For any set A ⊂ R λ, x ∈ R denote $dist(x,A)= inf_{a∈A} |x−a|$, $\lambda·A$={$\lambda$ a|a∈A}
For k ∈ N, we define $g_n(x) = dist (x, 2^{−n}Z)$. consider $f_N : [0,1] → R$,
$$f_N(x) = \sum_{n=0}^N g_n(x) $$
Show that f_N converges uniformly to a function $f_{\infty} ∈ C[0,1]$.
What I have is that Z is divided into 2^n intervals so gn(x) approaches 0 as n approaches infinity (since any x between [0,1] we choose $2^{−n}Z$ arbitrarily close to x.) so if gn converges to 0, then $\sum_{n=0}^{\infty} g_n$ converges uniformly thus $\sum_{n=0}^N g_n(x) $ coveres unfiromly. Am i on the right track and how would i formalize this? Thanks.
You've started off in the right direction, but you seem to get a bit lost. The question requires you to do three things:
HINT for step 1: the functions $g_n$ clearly shrink to $0$, but you still need to prove that they shrink fast enough for their sum to converge (if $g_n =1/n$ they would go to zero but their sum would slowly approach $\infty$ so you do need to prove this).
HINT for step 2: Once you have the function $f_\infty$ it's a very standard argument to show that $f_n \stackrel{u}{\rightarrow} f_\infty$
HINT for step 3: Once you know about $f_\infty$ it's a very standard $\varepsilon - \delta$ argument to show that it is continuous on $[0,1]$
To help you get started, let's look at step 1. For any $x\in {\mathbb R}$ we can find an interval of width $2^{-n}$ containing it, i.e. $\exists z\in {\mathbb Z}$ s.t. $x\in [2^{-n}z, 2^{-n}(z+1))$. So $g_n(x) \leq 2^{-n} \ \forall x \forall n$. Since $\sum_{i=1}^\infty 2^{-n} =1$ we have $f_N(x) \leq 1 \ \forall x \forall N$. So the $f_N$ converge to some function $f_\infty(x) := \sum_{i=1}^\infty g_n(x)$