Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / E / 2

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.