0

Is $\mathbb{R}$ nowhere dense in $\mathbb{R}^2$?. The definition I have of nowhere dense sets is that a set is nowhere dense if its exterior is dense (everywhere dense). From the definition I've deduced that a set $A$ is nowhere dense in a topological space $(X, \tau)$ if $\bar{A}^º=\emptyset$ (interior of the closure is empty). I now that $\mathbb{R}$ isn't a subset of $\mathbb{R}^2$, but that's how the question is stated, so I'll treat it as $\mathbb{R}\times\{0\}$, which is closed because its complement is the union of two open sets. However, no ball is contained in $\mathbb{R}\times\{0\}$, so the interior is empty and $\mathbb{R}\times\{0\}$ is nowhere dense in $\mathbb{R}^2$. Is this correct?

Valere
  • 1,153
  • 6
    Yes, $\Bbb R\times {0}$ is nowhere dense in $\Bbb R^2$ because it's a closed set with empty interior. – Sassatelli Giulio Oct 30 '22 at 10:45
  • 1
    If you're trying to learn topological notions, then of course you'll want to work with the closure and interior operations. However, if this question arose from some general pursuit of yours or in a course not directly connected with learning and gaining experience with these topological notions, then the characterizations given in this answer are much better to work with for your spaces -- less abstract and makes the result trivial, as well as easy to apply to many other subsets of ${\mathbb R}^2$ (and usually for metric spaces in general). – Dave L. Renfro Oct 30 '22 at 11:55

0 Answers0