constant. All possibilities are thereforeConversely every such function has the stated imaginary part. The data determine $f$ only up to this real additive constant.

The identity theorem states: if $V\subseteq \mathbb{C}$ is a nonempty connected open set and $p,q$ are holomorphic functions on $V$, and their equality set has a limit point in $V$, thenThe interior location of the limit point and connectedness are essential hypotheses.

For the proof put $h=p-q$. At any $z_0\in V$, its convergent Taylor series either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\ge 0$. In the latter case,where $k$ is holomorphic near $z_0$. By continuity, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior limit point therefore forces the neighbourhood alternative.

Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne \varnothing$, and $E$ is an open set. It is also a closed set relative to $V$: a point of its closure outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same Taylor series argument would place it in $E$. Since $V$ is connected, $E=V$. Thus $h=0$ on $V$, proving the theorem.

Define $F(z)=\overline{f(\overline{z})}$. To apply the identity theorem, we first verify that $F$ is holomorphic; mere complex conjugation of the values of $f$ would not suffice. For $h\ne 0$,Thus $F$ is an entire function, just as $f$ is. At each positive integer $n$, the point $1/n$ is real and $f(1/n)$ is real, so $F(1/n)=f(1/n)$. These distinct points accumulate at $0\in \mathbb{C}$. The identity theorem on the connected complex plane givesThis is conjugate reflection of a holomorphic function, with the reality data forcing equality to its reflection.

A function has an isolated singularity at $a$ if, for some $r>0$, it is defined and holomorphic on the punctured disc $0<|z-a|<r$, while a holomorphic extension through $a$ has not been assumed. In particular, its domain omits $a$, or its assigned value there fails to give such an extension. The key local condition isThe convention includes removable singularities, even when a suitable extension would eliminate the missing point.

An isolated singularity at $a$ is a removable singularity when there is a holomorphic function $F$ on a full disc $B(a,r)$ agreeing with $f$ away from $a$. Equivalently,The Riemann removable singularity theorem also characterizes this by boundedness of $f$ on a sufficiently small punctured disc. A preassigned incorrect value at $a$ is replaced by $L$.

A holomorphic function has a zero of order $n$ at $a$ when it is holomorphic on a neighbourhood of $a$ and factors asEquivalently, $f(a)=f^{\prime }(a)=\cdots =f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne 0$. The order of a zero of a holomorphic function is the degree of the first nonzero term of its Taylor series.

A function has a pole of order $n$ at $a$ when, on a punctured neighbourhood,Equivalently, $(z-a{)}^{n}f(z)$ has a removable singularity whose extension is nonzero at $a$. After shrinking the neighbourhood so $h\ne 0$, its reciprocal $1/f$ extends with a zero of order $n$. In particular, $|f(z)|\to \infty$ at a pole.

We prove the Casorati-Weierstrass theorem by contradiction. Fix an allowed $\rho$. If the stated image were not a dense set in $\mathbb{C}$, there would be a disc $B(w,ϵ)$ disjoint from it. HenceThe holomorphic function $h(z)=1/(f(z)-w)$ is bounded by $1/ϵ$ on that punctured disc. By the permitted Riemann removable singularity theorem, it extends to a holomorphic function $H$ on the full disc.

If $H(a)\ne 0$, then $w+1/H$ is holomorphic near $a$ and agrees with $f$ away from $a$. This makes $a$ a removable singularity of $f$.

If $H(a)=0$, then $H$ is not identically zero, since $h$ is nowhere zero away from $a$. Its Taylor series therefore has a finite first nonzero coefficient: otherwise the identity theorem would make it identically zero. Write $H(z)=(z-a{)}^{m}K(z)$ with $m\ge 1$ and $K(a)\ne 0$. Thenhas a pole of order $m$ at $a$, since multiplying by $(z-a{)}^m$ gives the nonzero limit $1/K(a)$.

Both cases contradict an essential singularity. Thus

Use the local factorization for a pole of order $n$:Shrink the disc so that the holomorphic function $h$ has no zeros there. Differentiating and dividing gives the logarithmic derivativeThe second term is holomorphic across $a$. Since $n\ge 1$, the first term cannot cancel. Hence $f^{\prime }/f$ has a simple pole, and the coefficient of $(z-a{)}^{-1}$ in its Laurent series isThis is the pole case of the integer residue of a logarithmic derivative.

Let $F=e^g$. It is holomorphic and nowhere zero on a punctured neighbourhood of $a$, and the chain rule givesIf $F$ had a pole of order $n\ge 1$, part (c) would imply ${\mathrm{Res}}_a(g^{\prime })=-n$. But the derivative of a holomorphic function on a punctured disc has residue zero. Indeed, write its Laurent seriesTermwise differentiation is valid on every smaller closed annulus. The only possible $(z-a{)}^{-1}$ term comes from $k=0$ and has coefficient zero. Thusa contradiction. Therefore $\boxed{e^g\text{ cannot have a pole at }a}$. This proves an exponential has no isolated pole, including when $g$ has an essential singularity; no choice of a holomorphic logarithm on the entire punctured disc was assumed.

The simple zero assumption gives the local factorizationwhere $k$ is holomorphic. In a small disc $k$ stays nonzero, so $q=g/k$ is holomorphic there. Expanding $q$ to first order,The residue is thereforeIf $g(a)=0$, the quotient has a removable singularity rather than a simple pole, and the same formula correctly gives zero. This is the residue formula for a quotient at a simple denominator zero.

The improper integral converges: its integrand is bounded near zero and at most $x^{-n}$ for $x\ge 1$, which is integrable because $n\ge 2$.

Set $\theta =\pi /n$, $\omega =e^{2i\theta }$ and $\zeta =e^{i\theta }$. Apply the residue theorem to $F(z)=1/(1+z^n)$ on the positively oriented sector contour consisting of the segment $0\to R$, the arc $R{e}^{it}$ for $0\le t\le 2\theta$, and the segment $R\omega \to 0$, with $R>1$. No pole lies on the boundary. Exactly one root of $z^n=-1$ is inside this sector, namely $\zeta$; it is a simple pole. Part (a) givesOn the returning ray, put $z=\omega x$ with $x$ decreasing from $R$ to zero. Since ${\omega }^n=1$, the two ray contributions areOn the arc, $|1+z^n|\ge R^n-1$, so the estimation lemma bounds its contour integral byThe residue theorem and the convergence already established now giveUsing $1-e^{2i\theta }=-2i{e}^{i\theta }\sin \theta$, we concludeThe sector also works for $n=2$, when it is the upper semicircle. Its angle is chosen so that $z^n$ has the same value on the two rays.

Factor the denominator as $(z^2+1)(z^2+4)$ and integrateover $[-R,R]$ followed by the upper semicircle, for $R>2$. The enclosed simple poles are $i$ and $2i$. The local quotient formula from part (a) givesOn this arc, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^2+1|\ge R^2-1$, $|z^2+4|\ge R^2-4$. Thus the estimation lemma givesOn the real line, $|F(x)|\le 1/[(x^2+1)(x^2+4)]$, so the improper integral is absolutely convergent. Hence the residue theorem yieldsTaking real parts is justified by absolute convergence. ThereforeThe imaginary part vanishes as a further check, since the sine integrand is odd.