Why $\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}$ has empty interior?

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.

• The interior is the largest open subset. No subset of $S$ is open in the reals. Including zero does not change this. Sep 29, 2016 at 2:39
• $\partial S = \{0\} \cup S$. Sep 29, 2016 at 2:54

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.

The interior, if not empty, is an uncountable set.

• I think this is misleading without more context. This is true just for some topologies on some sets. If you said that the OP can think that this is a general statement for any topology, what is not true. Sep 29, 2016 at 2:45
• The context is, of course, that of the question I am answering. Were it another, then surely I should have made it explicit. Sep 29, 2016 at 2:46
• To clarify this method of approaching the problem to the OP: Any open subset of the real numbers (or $\mathbb{R}^n$) is uncountable as a set, except of course the empty set. Sep 29, 2016 at 2:47
• Anyway I must insist that an answer as this is not good. You can extend the phrase to "... for the standard topology on R". If the Op dont understand what is an interior it is obvious that it knowledge on topology is not so much. Sep 29, 2016 at 2:47
• I am pretty sure that you can write an amazing answer which would satisfy you, @masacroso. Sep 29, 2016 at 2:51

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.

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.