Why the boundary set of a set can have more elements of the set in question?

Question

We know by definition, the boundary point of $S \subset \mathbb{R}$ is any point $x$ (in every neighborhood of $N$) such that $N \cap S \neq \emptyset$ and $N \cap (\mathbb{R} \setminus S)$. This definition tells us what points are in the boundary set but it does not tell us what is the boundary set. For this reason I am having difficulty to see why $\partial \mathbb{Q} = \mathbb{R}$. I can see at least $\partial \mathbb{Q} \subset \mathbb{Q}$ is true. To better show my confusion, I thought for any set $S$ a subset of $\mathbb{R}$, $\partial S$ is at most $S$.

Answer 1

Given a point $x$ a neighborhood of $x$ must include a ball of some radius $r$ around $x$, so it includes the interval $(x-r,x+r)$. Whether $x$ is rational or irrational, this interval will include both rationals and irrationals, so $x$ is part of the boundary of $\Bbb Q$. This shows that $\partial \Bbb Q$ is all of $\Bbb R$.

Your last two sentences are not correct. We have just shown that $\partial \Bbb Q \not \subset \Bbb Q$. Probably you meant $\Bbb Q \subset \partial \Bbb Q$. The boundary of $S$ is not at most $S$. It may not even be part of $S$. If $S=(0,1), \partial S = \{0,1\} \not \subset S$