Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 1

A homeomorphism $f:M\to N$ is a bijection for which both $f$ and $f^{-1}$ are continuous maps, using the respective metric topologies.Continuity of a bijection alone is insufficient: continuity of the inverse is a separate requirement.

The closure $\overline{A}$ is the smallest closed subset of $M$ containing $A$, equivalently the intersection of all such closed subsets. In a metric space it is characterized byIf some ball misses $A$, its centre lies outside a closed superset of $A$; conversely the complement of any closed superset is an open neighbourhood disjoint from $A$. This verifies the equivalence of the two descriptions.

False. Removing a point from the real line leaves two disjoint nonempty open rays, so the remainder is disconnected. Removing a point from the Euclidean plane leaves a path-connected space: join any two remaining points by a line segment if it avoids the removed point, or by two segments through a point not on either of the two lines through the removed point and the chosen endpoints.

A proposed homeomorphism would restrict to a homeomorphism between these punctured spaces and preserve connectedness, which is impossible. Hence

False: the target need not equal the image. Take the one-point compact space $M=\{0\}$, $N=\mathbb{R}$, and the continuous map $f(0)=0$. The target is not compact, for example because it is unbounded.

The valid theorem is that a continuous image of a compact space is compact. Thus $f(M)$ is compact; compactness of $N$ follows if surjectivity is added.

True. For $u,v\in f(M)$ choose preimages $x,y\in M$. A path $\gamma :[0,1]\to M$ joining $x$ to $y$ exists because $M$ is path-connected. The composition of continuous functions $f\circ \gamma$ is a path in $f(M)$ from $u$ to $v$.Surjectivity onto a larger target is unnecessary because the conclusion concerns the image itself.

True. For any continuous map, $f(\overline{A})\subseteq \overline{f(A)}$: every open neighbourhood of $f(x)$ pulls back to an open neighbourhood of $x\in \overline{A}$, and therefore meets $A$ in its preimage.

Apply the same inclusion to the continuous inverse of the homeomorphism:Applying $f$ gives the reverse inclusion, provingThe closures are taken in their respective ambient spaces.

True. For every finite partial sum,The nondecreasing partial sums on the left are bounded, so the square series converges. Hence ${\ell }^1\subseteq {\ell }^2$ andApplying this norm inequality to $x-y$ proves $\Vert f(x)-f(y){\Vert }_2\le \Vert x-y{\Vert }_1$. Thus the inclusion is a Lipschitz continuous map, in particular continuous, and it is well-defined on every element of l1 sequence space.