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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 $(X,d)$.

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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
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
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
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 $[C(\Bbb Q)]^\circ=[\Bbb R]^\circ=\Bbb R$".

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