Let $(U_i)_{i\in I}$ be an <open cover> of $f(X)$, with the sets open in $Y$. The <preimages> $f^{-1}(U_i)$ form an <open cover> of $X$, because $f$ is <continuous>. By <compactness>, finitely many of these preimages cover $X$. Their corresponding sets $U_i$ cover $f(X)$, providing a finite <subcover>. Hence \b[the <continuous> image is <compact>].

Solution

Past exam of the options I course of the University of Oxford / 2009 / E / E2 / a / ii / Solution

 Solution (source code)