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

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The discussion defines real numbers in terms of rational sequences using the Cauchy Criterion, establishing that real numbers are represented by equivalence classes of convergent sequences of rational numbers. The relation defined on sequences indicates that two sequences are equivalent if their difference converges to zero. It is emphasized that while real numbers include sequences that converge to limits like π, such convergence does not hold in the rationals. The conversation highlights that the real numbers serve as the completion of the metric space of rational numbers, clarifying that convergence in the reals differs from that in the rationals. Overall, the definition illustrates the foundational relationship between rational sequences and real numbers.
evinda
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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)
 
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Hi,

That means that $\Bbb{R}$ is the "complection" (I don't really know if complection is an english word :confused:) 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.
 

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