Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 1

A metric space $(X,d)$ is a complete metric space if every Cauchy sequence in $X$ converges to a point of $X$. Explicitly, the hypothesis isand the conclusion is that some $x\in X$ satisfies $d(x_n,x)\to 0$.

It is a sequentially compact space if every sequence in $X$ has a convergent subsequence whose limit of a sequence belongs to $X$.

It is a connected space if there are no disjoint nonempty open subsets $U,V$ of $X$ with $X=U\cup V$. Here openness refers to the metric topology of $X$.

It is a path-connected space if, for every $x,y\in X$, there is a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$. Such a map is a path. Every path-connected space is a connected space: a separation would pull back along a path joining its two sides to a separation of the interval $[0,1]$.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

The answers areDistinct natural numbers have $d(x,y)>1$. Taking $\epsilon =1$ in the Cauchy sequence condition forces every sufficiently late pair of terms to be equal. Every Cauchy sequence is therefore eventually constant and converges in $\mathbb{N}$, proving completeness.

The sequence $x_n=n$ has no convergent subsequence, since convergence would force its sufficiently late terms to lie within distance $1$ of its limit of a sequence, hence to equal that limit. Thus sequential compactness fails.

Every singleton is an open ball of radius $1$, so the metric topology is the discrete topology. For example, $\{1\}$ and $\mathbb{N}\setminus \{1\}$ form a separation into two nonempty open sets, so $X$ is not connected. It is consequently not path-connected either.

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.

The inverse tangent $T(x)=\arctan x$ is a homeomorphism from the real line with its Euclidean metric onto $I=(-\pi /2,\pi /2)$ with its subspace topology: $T$ and its inverse $\tan$ are continuous. It is also an isometry from $(\mathbb{R},\rho )$ onto $I$, because $\rho (x,y)=|T(x)-T(y)|$. Consequently, for $U\subseteq \mathbb{R}$,For a direct open ball interpretation, $B_{\rho }(x,r)=T^{-1}((T(x)-r,T(x)+r)\cap I)$ is $d$-open by continuity of $T$. Conversely, continuity of $\tan$ at $T(x)$ supplies a sufficiently small $\rho$-ball inside any prescribed $d$-ball about $x$. Thus both metrics generate exactly the same open sets.

The arctangent pullback metric is not complete. Take $x_n=n$. Since $\arctan n\to \pi /2$, its transformed sequence is Cauchy, and hence $(x_n)$ is a Cauchy sequence for $\rho$. If it converged in $\rho$ to $x\in \mathbb{R}$, thenwould force $\arctan x=\pi /2$, which is impossible. The missing endpoint of $I$ explains the failure of completeness, even though part (i) shows that the metric topologies agree.

Extend the coefficient lists of each polynomial by zeros. Then $\rho (p,q)={\sum }_{k\ge 0}|a_k-b_k|$ is finite because only finitely many terms are nonzero. It is nonnegative and symmetric. Moreover, $\rho (p,q)=0$ holds exactly when every coefficient agrees, which by uniqueness of polynomial coefficients is equivalent to $p=q$.

If $r(t)=\sum c_k{t}^k$, the triangle inequality for the absolute value givesThus $\rho$ satisfies all the metric axioms. It is the metric induced by the polynomial coefficient sum norm.

The space of polynomials is not complete for the supremum norm metric $d$. ConsiderThe geometric series has sum $F(t)=1/(1-t/2)$, and for $0\le t\le 1$ its tail satisfiesThus $(p_n)$ converges uniformly to $F$ and is a Cauchy sequence for $d$.

There is no polynomial $p$ agreeing with $F$ on $[0,1]$: otherwise the polynomial $(1-t/2)p(t)-1$ would vanish at infinitely many points and therefore be the zero polynomial. Evaluating this identity at $t=2$ gives $-1=0$. By uniqueness of the uniform limit, $(p_n)$ has no limit in $X$. This proves the claimed failure of completeness.

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.