Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / E / 1 (E1)

A topology $\mathcal{T}$ on $X$ contains $\varnothing$ and $X$, is closed under arbitrary unions, and is closed under finite intersections. Its members are the open sets. A subset $F\subseteq X$ is closed precisely when $X\setminus F$ is open.

The subspace topology on $Y$ isIf $Y$ is open in $X$, each such intersection is open in $X$. Conversely, an open subset of $X$ contained in $Y$ is its own intersection with $Y$. Thus the open sets of $Y$ are exactly the open subsets of $X$ contained in $Y$ in this case.

The closure $\overline{A}$ is the intersection of all closed subsets of $X$ containing $A$. It is the smallest such closed subset. Equivalently, $x\in \overline{A}$ exactly when every open neighborhood of $x$ meets $A$.

We prove (i)$\Rightarrow$(ii). If $B=U\cap F$ with $U$ open and $F$ closed, then $\overline{B}\subseteq F$. Henceso $B=U\cap \overline{B}$. By the subspace topology definition this makes $B$ open in its closure.

We prove (ii)$\Rightarrow$(iii). If $B$ is open in $\overline{B}$, its complement $\overline{B}\setminus B$ is closed in the subspace $\overline{B}$. A relatively closed subset of a closed subset of $X$ is closed in $X$: it is the intersection of that closed subset with another closed subset of $X$. Since a closure is closed, $\overline{B}\setminus B$ is closed in $X$.

We prove (iii)$\Rightarrow$(i). If $C=\overline{B}\setminus B$ is closed in $X$, set $U=X\setminus C$ and $F=\overline{B}$. Then $U$ is open, $F$ is closed, andTogether the three implications prove the equivalence. These are the standard characterizations of a locally closed subset.