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.

Show that in ($\mathbb{R},d_2$), $\mathbb{Q}^{o}=\emptyset$ but $C((\mathbb{Q})^{o})=\mathbb{R}$.
(C(A) meaning the closure of A, not sure how to do a macron in LaTeX).

My attempt: Since for any subset $S$ of a metric space $X$, $S^o=S/\partial S$, I want to show $\partial\mathbb{Q}=\mathbb{R}$ and thus have that $\mathbb{Q}^{o}=\mathbb{Q}/\mathbb{R}=\emptyset$.
$\partial\mathbb{Q}=${all $x\in\mathbb{R}|dist(x,\mathbb{Q}=\inf_{y\in\mathbb{Q}}d(x,y)$}. But $\mathbb{Q}$ is dense in $\mathbb{R}$ so $\forall x\in\mathbb{R}$, $d(x,\mathbb{Q})=0=\inf_{y\in\mathbb{Q}}d(x,y)\implies\partial\mathbb{Q}=\mathbb{R}$.

To show that $(C(\mathbb{Q})^{o})=\mathbb{R}$, I use the fact that for any set $S\subseteq X$ dense in $X$, $C(S)=X$. Thus we have $\mathbb{R}^o=\mathbb{R}$ since $X^o=X$ in $(X,d)$.

share|improve this question
    
What is $d_2{}$? –  Chris Eagle Feb 7 '12 at 22:58
    
the usual metric in $\mathbb{R}$ –  Emir Feb 7 '12 at 23:00
    
Maybe it's fitting into this family of norms. –  Dylan Moreland Feb 7 '12 at 23:06
1  
In the last sentence, you need to say "$\Bbb Q$ is dense in $\Bbb R$, thus we have $C(\Bbb Q)=\Bbb R$. So $[C(\Bbb Q)]^\circ=[\Bbb R]^\circ=\Bbb R$". –  David Mitra Feb 7 '12 at 23:09
1  
In the first line, it looks like the parentheses are in the wrong place. Also, you can get the closure symbol with either \bar{A} (which puts a fixed-width bar over a single letter) or \overline{A} (which puts a variable-width bar over an arbitrary expression). –  Nate Eldredge Feb 7 '12 at 23:57

1 Answer 1

up vote 1 down vote accepted

I'm not sure what you are doing for your first argument.

Here, you wish to show that every real number $x$ is a boundary point of $\Bbb Q$. But $x$ is a boundary point of $\Bbb Q$ if and only if every open set containing $x$ contains points of $\Bbb Q$ and points of $\Bbb R\setminus \Bbb Q$. But this latter statement is true for any $x$ because both $\Bbb Q$ and its complement are dense in $\Bbb R$. So, the boundary of $\Bbb Q$ is $\Bbb R$; whence $\Bbb Q^\circ=\emptyset$.

You could also argue more directly that $\Bbb Q^\circ=\emptyset$, by showing that $\Bbb Q$ contains no nonempty open set (every nonempty open set contains irrationals); thus, the interior of $\Bbb Q$ is empty (since a point $x$ is an interior point of $A$ if and only if there is an open set $O\subset A$ containing $x$).

For the last part, it seems you're on track; but, it's poorly phrased. Say something like " $\Bbb Q$ is dense in $\Bbb R$; thus $C(\Bbb Q)=\Bbb R$. Thus $[C(\Bbb Q)]^\circ=[\Bbb R]^\circ=\Bbb R$".

share|improve this answer

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.