If $A$ is <open> and contains $a$, it contains some $d_a$-ball about $a$, which is exactly the corresponding $d$-ball. Conversely, if $B_d(a,\varepsilon)\subseteq A$, 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 \b[necessary and sufficient].

Solution

Past exam of the options II course of the University of Oxford / 2009 / E / E3 / b / iv / β / Solution

 Solution (source code)