Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{R}^2$

share|improve this question
3  
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
1  
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

1 Answer 1

up vote 4 down vote accepted

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 $\partial S = \mathbb R^2 \cap \mathbb R^2 = \mathbb R^2$

share|improve this answer
    
Another nice answer with clear explanations and guidance. +1 –  Amzoti Apr 9 '13 at 0:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.