Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 4

In the subspace topology, $B$ is relatively open in $A$ if $B=A\cap U$ for an open subset $U$ of the ambient Euclidean space. Equivalently, every $b\in B$ has a radius $\epsilon >0$ with $A\cap B_{\epsilon }(b)\subseteq B$. It is relatively closed if its complement $A\setminus B$ is relatively open, equivalently $B=A\cap F$ for an ambient closed set $F$.

For the upper half of the open disk, $B$ is closed in $A$, but not open in $A$. It is the intersection with the closed half-plane $\mathrm{Im}z\ge 0$. A relative neighborhood of zero contains points of negative imaginary part, so relative openness fails.

The rational cut is both open and closed in $\mathbb{Q}$. It is relatively open as $\mathbb{Q}\cap (-\infty ,\sqrt{2})$. Since $\sqrt{2}$ is irrational, its relative complement is $\mathbb{Q}\cap (\sqrt{2},\infty )$, also relatively open. Thus this cut is a clopen set in the subspace topology of the rationals.

A set is compact if every open cover of it has a finite subcover. Let $K_1,\dots ,K_m$ be compact and let $\mathcal{U}$ cover their union. For each $j$, the same family covers $K_j$, so compactness supplies a finite subfamily ${\mathcal{U}}_j$ covering $K_j$. The union of these finitely many finite subfamilies is finite and covers ${\bigcup }_j{K}_j$. Hence a finite union of compact sets is compact.

An arbitrary union need not be compact. Each singleton $\{n\}$, $n\ge 1$, is compact, but their union $\mathbb{N}$ has the open cover $\{(n-1/3,n+1/3):n\ge 1\}$ with no finite subcover. Every member covers only one positive integer, so finitely many leave infinitely many integers uncovered.