# $h\left(\frac{m}{2^n}\right)=0$ $\forall m\in\mathbb{Z},n\in\mathbb{N}$ implies $h(x)=0$ $\forall x\in\mathbb{R}$ if $h$ is continuous.

Let $h:\mathbb{R}\to\mathbb{R}$ be a function that is continuous on $\mathbb{R}$ and has the property that

$$h\left(\frac{m}{2^n}\right)=0,\quad\forall m\in\mathbb{Z},n\in\mathbb{N}.$$

How can we show that this implies that $h(x)=0$, $\forall x\in\mathbb{R}$?

$$S:=\left\{\frac{m}{2^n}:\forall m\in\mathbb{Z},n\in\mathbb{N}\right\}$$

is dense in $\mathbb{R}$. It then follows that for any $c\in\mathbb{R}$ there exists a sequence $(x_n)$ that converges to $c$ such that all the terms are of the form $p/2^q$. Thus, the sequence $(h(x_n))$ converges to $0$ and so by the sequential criterion for continuity we must have $h(c)=0$.

Is there a simpler approach? Proving that $S$ is dense in $\mathbb{R}$ seems overly complicated for this problem. Or, can we deduce it from the density of $\mathbb{Q}$ in $\mathbb{R}$?

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Your argument is looks fine to me, and it's already a very simple argument. You do need to show that $S$ is dense in $\mathbb{R}$; it's not enough to use the fact that $\mathbb{Q}$ is dense because $S$ is a subset of $\mathbb{Q}$. (It would be enough to show that $S$ is dense in $\mathbb{Q}$, but that's not any easier.) – Brett Frankel Dec 17 '12 at 20:35

Proving $S$ is dense is not very hard. If you realize that increasing $n$ to very large can get you to a very small number and then multiplying by a sufficiently large number $m$ can get you close to any real number within the $\frac1{2^n}$ error margin.
A formal proof will involve using the Archimedian property of $\mathbb{R}$.