Cylinders and dyadic Intervals Let be $\Sigma_{+}=\{0,1\}^{\mathbb{N}}$ the space of sequences of $0^{'s}$ and $1^{'s}$. Consider the following surjective  application
$\phi:\Sigma_{+}\to [0,1]$ given  by
$$ \phi(\underline{x})=\sum_{i=1}^{\infty}\dfrac{x_i}{2^{i}} $$
where $\underline{x}=(x_i)$. 
My Problem: I want to see that there is a correspondence between the cylinders of $ \Sigma_{+}$ and the dyadic intervals of [0,1].
Obs 1:  For me this has more or less clear that the image of $ \phi $ of a cylinder is a dyadic interval.
Obs 2: For me, a dyadic interval is an interval of the form $[\dfrac{k}{2^n},\dfrac{k+1}{2^n}]$ where $k=1,\ldots 2^n$
 A: I’ve seen at least two definitions of cylinder set.


*

*For each finite, non-empty $F\subseteq Z^+$ and function $\sigma:F\to\{0,1\}$ let $$B(\sigma)=\left\{x\in\Sigma_+:x\upharpoonright F=\sigma\right\}\;;$$ the cylinders are the sets $B(\sigma)$.

*The cylinders are the sets $B(\sigma)$ defined in (1) for which $\operatorname{dom}\,\sigma$ is an initial segment of $\Bbb Z^+$.
Assuming that your definition is the second one, the result is straightforward.
Fix $n\in\Bbb Z^+$ and $m\in\{0,\dots,n-1\}$, and let $D=\left[\dfrac{m}{2^n},\dfrac{m+1}{2^n}\right]$. Let 
$$m=\sum_{k=0}^{n-1}\frac{b_{n-k}}{2^k}\;,$$ where each $b_k\in\{0,1\}$, so that $b_1\dots b_n$ is the $n$-bit binary representation of $m$, and let
$$\sigma:\{1,\dots,n\}\to\{0,1\}:k\mapsto b_k\;.$$
I claim that $D=\varphi\big[B(\sigma)\big]$, i.e., that $\varphi(x)\in D$ if and only if $x_k=b_k$ for $k=1,\dots,n$. 
If $x\in B(\sigma)$, then 
$$\varphi(x)=\sum_{k\ge 1}\frac{x_k}{2^k}=\sum_{k=1}^n\frac{b_k}{2^k}+\sum_{k>n}\frac{x_k}{2^k}=\frac{m}{2^n}+\sum_{k>n}\frac{x_k}{2^k}\;,$$
and
$$0\le\sum_{k>n}\frac{x_k}{2^k}\le\frac1{2^n}\;,$$
so $\varphi(x)\in D$, and $\varphi\big[B(\sigma)\big]\subseteq D$. On the other hand, we know that 
$$\left\{\sum_{k>n}\frac{x_k}{2^k}:x_k\in\{0,1\}\text{ for }k>n\right\}=\left\{\frac1{2^n}\varphi(x):x\in\Sigma_+\right\}=\left[0,\frac1{2^n}\right]\;,$$
so in fact $\varphi\big[B(\sigma)\big]=D$.
