Limit points of $\frac{p}{2^k}$

Question

Find all limit points of the set $S = \{\frac{p}{2^k} : p \in \mathbb{Z}, k \in \mathbb{N}\}$.
I could prove that all elements of $S$ are limit points of $S$. However, answer is all real numbers. Any help?

Answer 1

Take an arbitrary point on the reals, and write it in binary notation. For example, x = 101.111100001010...Then x can be approached by 101, 101.1, 101.11, 101.111, 101.1111, 101.11110, ...each of which is a fraction with denominator a power of 2.

Answer 2

Given $x \in \Bbb R$, consider the sequence $$x_n := \dfrac{\lfloor 2^n x\rfloor}{2^n}.$$ Show that $\{x_n\} \subset S$ and $x_n \to x$. Conclude from this.

Answer 3

Here's some intuition to get you started, in order to understand why your answer is wrong and the correct answer is all real numbers.

Look at an ordinary ruler marked in inches (using the imperial system of units). You will see each inch divided in half, each $1/2$ inch divided in half to give fourths, each $1/4$ inch divided in half to give eighths, each $1/8$ inch divided in half to give sixteenths... In the abstract, you can continue this forever, using arbitrarily short subdivision lengths $1/2^n$ as $n$ goes to $\infty$ (although I don't think I've ever seen an actual physical ruler go any shorter than $1/2^5$, and even that is rare).

Perhaps from this it will become evident to you that inside the 1 inch long interval from inch mark number $N$ to inch mark number $N+1$, the subdivision points are dense, and every real number in that interval is a limit point of the set of subdivision points. And then, letting $N$ vary over $\mathbb Z$, you can see that the set of subdivision points over the entire real line is dense.

Once you understand this picture, turning it into a mathematical proof is not actually very hard. The key is to notice that for each $r > 0$, if you choose a natural number $n$ so large that $2^n > \frac{1}{r}$, then $\frac{1}{2^n} < r$. Each real number $x$ is contained in one of the subdivision intervals $\left[\frac{k}{2^n},\frac{k+1}{2^n}\right]$ (i.e. the union of those intervals is the entire real line), and therefore $x$ has distance less than $r$ from either of the two subdivision points at the ends of that interval.