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I am trying to show that the rational interval $ S= \{x \in \mathbb Q : a \leq x \leq b\} $ is disconnected in the metric space $(X,d)$ where $X=\mathbb R$ and $d$ is the standard metric ( $ d(x,y)=|x-y| \; \forall \; x,y \in \mathbb R$ )

The definition of disconnected I would like to use is: a metric space $(X,d)$ is disconnected if there are non-empty, open, disjoint sets $A, B$ such that $X=A \cup B$.

If $a,b \notin \mathbb Q$ I think I have shown the above by what follows.

Any rational interval contains an irrational number (assume known). Let $c$ be an irrational number and $a<c<b$. Let $A=\{x\in \mathbb Q : a<x<c \}$ and $B=\{x\in \mathbb Q : c<x<b \}$. $A$ and $B$ are disjoint and open. $A$ and $B$ are non-empty as every interval contains a rational number (assume known). As $c \notin \mathbb Q$ $S=A \cup B$ and so S is disconnected.

If either $a$ or $b$ do belong to $\mathbb Q$ then I don't know how to construct the sets $A$ and $B$ so that they are both open, disjoint and their union is $S$. As if for example if $A=\{x\in \mathbb Q : a\leq x<c \}$ then this is neither open nor closed I think.

How could this problem be solved?

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It doesn't matter if $a,b\in\mathbb{Q}$ or not, since you're looking at the set $S$ with subspace topology from $\mathbb{R}$. –  Thomas E. Dec 17 '12 at 11:18
    
If you consider space $[a,b] \cap \mathbb{Q}$ then $[a,x)$ is open for any $x$, because it is a complement of closed set $[x,b]$. –  dtldarek Dec 17 '12 at 11:20
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2 Answers

As I mentioned in the comments, you don't have to worry about $a$ and $b$ being in $\mathbb{Q}$, since you're looking at the set $S$ with subspace topology from $\mathbb{R}$. So you want to show that $S$ is a disjoin union of non-empty open sets in this topology, and not in the topology of $\mathbb{R}$. Recall that a set $U\subset S$ is open in $S$ if there exists an open $V\subset \mathbb{R}$ so that $U=V\cap S$.

Everything else is fine in your proof. Just choose $A=\{x\in\mathbb{Q}:a\leq x<c\}$ and $B=\{x\in\mathbb{Q}:c<x\leq b\}$ instead. Why are these open in $S$? By choosing for example $V=]a-1,c[$, which is open in $\mathbb{R}$, it follows that \begin{equation*} ]a-1,c[\cap S=]a-1,c[\cap[a,b]\cap\mathbb{Q}=[a,c[\cap\mathbb{Q}=\{x\in\mathbb{Q}:a\leq x<c\}=A, \end{equation*} so $A$ is of the form $V\cap S$ for some open $V$ in $\mathbb{R}$. Similarly you can see that $B$ is an open set in $S$.

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Thank you. Your definition of open is different to what I saw in my metric spaces course. We had a set $A \subseteq X$ is open in $(X,d)$ if it contains its boundary which is equivalent to every point $x \in A$ is an interior point. An interior point of $A$ being an $x \in X$ where you can find an $r>0$ such that $B(x,r) \subseteq A$ where $B(x,r)$ is the open ball of radius $r$ centred at $x$. Your definition seems simpler. Not sure how I would prove $[a,c)$ was open with this definition. You proved that $[a,c)$ is open in $\mathbb Q$. Is it true it is not open in $\mathbb R$. –  Sam Forbes Dec 17 '12 at 16:27
    
@SamForbes In this exercise you are trying to show that $(S,d\,|_{S})$ is a disconnected metric space, where $d\,|_{S}$ is the subspace metric from $\mathbb{R}$ to $S$. Are you familiar with the definition of subspace topology? And yes, $[a,c[$ is open in $S$, but not open in $\mathbb{R}$. –  Thomas E. Dec 17 '12 at 16:45
    
The course I did was just on metric spaces and didn't mention topology (although I know they are related). Thanks for confirming my last statement. –  Sam Forbes Dec 17 '12 at 16:49
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Hint: Prove the following easy

Claim: A topological space $\,X\,$ is not connected iff there exists a continuous, surjective function $\,f: X\to \{0,1\}\,$ , with the latter space having the topology inherited from the usual one in $\,\Bbb R\,$ ( it is thus a discrete space)

Finally, just check

$$f:S\to\{0,1\}\,\,,\,f(x)=\begin{cases}0\;\;,&\text{ if}\,\;\;\;x\leq\frac{b-a}{2}\\{}\\1\;\;,&\text{ if}\,\;\;\;x>\frac{b-a}{2}\end{cases}\;\;\;\;\;\;\;\;,\;\;x\in S$$

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Thank you. I haven't actually seen the term topological space in my metric spaces course. I will give the proof of this a go soon. –  Sam Forbes Dec 17 '12 at 16:33
    
Ok, then erase "topological space" and write "metric space": it'll work just the same. –  DonAntonio Dec 17 '12 at 17:06
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