How would I show that a set is not dense?

Question

Consider the Cantor set. I want to prove that it is not dense in $\mathbb{R}$. Now it makes sense to me since the Cantor set misses entire open intervals in $\mathbb{R}$. So my idea was to choose some open interval in the cantor set and show their is no element of the cantor set between the open interval. Not sure how to prove that though.

Answer 1

The cantor set, $\mathcal{C}$, on say $[0, 1]$ is a closed set. Suppose that $\mathcal{C}$ was dense in $[0, 1]$, then $\overline{\mathcal{C}} = [0,1]$. But $\overline{\mathcal{C}} = \mathcal{C}$. A contradiction since $\mathcal{C} \neq [0, 1]$.

Answer 2

The Cantor set is closed. A closed set is dense in the space iff it is the whole space.

In fact, the Cantor set is nowhere dense, since $C^{\circ}=\varnothing$ and $\bar C=C$, we have $(\bar C)^\circ =\varnothing$.

Answer 3

So my idea was to choose some open interval in the cantor set and show their is no element of the cantor set between the open interval.

Something went a bit haywire here. If you were to choose an "interval in the Cantor set" then that interval (being "in" the Cantor set) would obviously contain points of the Cantor set. Carefully review what "dense" means. If you know what "dense" means, then you'll see it isn't even dense in $[0,1]$.

$(\frac12,\frac{12}{20})$ is a nonempty open set that does not intersect the Cantor set.

Of course, there are many other choices for such an interval.