Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / E / 2 (E2)

A subset is compact if every family of open sets covering it has a finite subfamily still covering it. Equivalently, its subspace topology has the finite-subcover property.

Let $\{U_i\}$ be an open cover of $f(X)$. By continuity the inverse images $f^{-1}(U_i)$ are open and cover $X$. Compactness provides finitely many of them covering $X$, and the corresponding $U_i$ then cover $f(X)$. Thus a continuous image of a compact space is compact.

The family contains the whole space and the empty set. A finite intersection of nontrivial members is $({\max }_j{a}_j,\infty )$. A nonempty union of rays is $({\inf }_j{a}_j,\infty )$ when the infimum is finite, and is the whole line when the infimum is $-\infty$. Empty unions and unions involving the whole space are already included. Hence $\tau$ is a topology. This is the upper-ray topology on the real line.

If $A$ has a smallest element $a_0$, any open cover has a member containing $a_0$. This member is either the whole line or a ray $(c,\infty )$ with $c<a_0$, so it already contains all of $A$. Thus a single member suffices and $A$ is compact.

If $A$ has no smallest element, the family $\{(a,\infty ):a\in A\}$ covers $A$: each point has another point of $A$ below it. A finite subfamily has union $({\min }_j{a}_j,\infty )$, which misses its smallest listed endpoint, itself a point of $A$. No finite subcover exists. Therefore a nonempty subset is compact exactly when it has a smallest element in this topology.