Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / E (Topology)

A subset $U$ of a metric space is open if for every $x\in U$ there exists $r>0$ with $B_d(x,r)\subseteq U$, where $B_d(x,r)=\{y:d(x,y)<r\}$. The radius may depend on the point. The empty set is open vacuously.

Continuity at $x$ means that for every $\epsilon >0$ there is $\delta >0$ such that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\epsilon$ for every $y\in X$. Equivalently, the inverse image of each neighborhood of $f(x)$ is a neighborhood of $x$.

The map is continuous if it is continuous at every point of its domain. Equivalently the inverse image of every open subset of $Y$ is open in $X$. These equivalent formulations use the given metrics' induced topologies.

Write $r_x=d(x,a)$. The prescribed function is symmetric and nonnegative. If $x\ne y$, both $r_x,r_y$ could be zero only if $x=y=a$, so $d_a(x,y)>0$. For the triangle inequality, if any two of $x,y,z$ coincide it is immediate. If all three are distinct,Thus $\boxed{d_a\text{ is a metric}}$. Also $d_a(a,x)=d(a,x)$, a fact useful in the following parts.

If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$. The strict open-ball inequality therefore excludes every such $x$, while including $b$. Hence $\boxed{B_{d_a}(b,d(b,a))=\{b\}}$. Every point other than $a$ is isolated in this metric.

If continuity holds everywhere, it holds at $a$. Conversely, if it holds at $a$, it holds at every other point automatically: the singleton neighborhood from (ii) maps entirely to the single value at that point. Thus continuity everywhere in $d_a$ is equivalent to continuity at $a$ in $d_a$.

For every $r>0$ one has $B_{d_a}(a,r)=B_d(a,r)$ because $d_a(a,x)=d(a,x)$. The epsilon–delta definition at $a$ is therefore identical for the two domain metrics. In particular continuity at $a$ in $d_a$ implies continuity there in $d$.

The same equality of balls gives the reverse implication: continuity at $a$ in $d$ implies continuity at $a$ in $d_a$. The latter implies continuity everywhere by the first part. These implications complete the equivalence of all three printed conditions.

If $a\notin A$, every point of $A$ is isolated by (ii). The set $A$ is the union of these open singletons, so $\boxed{A\text{ is open in }(X,d_a)}$, including the empty case.

If $A$ is open and contains $a$, there is an epsilon ball at $a$ contained in it. This ball is exactly $B_d(a,\epsilon )$ by the equality of balls proved above. Conversely such a ball supplies a neighborhood at $a$, and all other points of $A$ have their singleton neighborhoods inside $A$. Thus the stated ball condition is necessary and sufficient.

An isolated point outside $B$ cannot belong to its closure. Thus the only possible extra closure point is $a$. The balls at $a$ meet $B$ exactly when arbitrarily small original-metric balls at $a$ do so. ConsequentlyIn the first case $a$ is already included. In the second case a ball smaller than the positive distance separates it from $B$; in the last case every ball meets $B$. If $B$ is empty its closure is empty, with distance convention $d(a,\varnothing )=+\infty$.

A subset is connected if it has no decomposition into two disjoint nonempty sets both relatively open in it. Equivalently its only relatively clopen subsets are itself and the empty set. Connectedness here refers to the subset's induced topology.

Suppose $f(X)=U\cup V$ were a separation into two disjoint nonempty relatively open sets. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$, and they are disjoint, nonempty and cover $X$. This contradicts connectedness of $X$. Hence $\boxed{f(X)\text{ is connected}}$.

The set $X/R$ consists of the equivalence classes, with canonical surjection $q:x\mapsto [x]$. The quotient topology declares $V\subseteq X/R$ open exactly when $q^{-1}(V)$ is open in $X$. This makes $q$ continuous and is the finest topology on the quotient set doing so.

Let $A$ be clopen and $C$ an equivalence class. The intersections $C\cap A$ and $C\setminus A$ are complementary relatively open subsets of $C$. Connectedness of $C$ means one must be empty. Thus every equivalence class is entirely in $A$ or entirely outside it, proving $\boxed{q^{-1}(q(A))=A}$. This is precisely saturation.

If $X$ is connected, its continuous surjective image under $q$ is connected by (a)(ii). Conversely suppose $X/R$ is connected and $A\subseteq X$ is clopen. By (i) both $A$ and its complement are saturated. The quotient definition then makes $q(A)$ open, and its complement $q(X\setminus A)$ open too. Thus $q(A)$ is clopen in the connected quotient, so is empty or the whole quotient. Pulling back gives $A=\varnothing$ or $X$. By the clopen criterion $X$ is connected. Hence $\boxed{X\text{ connected}⟺X/R\text{ connected}}$.

The difference map $(z,w)\mapsto z-w$ is continuous and $X$ is its inverse image of the open set $\mathbb{C}\setminus \{0\}$, so $X$ is open in ${\mathbb{C}}^2$. For fixed $z$, the equivalence class is $\{z\}\times (\mathbb{C}\setminus \{z\})$, homeomorphic to a punctured plane. It is path-connected: two points can be joined by a polygonal path avoiding the missing point. Thus all equivalence classes are connected.

The projection $\pi$ is continuous and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in ${\mathbb{C}}^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi (U)$. Hence $\pi (U)$ is open.

Its fibers are precisely the equivalence classes, so it induces a bijection $\bar{\pi }:X/R\to \mathbb{C}$. Continuity follows from the quotient topology and $\pi =\bar{\pi }\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar{\pi }(V)=\pi (q^{-1}(V))$ is open. Thus the inverse is continuous too, proving $\boxed{X/R\cong \mathbb{C}}$ as topological spaces. The plane is connected, so the connected-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.