Let $h:R\to R$ be continuous on$ $R satisfying $h(m/2^{n}) =0$ for all $ m \in Z $, $ n \in N $. Show that $h(x)=0$ for all $ x \in R $. Let $h:R\to R$ be continuous on $R$ satisfying $h(m/2^{n}) =0$ for all $ m \in  Z $, $ n \in N $ .Show that $h(x)=0$ for all $ x \in R $.
I think sequential criterion of continuity  .but how to show $(m/2^{n})$ converges to every real number .
 A: 

Hint $:$ $S:=\left \{ \dfrac {m} {2^n} : m \in \Bbb Z, n \in \Bbb N \right \}$ is a set of dyadic rationals which is dense in $\Bbb R.$
Proof $:$ It suffices to show that $S$ intersects any open interval or basic open sets of $\Bbb R$ with usual topology. Let $(a,b)$ be an open interval. Then $a<b$ i.e. $b-a>0.$ So by Archimedean property of real numbers we can find a natural number $n$ such that $0 < \dfrac 1 n < b-a \implies 0< \dfrac 1 {2^n} < \dfrac  1 n < b-a.$ So we have $(b-a)2^n > 1 \implies b2^n - a2^n > 1.$ Since the distance between $a2^n$ and $b2^n$ is greater than $1$ so $\exists$ an integer $m$ such that $a2^n < m < b2^n \implies a< \dfrac m {2^n} < b.$ This shows that $(a,b) \cap S \neq \varnothing,$ as claimed.


Now let $x \in \Bbb R.$ Since $S$ is dense in $\Bbb R$ we can a find a sequence $\{x_n \}$ in $S$ converging to $x.$ Since $h$ is continuous by sequential criterion for continuous functions we have $h(x) = \lim\limits_{n \rightarrow \infty} h(x_n) = 0$ (Since $h(x_n) = 0,$ for all $n \in \Bbb N$ because $x_n \in S$ for all $n \in \Bbb N$).
Since $x \in \Bbb R$ was arbitrarily taken so we find that $h(x) = 0,$ for all $x \in \Bbb R,$ as required.
QED
