Dense Countable Subsets of the Unit Disk

Question

Consider a countable dense subset A of the unit disk D in R. Is it true that the sum over elements of A vanishes? Intuitively if we choose a neighborhood U of a point x in D we may find points in A arbitrarily close to x. Similarly we may do this with -x hence we may find points in A whose sum is arbitrarily close to zero. Doing so for each x in D should lead us to the conclusion that the sum over A is zero (or as close as we'd like).

Suppose that for every countable, dense subset $A \subseteq D \subseteq \mathbb{R}$ we had $\sum_{x \in A} x = 0$. Then we could just pick any $z \in D(A \cup {0})$ and end up with a countable, dense subset of $D$ such that $\sum_{x \in A \cup {z}} x = z + \sum_{x \in A} x$, which witnesses that we cannot estimate a general value of such a sum. — user79202, May 25 '13 at 19:35
If you consider everywhere dense subset $A$ of unit SPHERE in $\mathbb{R}$, then the answer is positive :) — Antoine, May 25 '13 at 19:40

score 1 · Answer 1 · answered May 25 '13 at 19:34

You say "unit disk in $\mathbb{R}$", which is just the interval $[-1,1]$, but the same argument works in $\mathbb{R}^n$ too. A countable dense subset $S\subseteq[-1,1]$ will necessarily have infinitely many terms of absolute value greater than $\frac{1}{2}$, so the sum $$\sum_{x\in S}|x|$$ diverges. Riemann's rearrangement theorem implies that the sum $$\qquad\qquad\sum_{n=1}^\infty x_i,\qquad S=\{x_1,x_2,\ldots\}$$ can therefore be reordered to give any real number whatsoever, and so in particular, any non-zero number. But just to give an example where the "intuitive" sum is not zero, let $S$ be $$S=(\mathbb{Q}\cap[-1,1])\cup\left\{\tfrac{1}{\sqrt{2}}\right\}.$$ All the rationals "cancel out with their negatives" (metaphorically) and $\tfrac{1}{\sqrt{2}}$ is left over.

score 0 · Answer 2 · answered May 25 '13 at 19:34

If we're thinking of the unit disk $D$ in $\mathbb{R}$, then let's take $S = (0,1) \cap \mathbb{Q}$ as our dense countable subset (everything I'm saying works for any $S$, but this is nice for examples). Let's assume we're summing $S$ in some order and that the sum converges. It's pretty clear that the sum is not absolutely convergent, so by Riemann's rearrangement theorem we can change the order of summation to sum to anything at all. In specific we can order it in such a way that the sum is zero (for instance, $0 + \frac{1}{2} - \frac{1}{2} + \ldots)$. But we can also have it sum to $1$ (try to come up with this one, it's not hard), or $2$, or $\infty$. Basically, there isn't really a notion of just summing a set, unless it's a very well-behaved set (for instance, is composed only of finitely many number of one sign, and the rest of the other).

score 0 · Answer 3 · answered May 25 '13 at 19:35

Even ignoring any convergence/divergence questions, while it is true that for each $x$, you can find another point arbitrary close to $-x$, there's no guarantee that the points are uniformly distributed in the disc. If there are more points near $x$ than near $-x$, then you can't count the ones near $-x$ twice, so it's possible the sum is not zero.

Dense Countable Subsets of the Unit Disk

3 Answers3