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

A subset $A$ is compact when every open cover of $A$ has a finite subcover. This may be stated using sets open in the ambient space or using the subspace topology on $A$; the definitions agree because every subspace-open set is the intersection of $A$ with an ambient-open set.

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 the continuous image is compact.

The family contains the empty set and the entire real line. The intersection of two proper nonempty members isso it is closed under finite intersection. For an arbitrary nonempty family of proper rays, put $c={\inf }_i{a}_i$. If the thresholds are unbounded below, the union is $\mathbb{R}$; otherwise the union is $(c,\infty )$. In the latter case $c$ itself is in no ray, and every point greater than $c$ lies above some threshold, proving the equality. Empty unions and unions involving $\mathbb{R}$ cause no difficulty. Thus arbitrary unions also belong to the family, establishing the topology axioms.

If $m=\min A$, any open cover contains a member containing $m$. Such a member is either $\mathbb{R}$ or a ray $(a,\infty )$ with $a<m$. It therefore contains every element of $A$, so that single member is already a subcover.

Conversely suppose $A$ is nonempty and has no smallest element. Thenis an open cover of $A$: for any $x\in A$, the lack of a minimum gives an $a\in A$ below $x$. A nonempty finite subcollection has union $(a_0,\infty )$, where $a_0$ is the smallest of its thresholds. It misses $a_0\in A$. The empty subcollection cannot cover nonempty $A$ either. Hence there is no finite subcover, and $A$ is not compact. Thus compactness is equivalent to having a smallest element.