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I know that this is true and is used to prove that $\mathbb{Q}$ is not a discrete metric space, but I can't figure out, why is it true ?

math
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johny
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3 Answers3

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A set $U$ in a metric space $(X,d)$ is open if, and only if, it contains a neighborhood around each of its points. That is, for any $a\in U$, you can find a radius $r>0$ so that $B(a,r):=\{y\in X\mid d(a,y)<r\}$ is contained in $U$.

So, for $\{a\}$ to be an open set in $\mathbb{Q}$ (under the usual metric), it must be true that there is a positive radius $r$ so that $B(a,r)\subseteq\{a\}$ -- that is, there must be a ball around the point $a$ which contains no other rational numbers.

But this is impossible: if you give me any radius $r>0$, I can find $n\in\mathbb{N}$ such that $0<\frac{1}{n}<r$; the number $a+\frac{1}{n}$ is then contained in $B(a,r)$, but not contained in $\{a\}$.

Nick Peterson
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  • @R.Peterson I just re read your answer and even though its perfect, i was wondering if there was an easier way of showing this. i figured that a singleton set would be open in $\mathbb{Q}$ if and only if it would be an isolated point of $\mathbb{Q}$ but $\mathbb{Q}$ has no isolated points and hence any singleton set cannot be its itolated point and so cannot be open on $\mathbb{Q}$. Do you think that this is good enough ? – johny Oct 11 '13 at 06:05
  • @johny Well, that depends on your instructor, and on what path you took to get to this exercise. Personally, I would argue that "$\mathbb{Q}$ has no isolated points" is exactly what you are asked to show; so, I certainly wouldn't give you feel credit for that answer on a homework assignment. I wouldn't give you 0 credit either, though. :-) – Nick Peterson Oct 11 '13 at 11:04
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    @R.Peterson Alright, it's better to opt for your argument then ! – johny Oct 11 '13 at 11:11
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As you are viewing $\mathbb{Q}$ as a metric space, I assume with the metric $d(x, y) = |x-y|$, $A \subseteq \mathbb{Q}$ is open if for every $a \in A$, there is $r > 0$ such that $B(a, r) \subseteq A$. So $\{a\}$ is open if there is $r > 0$ such that $B(a, r) \subseteq \{a\}$; actually, as $\{a\} \subseteq B(a, r)$, we would have $B(a, r) = \{a\}$. Can you see why no such open ball exists?

Michael Albanese
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Any subset of $\mathbb{Q}$ is open if it is of the type $G\cap \mathbb{Q}$, where $G$ is open in $\mathbb{R}$. If you choose any singleton $\{a\}$ in $\mathbb{Q}$, then it can not be written in the above form.

Anupam
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