MHB What is the definition of real numbers in terms of rational sequences?

[evinda]

Hi! (Smile)

We define the set $U=\mathbb{Z} \times (\mathbb{Z}-\{0\})$ and over $U$ we define the following relation $S$:

$$\langle i,j \rangle S \langle k,l \rangle \iff i \cdot l=j \cdot k$$

$$\mathbb{Q}=U/S=\{ [\langle i, j \rangle ]_S: i \in \mathbb{Z}, j \in \mathbb{Z} \setminus \{0\} \}$$

Constitution of real numbers (Cantor)

We consider the space $X$ of sequences $(a_n)_{n \in \mathbb{N}}$ of rational numbers that satisfy the following Cauchy Criterion:"for each rational number $\epsilon>0$ there is a $n_0 \in \omega$ such that for all $n \in \omega$ and for all $m \in \omega$ with $n>n_0$ and $m>n_0$ it holds that $|a_n-a_m|< \epsilon$"

At $X$ we define the relation $\sim$ for $(x_n)_{n \in \omega}, (y_n)_{n \in \omega} \in X$
$(x_n)_{n \in \omega} \sim (y_n)_{n \in \omega}$ iff $\lim (x_n-y_n)=0$

(Definition of $\lim a_n=0$ for $(a_n)_{n \in \omega}$ sequence with $a_n \in \mathbb{Q}, n \in \omega$:
$\lim a_n=0$ iff for each rational $\epsilon>0, \exists n_0 \in \omega$ such that for all $n \in \omega$ with $n>n_0$ it holds that $|a_n|< \epsilon$)It can be proven that the relation $\sim$ is an equivalence relation on $X$ and $X/ \sim$ is defined to be the set of real numbers $\mathbb{R}$.So that means that the set $\mathbb{R}$ is a set that contains the rational sequences that converge, right?
If so, could you explain me why it is like that? (Thinking)

 

[Fallen Angel]

Hi,

That means that $\Bbb{R}$ is the "complection" (I don't really know if complection is an english word ) of the metric space given by the rationals with the absolute value.

I mean, when you say "contains the rational sequences that converge" is right, but it's not the same convergence on $\Bbb{Q}$ than $\Bbb{R}$. A sequence of rational numbers convergent to $\pi$ over the reals is not convergent over $\Bbb{Q}$.

This is just a definition so it's like that because we want it to be like that.