Past exam of the options II course of the University of Oxford / 2006 / E / 3

A subset is compact if every cover of it by open sets of the ambient space has a finite subcover; equivalently use open sets of its subspace topology. A cover of $A\cup B$ separately covers $A$ and $B$. Compactness of each supplies two finite collections, whose union is finite and covers $A\cup B$. If $A$ is closed in a compact space, add the open complement of $A$ to any open cover of $A$. Compactness supplies a finite subcover of the whole space, and discarding the complement leaves a finite cover of $A$.

For the enlarged space, every old open set is open in $C$ and has the same trace on $X$. The trace on $X$ of a new open set $V\cup \{\infty \}$ is $V$, which is open in $X$ because its complement is closed. Thus the induced topology on $X$ is exactly the original one.

Given an open cover of $C$, choose a member containing infinity. It has the form $V\cup \{\infty \}$ with $K=X\setminus V$ compact. The remaining cover restricted to $K$ has a finite subcover. Together with the chosen infinity-neighborhood, these finitely many sets cover $C$. Hence $C$ is compact, with no Hausdorff assumption needed.

Finally put $a=f(\infty )$. Continuity makes $U=f^{-1}((a-ϵ,a+ϵ))$ open and containing infinity. Therefore $U=V\cup \{\infty \}$ with $W=X\setminus V$ compact and closed in $X$. For every $x\in X\setminus W=V$ we haveThis expresses convergence to the value at infinity outside compact closed subsets.