Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know why this is true? I know that the closure of a set in a metric space is the set of points a distance of 0 away from the original set. Also, a discrete set is one with a neighborhood about any point whose union with the discrete set is that point.

share|improve this question

2 Answers 2

Let $d\colon X \times X \to \mathbf R$ be the metric, and for $x \in X$ and $r > 0$ define $$ B(x, r) = \{y \in X : d(x, y) < r\} $$ as usual. Writing $D$ for this discrete subset, suppose we have a non-empty open set $U \subset \overline{D}$. There must be an $a \in U \cap D$; find an $r > 0$ such that $B(a, r)$ is contained in $U$ and contains no other points of $D$. Since there are no isolated points, there is a $y \neq a$ in $B(a, r)$. Can you find a ball around $y$ that doesn't intersect $D$?

share|improve this answer

If $D$ is the discrete set in question and $x$ is in $D$, let $U$ be an open neighborhood of $\{x\}$ disjoint from $D\setminus\{x\}$. Then $U$ is also disjoint from $\overline{D\setminus\{x\}}$. In particular, this means that $\{x\}$ is not in $\overline{D\setminus\{x\}}$, which implies that $\overline{D\setminus\{x\}}=\overline D\setminus\{x\}$. Thus $U$ is disjoint from $\overline D\setminus\{x\}$. Since $x$ is not an isolated point, this implies that $U$ is not contained in $\overline D$. Considering that $U$ could be taken to be any sufficiently small ball centered at $\{x\}$, this shows that $x$ is not in the interior of $\overline D$.

Suppose that $y$ is an element of $\overline D$, and $V$ is a neighborhood of $y$. Then there is an $x$ in $V\cap D$. Since $V$ is a neighborhood of $x$, $V$ is not contained in $\overline D$ by the previous paragraph. Therefore $y$ is not an interior point of $\overline D$. Since a closed set is nowhere dense if and only if it has empty interior, this shows that $\overline D$ is nowhere dense.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.