Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / E (Topology)

A continuous map $f:X\to Y$ is one for which $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$. This defines continuity using the two topologies, without a metric or a coordinate description.

For metric spaces, continuity at $x_0$ means that for every $ϵ>0$ there is $\delta >0$ such thatThe choice of $\delta$ may depend on both $ϵ$ and $x_0$.

A basis of a topology is a collection $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, for every open $U$ and $y\in U$ there is $B\in \mathcal{B}$ with $y\in B\subseteq U$.

A map $f:X\to Y$ is continuous exactly when $f^{-1}(B)$ is open in $X$ for every basic set $B$ of $Y$. Necessity follows from the definition of a continuous map. For sufficiency, express an arbitrary open set as a union of basic sets and use preservation of unions by inverse images.

Every open interval is open in the usual metric topology. Conversely, if $U\subset \mathbb{R}$ is open and $x\in U$, there exists ${ϵ}_x>0$ with $(x-{ϵ}_x,x+{ϵ}_x)\subseteq U$. Taking the union of these intervals over all $x\in U$ recovers $U$. Thus open intervals form a basis of the usual topology.

If $f$ is continuous, the two sets are inverse images of the open rays $(-\infty ,\lambda )$ and $(\lambda ,\infty )$, so they are open. Conversely, suppose both families of inverse images are open. For any interval $(a,b)$,which is open. By part (c)(i) and the basis of a topology criterion from part (b), all inverse images of open sets are open. Hence $f$ is continuous.

A connected set is one whose relative topology admits no decomposition into two disjoint nonempty open sets. A path-connected set permits, between any two points, a continuous map $\gamma :[0,1]\to A$ with the prescribed endpoints.

Path connectedness implies connectedness, because a continuous image of an interval is connected. The converse can fail, for example for the topologist's sine curve. On open subsets of Euclidean spaces ${\mathbb{R}}^n$, the two properties are equivalent; more generally this holds in locally path-connected spaces.

If $f(A)$ had a separation into two nonempty relatively open disjoint pieces, their inverse images under the restriction $f{|}_A$ would be two nonempty relatively open disjoint subsets covering $A$. This would contradict connectedness of $A$. Thus a continuous image of a connected set is connected.

The unit circle is path-connected, since it is the continuous image of the interval $[0,2\pi ]$ under $\theta \mapsto (\cos \theta ,\sin \theta )$. It is therefore connected, and part (ii) makes $f(C)$ connected. A connected subset of $\mathbb{R}$ is an interval: omission of a point lying between two of its points would separate it by the two open rays about that point.

Moreover, the circle is compact, so the continuous image of a compact space is compact. Hence here the image is a closed bounded interval, possibly a singleton.