# Correctness of topological reasoning (interior, closure and boundary of sets)

I have a set $S= \{(x, y) | x\in\mathbb{Q} \land y\in\mathbb{R}\}$. I am pretty sure I am correct, however please comment on my reasoning. I do not want to be terribly formal (proving each statement) I just want the correct mathematical intuitive argument. Here are my propositions:

1) $S$ is a closed set i.e. the closure $\bar{S}=S$

Take any Ball $B(x,\epsilon)$ (center at $x\in X$ with radius $\epsilon >0$) then $B(x, \epsilon) \cap B \neq \emptyset$ as $\mathbb{Q}$ is dense.

2) The interior of $S$, $S^\circ = \emptyset$

Take any Ball $B(x, r)$ for some $r>0$. $B(x, r) \nsubseteq B$ as there are infinitely many irrational numbers close to $x$

3) The boundary

\begin{align} \partial S =& \bar{S}\cap\bar{(S^c)} \\ =& \bar{S}\cap(S\cup S^c)\\ =& \bar{S}\cap\mathbb{R}^2 \\ =& \bar{S} \end{align} This is because the complement of $S$ is the points such that the $x$ coordinate is irrational. Again, to find the closure of this set take any ball $B(x, \epsilon)$ then the points that intersect the balls will be the set $S$ and $S^c$ i.e $\mathbb{R}^2$

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Your set $S$ is $\Bbb Q\times\Bbb R$, which is not a closed set in $\Bbb R^2$, simply because $\Bbb Q$ is not closed in $\Bbb R$. In fact $S$ is dense in $\Bbb R^2$: its closure is all of $\Bbb R^2$. – Brian M. Scott Apr 8 '13 at 22:00
To say anything meaningful about topological properties of a set you have to specify a topological space in which you consider that set. – Godot Apr 9 '13 at 0:28
Yes, I see now that it is closed in $\mathbb{R}^2$ For every $x$ ball $B(x, \epsilon)$ will intersect $S$ making a non-empty set. – CodeKingPlusPlus Apr 9 '13 at 1:37

Your set, $S= \{(x, y) | x\in\mathbb{Q} \land y\in\mathbb{R}\}$ is precisely the set $S = \mathbb Q \times \mathbb R$.
(1) As Brian points out, $S = \mathbb Q \times \mathbb R$ is NOT closed, essentially because $\mathbb Q$ is not closed in $\mathbb R$. Indeed, $S$ is dense in $\mathbb {R}^2$ since its closure is all of $\mathbb R^2.$
(2) Noting the above, your argument for the boundary of $S$ should change: if $\partial S = \bar S \cap \mathbb R^2$, then $\partial S = \mathbb R^2 \cap \mathbb R^2 = \mathbb R^2$