Let $$\rm{A}=\left\{x\in [0.1]\mid \exists\, n \in \mathbb{N}\; \exists\, k\in \{0,\ldots, 2^{n} \}, \; x=\dfrac{k}{2^n} \right\} $$ Show that A is dense in $[0.1]$
$A$ is dense in $[0.1] \iff \begin{cases} A\subset [0.1] \\ \\ \forall x,y\in [0.1] \quad \left(x<y \implies \exists\; a\in A ,\; x<a<y \right) \end{cases}$
- $A\subset [0.1]$
- Let $n\in\mathbb{N}$ and Let $x,y\in [0.1]$ such that $x<y$ then $y-x>0$ thus
$$\lim_{n\to +\infty}\left( 2^{n}(y-x)\right)=+\infty$$ so $$\lim_{n\to +\infty}\left( 2^{n}(y-x)\right)=+\infty\iff \forall R\in \mathbb{R}\;\exists N\in\mathbb{N}\;\forall n\in\mathbb{N}\quad \left(n\geq N\implies 2^{n}(y-x)> R \right) $$
In particular $$\exists n \in\mathbb{N}\quad 2^{n}(y-x)>2 $$ i.e. $$\exists n \in\mathbb{N}\quad 2^{n}y-2^{n}x>2 $$
then there exists at least $k\in (2^{n}x,2^{n}y)$ In particular $$\exists n \in\mathbb{N}\quad x<\dfrac{k}{2^{n}} <y $$
since $x,y\in[0.1]$ then $$0\leq x<\dfrac{k}{2^{n}} <y\leq 1 $$ i.e. $$0\leq k \leq 2^{n}$$ Thus $$\exists n\in\mathbb{N} \quad k\in \{0,\ldots, 2^{n} \}\quad x<\dfrac{k}{2^{n}} <y $$ thus $$\forall x,y\in [0.1] \quad \left(x<y \implies \exists\; a=\dfrac{k}{2^{n}} \in A ,\; x<a<y \right)$$
- Is my proof correct
- could someone prove it by reasoning called analysis synthesis Raisonnement par analyse-synthèse