= Solution
{c}
A point is in the <closure> $\overline E$ exactly when every <neighbourhood> meets $E$. This is equivalent to belonging to $E$ or being a <limit point> of $E$.
If $p\notin\overline E$, choose $\varepsilon>0$ with $(p-\varepsilon,p+\varepsilon)\cap E=\varnothing$. Every $q$ with $|q-p|<\varepsilon/2$ has a smaller <neighbourhood> contained in this interval and disjoint from $E$. Thus $(p-\varepsilon/2,p+\varepsilon/2)$ misses $\overline E$, and the complement of $\overline E$ is <open>. Hence $\overline E$ is <closed>.
Every <closed set> $F$ containing $E$ contains all <limit points> of $E$: a <limit point> outside $F$ would have a <neighbourhood> missing $F$, and hence missing $E$. Therefore $\overline E\subseteq F$. Since $\overline E$ is itself one of these <closed sets>,
$$
\boxed{\overline E=\bigcap\{F:E\subseteq F,\ F\text{ closed in }\mathbb R\}.}
$$
Back to article page
