Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / E / E3

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 $d(a,\varnothing )=+\infty$.