Past exam of the options I course of the University of Oxford / 2010 / E (Topology)

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.

A space is Hausdorff if any two distinct points have disjoint open neighborhoods: for $x\ne y$, there are open $U\ni x$ and $V\ni y$ with $U\cap V=\varnothing$.

The map $f:X\to Y$ is continuous ifThis condition concerns inverse images of open sets; images of open sets need not be open.

Let $p:X\to X/R$ send a point to its equivalence class. The quotient topology isIt is the finest topology on the quotient making $p$ continuous.

For each continuous map to a Hausdorff target, equality of its values is reflexive, symmetric and transitive. The relation $R$ requires all these equalities simultaneously. Explicitly, $f(x)=f(x)$ gives reflexivity; $f(x)=f(y)$ implies $f(y)=f(x)$; and $f(x)=f(y)=f(z)$ gives $f(x)=f(z)$. These arguments apply to every map in the defining family, proving $R$ is an equivalence relation.

If $x\in A=g^{-1}(B)$ and $yRx$, then $g(y)=g(x)\in B$, so $y\in A$. Thus $A$ contains every equivalence class it meets: it is a saturated subset. Always $A\subseteq p^{-1}(p(A))$; conversely, a point of the latter set is equivalent to some $x\in A$ and hence belongs to $A$. Therefore

Take distinct classes $p(x)\ne p(y)$. The points are not related by $R$, so by the defining negation there is a Hausdorff space $Z$ and a continuous map $f:X\to Z$ with $f(x)\ne f(y)$. Choose disjoint open neighborhoods $U$ and $V$ of those two values. Their inverse images $A=f^{-1}(U)$ and $C=f^{-1}(V)$ are disjoint open subsets of $X$ and are saturated by part (ii).

Hence $p^{-1}(p(A))=A$ and $p^{-1}(p(C))=C$, making $p(A)$ and $p(C)$ open in the quotient topology. They contain the chosen classes and are disjoint: an equivalence class meeting both would contradict saturation and $A\cap C=\varnothing$. Thus $X/R$ is Hausdorff. This quotient is the Hausdorff reflection by continuous maps; every continuous map from $X$ to a Hausdorff space factors uniquely through it.