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

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It is no so hard to prove that the set $S$ is dense in $\Bbb R$. To do so you can argue as follows.
Let $x\in\Bbb R$.
Then $x\in I_1=[m,m+1]$ for some $m\in\Bbb Z$, namely $m=\lfloor x\rfloor$. Divide $I_1$ in two closed subintervals of the same length, then $$[m,m+1]=\left[m,m+\frac1{2}\right]\cup\left[m+\frac1{2},m+1\right]$$ If $x=m+\frac1{2}$,let $x_2=x$. In this case we are done. If not, Call $I_2$ the subinterval that contains $x$ and Let $$x_2=\begin{cases} \text{right endpoint of I_2} &\text{if } x=m\\ \text{left endpoint of I_2} &\text{otherwise} \end{cases}$$ Once you have defined $I_n$, divide $I_n$ in two sbintervals of the same length. If $x_n$ is the middle point of $I_n$, let $x_n=x$. Otherwise, let $I_{n+1}$ be the subinterval that contains $x$ and define $$x_n=\begin{cases} \text{right endpoint of I_n} &\text{if } x=m\\ \text{left endpoint of I_n} &\text{otherwise} \end{cases}$$
Notice that the sequence $(x_n)$ is in $S$ and $$|x_n-x|\leq \frac1{2^{n-1}},\quad\forall n\in\{2,3,4,\ldots\}$$ which implies $$x_n\to x.$$