There is this statement in Wikipedia, that states $S=\{\frac{1}{n}:n\in\mathbb{N}\}$ is nowhere dense in $\mathbb{R}$. Although the points get arbitrarily close to $0$, the closure of the set is $S\cup\{0\}$, which has empty interior.
My question is why $S\cup\{0\}$ has empty interior? Is $\{0\}$ the boundary of $S\cup\{0\}$? And $S$ is the interior of $S\cup\{0\}$ which is not empty?
I am a bit confused, could somebody please give some explanation on this? Thanks.
On
The interior point $x_0$ of set $S$: there exists a $\epsilon-$interval $(x_0-\epsilon,x_0+\epsilon)$ which is subset of $S$. But you can see that for all point $\frac{1}{n}$ or $0$, there is no such interval because in the arbitrary interval, there is a irrational number, which is not an element of $S$.
By the definition of boundary, you have $S \cup \{0\}$ is its boundary.
On
The standard topology on $\Bbb R$ is the collection of subsets that is the arbitrary union of open intervals $(a,b)$ where $a<b$ and $a,b\in\Bbb R\cup\{-\infty,+\infty\}$, more the emptyset. These sets are called open sets.
Now, the interior of a set is the union of all open sets contained in the set. But the open sets, in this topology, are composed by open intervals and the empty set. So the unique open set contained in the set of the question is the empty set.
Then the interior is the empty set.
A point $x$ is in the interior of this set if you can find some small $\epsilon$ such that the entire interval $(x-\epsilon,x+\epsilon)$ is contained in the set. However, $S$ contains no intervals.