Vtyjkyuk

Is the following statement true or false?

Let $(X,d)$ be a metric space. Then for every $x in X$, there exists $y in X$ such that $d(x, y)$ is a non-zero rational number.

I'm not able to find a counter example.

False. Take $X=mathbbR$ and $d(x,y)=pi$ for $xneq y$.

Here are some examples that are subsets of $mathbbR$ (or $mathbbR^n$), which are counterexamples if $d$ is taken to be the Euclidean metric.

• The two-point set $0, sqrt2$.




• Take a Vitali set in $mathbbR$. By construction, the distance between any two points is irrational.




• Take any continuous probability distribution on $mathbbR^n$ (e.g. uniform in a region, Gaussian, etc), and choose a finite or countable number of points $X_1, X_2, dots$ independently according to this distribution. With probability one, the resulting set has no two points that are a rational (Euclidean) distance apart. (For any given pair $X_i, X_j$, there is probability zero that $|X_i - X_j|$ is rational. There are a countable number of pairs, so by countable additivity, there is probability zero that there exists a pair a rational distance apart.)

Sometimes extreme cases are a good source of counterexamples. Consider the metric space on a set of one element, $(star, d)$, with $d(star, star)=0$. Clearly for $x=star$, there is no $y$ such that $d(star, y)$ is non zero.

The statement in the text of the question is almost true (would be true without the word "nonzero"). For every x, there exists a point y, namely the point y=x, such that y is at a rational distance (zero) from x. The statement in the title of the question ("another point") is false, however, as the word "another" implies that y must differ from x and as the other answers already show. So you have to read this sort of question very carefully to answer it correctly.