Density of a set that is closed under halving and under Euclidean sum This is a problem that one of my neighbors (a sophomore Physics major) had in her last  Real Analysis test and was not able to solve. She posed it to me and I am also unable to bring and end to it.
Suppose $E$ is a non-empty subset of $(0,\infty)$ and that

*

*For any $x\in E$, $\frac{x}{2}\in E$ (i.e. $E$ is closed under halving)

*For any $x,y\in E$, $\sqrt{x^2+y^2}\in E$ (i.e. $E$ is closed under Euclidean sum (sorry for the rather pedantic name, that is totally on me.)

Problem: Show that that the $\overline{E}=[0,\infty)$.

So far, we can establish the following basic facts:
(a) $E$ has arbitrarily small elements ($x\in E$ implies that $2^{-n}x\in E$)
(b) For any $x\in E$ and $n\in\mathbb{N}$, $x\sqrt{n}\in E$ (by induction $x\sqrt{2}=\sqrt{x^2+x^2}\in E$. Once, we have  $x\sqrt{n}\in E$, then $x\sqrt{n+1}=\sqrt{(\sqrt{n}x)^2+x^2}\in E$.
After this, I can't see how to kill the problem. hopefully other undergraduate students out there (others are of course welcome) may have some hints or a solution to this puppy.
 A: Hint:  Suppose $y \in (0,\infty) \backslash \overline{E}$.  Let $z = \sup(E \cap [0,y))$ and get a contradiction.
A: Here's another answer, not as quick as Robert Israel's but maybe following a little more obviously from your observations in the question. Fix $a \in E$. Note that the set $F_a := \{ am2^{-n} | m, n \in \mathbb{N} \}$ is dense in $[0, \infty)$. Now, your observation about square roots implies (by taking roots of perfect squares) that $am \in E$ for every $m \in \mathbb{N}$. Then your observation about successive halving implies that $am2^{-n} \in E$ for every $n \in \mathbb{N}$; that is, $F_a \subseteq E$. So $[0, \infty) = \overline{F_a} \subseteq \overline{E} \subseteq [0, \infty)$, assuming that we are considering everything as a subspace at least of $[0, \infty)$ with its usual topology.
A: From your observations, it suffices to show that the numbers of the form $2^{-k}\sqrt n$ are dense in the positive reals.
Notice that $\sqrt{n+1}$ and $\sqrt n$ differ by at most $1$, so $2^{-k}\sqrt{n+1}$ and $2^{-k}\sqrt n$ differ by at most $2^{-k}$. The set $A_k=\{2^{-k}\sqrt n:n\in\Bbb N\}$ is also unbounded, so for any $r\in[0,\infty)$ we can find an element of $A_k$ that lies within $2^{-k}$ of $r$.
