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for which values of $x,y$ is $[x,y]\cap \mathbb{Q}$ closed in the metric space $(\mathbb{Q},d)$ where $d(x,y) = |x-y|$

my attempt:

I suspected it's closed for all real numbers:

let $x,y \in \mathbb{Q}$ then if $[x,y]\cap \mathbb{Q}$ is closed it means the compliment is open i.e. $(-\infty,x) \cup (y,\infty)$ is open. The set $(-\infty,x) \cup (y,\infty)$ is obviously open as between any two rational numbers there is another rational number so I can find a ball with radius $r>0$ in the set. (Is this logic correct?)

if $x,y$ are irrational however I don't know how to proceed - will the complement on the set still be $(-\infty,x) \cup (y,\infty)$? and if so how do I argue it is open (or if it isn't)

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  • $\begingroup$ The compliment of the set under consideration is not what you say it is. $\endgroup$ – Paul Jan 26 '15 at 19:14
  • $\begingroup$ I think a better way to approach this problem is to ask whether or not the set in question contains all its limit points. $\endgroup$ – Tim Raczkowski Jan 26 '15 at 19:18
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In all cases the complement will be

$$\Bbb Q\cap\Big((\leftarrow,x)\cup(y,\to)\Big)$$

(note that you really should include the intersection with $\Bbb Q$, unless you’ve previously established a convention that your interval notation is to be understood to mean intervals in $\Bbb Q$ rather than in $\Bbb R$), and your argument for this being open works equally well for all $x$ and $y$, not just the rational ones. All you need is the fact that between any two real numbers there is a rational.

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  • $\begingroup$ Could you explain why in all cases the complement would be in that form? why for instance could I not have $(\leftarrow, x] \cup [y, \rightarrow)$ $\endgroup$ – hellen_92 Jan 26 '15 at 19:21
  • $\begingroup$ I say this because, if $x,y$ where irrational, then wouldn't the set $\mathbb{Q} \cap [x,y]$ infact just be $\mathbb{Q} \cap (x,y)$ and then finding the complement of this would be what I wrote? $\endgroup$ – hellen_92 Jan 26 '15 at 19:25
  • $\begingroup$ @hellen_92: In fact if $x$ and $y$ are irrational, then $$\Bbb Q\cap\Big((\leftarrow,x)\cup(y,\to)\Big)=\Bbb Q\cap\Big((\leftarrow,x]\cup[y,\to)\Big)\;,$$ so it makes no difference which you use. The point is that $(\leftarrow,x)\cup(y,\to)$ is the complement of $[x,y]$ in $\Bbb R$, so its intersection with $\Bbb Q$ is the complement in $\Bbb Q$ of $\Bbb Q\cap[x,y]$. $\endgroup$ – Brian M. Scott Jan 26 '15 at 19:25
  • $\begingroup$ ah yes of course - is my explanation of why that set is open correct? Also - if say we have the set $[x,y] \cap \mathbb{Q}$ if $x,y$ were irrational would this set be open because of the same reasons above? $\endgroup$ – hellen_92 Jan 26 '15 at 19:26
  • $\begingroup$ @hellen_92: Yes, if $x$ and $y$ are irrational, $(x,y)$ and $[x,y]$ have the same intersection with $\Bbb Q$, and that intersection is both open and closed in $\Bbb Q$ (or for short, clopen). $\endgroup$ – Brian M. Scott Jan 26 '15 at 19:26
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Recall the following theorem.

Theorem. Let $X$ be a subspace of a topological space $Y$ and let $E\subset X$. Then $E$ is closed in $X$ if there exists a set $W$ closed in $Y$ such that $E=X\cap W$.

The proof of this theorem is not difficult and worth writing down yourself and good practice for thinking about subspaces.

If you accept this theorem, your problem becomes easier.

In your problem, we have \begin{align*} Y &= \Bbb R & X &=\Bbb Q & E &= [x,y]\cap\Bbb Q \end{align*} So your question translates to: Does there exist a set $W$ closed in $\Bbb R$ such that $E=W\cap\Bbb Q$?

The answer is quite obvious when phrased this way. Do you see how to find $W$?

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  • $\begingroup$ Be aware that this definition is not universal. In some treatments relatively open sets are defined as intersections with the subspace of an open set in the ambient space, and then relatively closed sets are defined as relative complements of relatively open sets. In that approach your definition becomes a theorem. $\endgroup$ – Brian M. Scott Jan 26 '15 at 19:18
  • $\begingroup$ @BrianM.Scott Good point. Edited to reflect this. $\endgroup$ – Brian Fitzpatrick Jan 26 '15 at 19:22

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