= Solution
{c}
A <limit point> $p$ of $E\subseteq\mathbb R$ is a point such that, for every $\varepsilon>0$, the punctured <neighbourhood> $(p-\varepsilon,p+\varepsilon)\setminus\{p\}$ meets $E$. A <closed set> contains all of its <limit points>.
Every nonempty <open interval> contains a <rational number>; applying this to $(p,p+\varepsilon)$ shows that every $p\in\mathbb R$ is a <limit point> of $\mathbb Q$. Thus \b[the <limit points> of $\mathbb Q$ form $\mathbb R$, and $\mathbb Q$ is not closed].
If $p\in\mathbb Z$, radius $1/2$ excludes all other <integers>. If $p\notin\mathbb Z$, its distances to the two neighbouring <integers> have a positive minimum, and a smaller radius excludes every <integer>. Thus \b[$\mathbb Z$ has no <limit points> in $\mathbb R$ and is closed].
Back to article page
