A metric space on the rationals would mean every element was an isolated point?

Question

Would this be true? Every element of the metric space on the rationals with the standard metric would mean that every point was an isolated point. Would this be a metric space?

Answer 1

Every point of $\mathbb{Q}$ with the standard metric $d$ will not be isolated. If $x \in \mathbb{Q}$ were an isolated point, we could find an open ball (relative to the topology on $\mathbb{Q}$) $B(x,\epsilon)$ not containing any other points in $\mathbb{Q}$. Now, $B(x,\epsilon)$ is precisely the set $$ B(x,\epsilon) = \left\{ r \in \mathbb{Q} : |x-r| < \epsilon \right\}. $$ However, it is well known that one can always find a rational number $r \neq x$ with $|x-r| < \epsilon$. Hence, every neighbourhood of $x$ (relative to the metric structure of $\mathbb{Q}$) will contain a rational number different than $x$.

It is true that $\mathbb{Q}$ is a metric space (and a metric subspace of $\mathbb{R}$) when equipped with the metric $d(x,y) := |x-y|$. More generally, let $(X,d)$ be a metric space and $Y$ a subset of $X$. If $d_Y$ denotes the restriction of $d$ to $Y \times Y$, then the pair $(Y,d_Y)$ will itself be a metric space.

Now, say we give $\mathbb{Q}$ the discrete metric $$ d(x,y) := \begin{cases} 1 & \text{if } x \neq y,\\ 0 & \text{if }x =y. \end{cases} $$ (I will leave it to you to verify that $d$ is indeed a metric on $\mathbb{Q}$). With this metric, $\mathbb{Q}$ will consist only of isolated points. Indeed, for each $x \in \mathbb{Q}$, the open ball $B(x,1)$ will contain only the point $x$.