The circle is <path-connected>, using circular arcs, and hence <connected>. Its <continuous> image is <connected> by part (ii). A <connected> subset of the <real line> is an interval: if it contained values on opposite sides of a missing value $c$, <intersection> with $(-\infty,c)$ and $(c,\infty)$ would separate it. Thus \b[$f(C)$ is an interval]. It is also <compact>, so it is a closed bounded interval, possibly reduced to one point.

Solution

Past exam of the options I course of the University of Oxford / 2008 / E / E2 / iii / Solution

 Solution (source code)