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

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.