Past exam of the options II course of the University of Oxford / 2009 / E / E3

A subset $U$ of a metric space is open if every $x\in U$ has a radius $r>0$ with the open ball $B_d(x,r)\subseteq U$.

Continuity at a point means: for every $\epsilon >0$ there exists $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ implies $d_Y(f(x),f(x^{\prime }))<\epsilon$.

The function is continuous if it is continuous at every point of $X$. Equivalently, the preimage of each open set in $Y$ is open in $X$.

Nonnegativity and symmetry follow from $d$, and $d_a(x,y)=0$ with $x\ne y$ would force both $x=a$ and $y=a$, a contradiction. For the triangle inequality, when the intermediate point $z$ is different from both endpoints,If $z$ equals an endpoint the inequality is immediate, and if $x=y$ its left target is zero. These cover every case, so $d_a$ is a metric.

This is the star metric centered at a point.

If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$, so no such $x$ lies in the strict open ball. Its centre always does. HenceThus every point other than $a$ is isolated.

The singleton calculation shows that every map out of $(X,d_a)$ is continuous at each point other than $a$: use the domain ball consisting only of that point. Thus global continuity in this metric is equivalent to continuity at $a$, proving $[\alpha ]⟺[\beta ]$.

For every $x$, including $x=a$, one has $d_a(a,x)=d(a,x)$. Therefore the open balls centred at $a$ are identical for $d_a$ and $d$. The epsilon-delta definitions of continuity at $a$ are identical, giving $[\beta ]⟺[\gamma ]$. Hence all three conditions are equivalent.

Every point of $A$ differs from $a$, so its singleton is an open ball by part (ii). The union of these singleton open sets is $A$. Thus $A$ is open whenever $a\notin A$.

If $A$ is open and contains $a$, it contains some $d_a$-ball about $a$, which is exactly the corresponding $d$-ball. Conversely, if $B_d(a,\epsilon )\subseteq A$, this is a $d_a$-neighbourhood of $a$, while every other point of $A$ has its open singleton inside $A$. Hence $A$ is open. Thus the stated ball-containment condition is necessary and sufficient.

Every point other than $a$ is isolated, so no such point can enter the closure unless it was already in $B$. The only possible extra point is $a$, and it lies in the closure precisely when every $d$-ball about $a$ meets $B$. Therefore the three answers, in the printed order, areIn the middle case, a ball of radius below $d(a,B)$ misses $B$; in the last case the infimum zero makes every positive-radius ball meet $B$. If $B$ is empty its closure is empty as usual, with the convention $d(a,\varnothing )=+\infty$.