A question about dense subsets of Euclidean spaces and Hilbert space Let $S$ be the Euclidean plane and let $p(S)$ be a fixed point of $S$. Does there exist a dense subset $D(S)$ of $S$, such that no pair of distinct points of $D(S)$ are at the same distance from $p(S)$? If the answer is "yes", is it still "yes" when $S$ is a higher dimensional Euclidean space or even when $S$ is a separable and infinite dimensional Hilbert space?
 A: Let $S=\mathbb R ^2$ and assume that $p=\langle 0,0 \rangle$ is the origin. Let $$\{U_n:n\in\omega\}$$ countable basis for $S$ consisting of nonempty open sets. We may recursively pick points $r_n\in \mathbb R$ such that $U_n\cap \partial B(p,r_n)\neq\varnothing$ and $r_n\neq r_m$ for all $n\neq m$ (at any given stage we have only defined finitely many $r$'s, while infinitely many cirles centered at $p$ must pass through $U_n$). Pick $$d_n\in U_n\cap \partial B(p,r_n)$$ for each $n\in\omega$. Then $D:=\{d_n:n\in\omega\}$ is as desired.
This is a rather non-constructive approach, but I think you can see that the same argument would work in any separable metric space $S$ as long as enough of the "circles" $\partial B(p,r)$ are not open.  So for instance it would work if $S$ is connected or locally connected.  In particular it works in a Hilbert space.
A: A kind of more constructive way for a separable normed space: assume $p$ is the origin, consider all positive rational radii $\{r_n\}_n$ and write $r_n=\frac{p_n}{q_n}$ in lowest common terms. Next enumerate a countable dense set $\{d_m\}_m$ in the sphere. Then take
$$D=\{r_nd_{q_n}\}_n.$$
A concrete example in two dimensions is 
$$D=\{r_n e^{iq_n\pi\alpha}\}_n$$
with irrational $\alpha$.
A: Let $X$ be $\mathbb R^n$ or $l^2.$ I'll take the designated point to be the origin. Let $S$ be the unit sphere in $X.$ Then there exists a function $f:[0,\infty)\to S$ such that for every interval $I\subset [0,\infty)$ of positive length, $f(I) = S.$ The set $D= \{rf(r): r \in [0,\infty)\}$ then fits the requirements.
The set $D$ is uncountable, but since every dense subset of a separable metric space contains a countable dense subset, we can find a countable dense subset that does the job.
So some of you are probably asking: How do you know that this function $f$ exists? It's really not that hard, but I'll omit the proof of this for now.
A: Let $D$ be a countable dense subset of the Euclidean plane $S.$
For each pair of distinct points $P,Q\in D,$ the set of points equidistant from $P$ and $Q$ is a line $\ell_{P,Q},$ the perpendicular bisector of the line segment $PQ.$
Let $L=\{\ell_{P,Q}:P,Q\in D,\ P\ne Q\},$ a countable set of lines in the plane.
The lines in $L$ do not cover the whole plane. To see this, we can use the fact that $\bigcup L$ is a set of the first category and Lebesgue measure zero; or more simply, choose a line $\ell\notin L,$ and observe that each line in $L$ meets $\ell$ in at most one point, while there are uncountably many points on $\ell.$
Choose a point $p\in S\setminus\bigcup L.$ Without loss of generality, we may assume that $p(S)=p.$ Let $D(S)=D.$ Then $D(S)$ is a dense set of points in $S,$ and no pair of distinct points of $D(S)$ are at the same distance from $p(S).$
For the higher-dimensional spaces it may be simplest to use a Baire category argument: the locus of points equidistant from two given points is a nowhere dense closed set; the union of countably many such sets is a set of the first category.
A: "Square-free" radicals, and in particular the square roots of primes are linearly independent over the rationals. See for example page 87 in http://www.math.harvard.edu/hcmr/issues/2.pdf and The square roots of different primes are linearly independent over the field of rationals 
Consider the set D defined by the Cartesian product 
$(\{q: q  \in Q \text{ and } q \le 0\} \cup \{2^{1/4}.q: q  \in Q \text{ and } q \ge 0\}) \times (\{3^{1/4}q: q  \in Q \text{ and } q \le 0\} \cup \{5^{1/4}.q: q  \in Q \text{ and } q \ge 0\})$
The set D is clearly dense in the plane .
Take the point $p$ at the origin (wlog as noted in other posts), then the Euclidean distance $d(p, x)$ between  $p$ and a point $x \in D$ is of the form 
$(q_{1, x} + q_{2, x}\sqrt(2) + q_{3, x}\sqrt(3) + q_{5, x}\sqrt(5))^{1/2}$
So if $d(p, x) = d(p, y)$ then  $d(p, x)^2 - d(p, y)^2 = 0$ and you are left with an expression of the form $q_{i} + q_{2}\sqrt(2) + q_{3}\sqrt(3) + q_{5,}\sqrt(5) = 0$ which has no solution other than $q_1 = q_2 = q_3 = q_5 = 0$, i.e. $x = y$.
The solution easily extends to all finite dimensional spaces by allocating pairs of primes to the $-, +$ directions in each dimension. A infinite dimensional separable space is another matter, since the linear independence of prime roots is only proven for finite combinations (I think).
