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

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.

A Hausdorff space is one in which any distinct points $x,y$ have disjoint open neighborhoods: there exist open sets $U,V$ with $x\in U$, $y\in V$ and $U\cap V=\varnothing$.

A map $f:X\to Y$ is continuous if the inverse image of every open subset of $Y$ is open in $X$:

For the natural projection $p:X\to X/R$, the quotient topology declares $U\subseteq X/R$ open exactly when $p^{-1}(U)$ is open in $X$. It is the finest topology on the quotient making $p$ continuous.

Every map has $f(x)=f(x)$, so the relation is reflexive. Equality is symmetric, so $f(x)=f(y)$ for all the specified maps is equivalent to $f(y)=f(x)$; the relation is symmetric. If $x\mathcal{R}y$ and $y\mathcal{R}z$, every continuous map to every Hausdorff space has $f(x)=f(y)=f(z)$, proving $x\mathcal{R}z$. Thus $\mathcal{R}$ is an equivalence relation.

If $x\in A=g^{-1}(B)$ and $y\mathcal{R}x$, the definition of the relation applies in particular to the fixed continuous map $g$ into the fixed Hausdorff space $Y$. It gives $g(y)=g(x)\in B$, hence $y\in A$. Thus $A$ is a union of whole equivalence classes, also called a saturation of a subset.

Always $A\subseteq p^{-1}(p(A))$. Conversely, a point in $p^{-1}(p(A))$ shares its class with some $x\in A$, and the preceding argument places it in $A$. Hence

Take distinct classes $p(x)\ne p(y)$. The failure of $x\mathcal{R}y$ supplies a Hausdorff space $Y$ and a continuous map $g:X\to Y$ with $g(x)\ne g(y)$. Choose disjoint open neighborhoods $V,W$ of these two images. Their inverse images $A=g^{-1}(V)$ and $B=g^{-1}(W)$ are disjoint open subsets of $X$, and part (ii) says both are saturated.

The quotient topology therefore makes $p(A)$ and $p(B)$ open, since their inverse images are precisely $A,B$. They are disjoint as well: an equivalence class meeting both would, by saturation, be contained in both. They contain $p(x)$ and $p(y)$ respectively. Thus $X/\mathcal{R}$ is Hausdorff. This construction is the Hausdorff reflection by continuous maps.