Let $f:\mathbb{R}→\mathbb{R}$ function given by $f(x)=\sum_{k=0}^{[|x|]}\frac{1}{2^k}$ , where $[y]$ denotes the greatest integer less or equal to $y$. Examine the following: $1.$ $f$ is right continuous function. $2.$ $f$ is left continuous function. $3.$ $f$ is a continuous function. $4.$ $f$ is uniformly continuous.
So to prove right and left continuity of $f$ in $\mathbb{R}$ it suffices to show that $f$ is continuous, then automatically $f$ will attain $1$ and $2$ both simultaneously. Now to prove $f$ is continuous on $\mathbb{R}$, let us take a point $x$ in $\mathbb{R}$ and for a chosen $ε>0$ we have to show that $|f(y)-f(x)|<ε$ whenever $|y-x|<d$ for a positive $d$. Now, If we take $d$ very small such that $[y]=[x]$ for every $y$ satisfying $|y-x|<d$ then we are done trivially. But otherwise I failed to show continuity of $f$. Is it from the consequence of the convergent of the series $\sum_{k=0}^{\infty}\frac{1}{2^k}$? I have a problem in this area. Please help me to solve this problem.