Past exam of the options I course of the University of Oxford / 2010 / E / E1

A topology 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 is closed precisely when its complement in $X$ is open.

The subspace topology on $Y$ isIf $Y$ is open in $X$, every such intersection is open in $X$. Conversely, every ambient open set contained in $Y$ equals its intersection with $Y$. Thus the relatively open sets are exactly the ambient open subsets contained in $Y$.

The closure of $A$ is the smallest closed subset of $X$ containing it:Equivalently, a point belongs to $\overline{A}$ exactly when every open neighborhood of it meets $A$.

For $(i)\Rightarrow (ii)$, suppose $B=U\cap F$ with $U$ open and $F$ closed. Since $B\subseteq F$, also $\overline{B}\subseteq F$. Therefore $U\cap \overline{B}=B$, showing that $B$ is relatively open in its closure.

For $(ii)\Rightarrow (iii)$, write $B=U\cap \overline{B}$ with $U$ open in $X$. Then $\overline{B}\setminus B=\overline{B}\cap (X\setminus U)$ is an intersection of closed sets, hence closed in $X$.

For $(iii)\Rightarrow (i)$, set $F=\overline{B}$ and $U=X\setminus (\overline{B}\setminus B)$. The assumed condition makes $U$ open; $F$ is closed and $U\cap F=B$. Thus the three conditions are equivalent. They characterize a locally closed subset.