Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / E (Topology)

A connected space has no separation into two disjoint nonempty open subsets whose union is the space. A continuous surjection to the discrete space $\{0,1\}$ produces such a separation by the two point preimages. Conversely, from a separation define a function equal to zero on one open piece and one on the other; it is continuous and surjective. Thus these formulations of connectedness are equivalent.

An interval in $\mathbb{R}$ is connected: if a continuous map from it to $\{0,1\}$ took both values, the intermediate value theorem would force value $1/2$, impossible. Conversely, a subset of $\mathbb{R}$ that is not an interval has points $a<b$ in it and a point $c\in (a,b)$ outside it. Intersecting it with $(-\infty ,c)$ and $(c,\infty )$ separates it. The empty set is connected, and singleton intervals are included.

For the union assertion, any separation of $X$ must put each nonempty connected $A_i$ entirely on one side, since intersecting the separation with $A_i$ would otherwise disconnect it. Pairwise nonempty intersections force all the $A_i$ onto the same side. Their union is $X$, contradicting a nonempty second side. The empty-union case is already connected.

The plane is the union of all lines through the origin. Each line is homeomorphic to $\mathbb{R}$ and connected, and every pair meets at the origin, so ${\mathbb{R}}^2$ is connected. For the punctured plane, let $R_{+}$ be the positive horizontal ray, and for each $r>0$ let $C_r$ be the circle of radius $r$. Each circle is a continuous image of $[0,2\pi ]$ and hence connected; $C_r\cup R_{+}$ is connected because its two connected pieces meet at $(r,0)$. These unions all contain $R_{+}$ and cover the punctured plane, proving it connected as well.

Finally suppose a continuous injection $g:{\mathbb{R}}^2\to \mathbb{R}$ existed. Its image is a connected interval, since a continuous image of a connected space is connected. It contains at least three distinct values; choose $p$ whose value lies strictly between two others. The punctured plane ${\mathbb{R}}^2\setminus \{p\}$ is connected by translation of the preceding result. By injectivity its image is precisely $g({\mathbb{R}}^2)\setminus \{g(p)\}$, which is disconnected by the missing interior point. This contradicts connectedness of a continuous image. Therefore

A sequentially compact space is one in which every sequence has a subsequence converging to a point of the space. A limit point $\ell$ of $S$ means every open ball around $\ell$ meets $S\setminus \{\ell \}$.

In a compact metric space, a sequence with a finite set of values has a constant subsequence. If its value set is infinite and had no limit point, each point of the space would have a neighbourhood containing at most one point of that value set. A finite subcover would then make the value set finite, a contradiction. It therefore has a limit point in the space, and the permitted subsequence result gives a convergent subsequence. This proves sequential compactness.

For the remaining assertions assume the compact space is nonempty, as required for a fixed-point claim. The displacement function $q(x)=d(f(x),x)$ is continuous by the stated continuity of the metric and of $f$. The extreme value theorem gives a minimum $\delta =q(p)$. If $\delta >0$, then $p\ne f(p)$, and strict decrease of distances impliescontradicting minimality. Thus $\delta =0$ and $p$ is a fixed point. Two distinct fixed points would satisfy $d(p,q)=d(f(p),f(q))<d(p,q)$, also impossible. Hence the fixed point is unique.

On the noncompact real line, takeIts derivative $e^x/(1+e^x)$ is strictly between zero and one. The mean value theorem therefore gives $|f(x)-f(y)|<|x-y|$ for each distinct pair. Yet $f(x)-x=\log (1+e^{-x})>0$, so it has no fixed point. Compactness, rather than merely pointwise strict shrinkage, supplies attainment of a zero displacement in the preceding proof.