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I tried to prove the following theorem whilst having $\mathbb{Z}$ in mind. However, it is by no means a rigorous proof, and because of that, I have the feeling that I may be missing something that cannot be shown by intuitive arguments:

Let $G\subseteq\mathbb{R}$, let $\bar{G}$ stand for the closure of $G$, and let $G^c$ stand for the complement of $G$.


  • $G$ is dense in $\mathbb{R}$ if and only if $\bar{G}=\mathbb{R}$.
  • $G$ is nowhere-dense in $\mathbb{R}$ if $\bar{G}$ contains no nonempty open intervals.

Theorem. A set $G$ is nowhere-dense in $\mathbb{R}$ if and only if the complement of $\bar{G}$ is dense in $\mathbb{R}$.

Proof. Let $G$ be nowhere-dense in $\mathbb{R}$. Then $\bar{G}$ contains no nonempty open intervals. This implies that $G$ must be composed of isolated points, and that $\bar{G}^c$ must contain infinitely many nonempty open intervals separated by these isolated points. Therefore, $\bar{\bar{G}^c}=\mathbb{R}$, and $\bar{G}^c$ is dense in $\mathbb{R}$. Conversely, let $\bar{\bar{G}^c}=\mathbb{R}$. It follows that either $\bar{G}^c=\mathbb{R}$ or $\bar{G}^c$ is composed of infinitely many open intervals that are separated by isolated points. If $\bar{G}^c=\mathbb{R}$, then $\bar{G}=G=\emptyset$, which is nowhere-dense in $\mathbb{R}$. If $\bar{G}^c$ is composed of infinitely many open intervals separated by isolated points, then $\bar{G}=G$ is a set composed of isolated points, which is nowhere-dense in $\mathbb{R}$. $\square$

What do you guys think?

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The proof is flawed. See my edited answer. – Alex Becker Feb 23 '12 at 16:18
up vote 1 down vote accepted

The statement "This implies that G must be composed of isolated points" is false. To see this, consider the set $\{1,\frac{1}{2},\frac{1}{3},\ldots\}\cup\{0\}$ which is nowhere dense, yet $0$ is not an isolated point.

However, the statement you are trying to prove is correct. Proof: Note that a subset is dense iff it contains points arbitrarily close to any point in the space. Suppose $G$ is nowhere dense in $\mathbb R$, so $\bar{G}$ contains no interval. Then $\bar{G}^c$ intersects every open interval, so for any $r\in\mathbb R$ and $\epsilon>0$ we have $\bar{G}^c\cap(r-\epsilon,r+\epsilon)\neq \emptyset$, thus $\bar{G}^c$ is dense. Suppose $G$ is not nowhere dense in $\mathbb R$, so $\bar{G}$ contains some open interval $I$. Then $I\cap \bar{G}^c=\emptyset$, so $\bar{G}^c$ is not dense.

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