Well Define Of Inequality of Real Number

Question

Definition

For two sequences $\alpha,\beta : \Bbb N \rightarrow \Bbb Q$

Define $\alpha \sim \beta$ when one can pick $N$ s.t. $\forall i \ge N$, $\lvert \alpha(i)-\beta(i)\rvert \lt e$

Since $\sim$ is equivalence relation of the set $\Bbb F = \{\alpha: \Bbb N \rightarrow \Bbb Q$}, call the element of Quotient set $\Bbb R = \Bbb F/\sim$ real number.when

Now

Define $[\alpha], [\beta] \in \Bbb R$, if there $\exists d>0 \in \Bbb Q $ and $ N \in \Bbb N$ s.t. $\forall i \ge N$, $\lvert \alpha(i)-\beta(i)\rvert \lt d$ , $[\alpha]\gt [\beta]$.

Question

From above, I have been requested to check whether the inequality of $[\alpha], [\beta]$ is well-defined.

Well-defined is still ambiguous term to me.

What do I have to check?

Answer 1

$[\alpha]$ denotes the equivalence class that includes $\alpha$. However, there are other ways we could have represented this equivalence class. For example, if $\beta \sim \alpha$, we could have istead written it as $[\beta]$. Thus, whenever we define a relation between two equivalence classes $[\alpha]$ and $[\beta]$ where the definition depends explicitly on $\alpha$ and $\beta$ (as does your definition for inequality), we face a potential problem: what if the relation may or may not hold, depending on how we choose to represent the equivalence classes? If this happens, then the relation is not well-defined.

For example, if I attempted to define a function $F: \mathbb{R}\to \mathbb{Q}$ given by $F([\alpha]) = \alpha(1)$, this would not be a well defined function for the above reason.

So, when the problem asks you to prove that $<$ is well defined, it wants you to prove the following: If $\alpha \sim \alpha'$ and $\beta \sim \beta '$ and $[\alpha]<[\beta]$, show that $[\alpha'] < [\beta']$.

EDIT- To show that inequality is well defined, suppose $\alpha \sim \alpha'$, $\beta \sim \beta'$, and that $[\alpha] < [\beta]$. Then there exists $N\in \mathbb{N}$ and $d\in \mathbb{Q}^{> 0}$ such that whenever $n>N$ we have $\beta(n) - \alpha(n) > d$.

Since $\alpha' \sim \alpha$ and $\beta' \sim \beta$, there exists $N'\in \mathbb{N}$ such that $N'> N$ and such that whenever $n>N'$ we have $\vert \alpha(n)-\alpha'(n)\vert < d/4$ and $\vert\beta(n)-\beta'(n)\vert < d/4$. Then $$\beta'(n) - \alpha'(n) = (\beta'(n) - \beta(n))+(\beta(n)-\alpha(n))+(\alpha(n) - \alpha'(n))$$ $$\geq -d/4 + d - d/4 = d/2$$ So we have that $\beta'(n) > \alpha'(n)$, and that $\vert \beta'(n)-\alpha'(n)\vert > d/2$, and so $[\alpha'] < [\beta]'$.

$\;$

Answer 2

$\forall \alpha, \beta \in \Bbb R$ take $\alpha',\beta'$ s.t. $\alpha'\sim\alpha,\; \beta' \sim \beta$ and

Suppose $[\alpha] \gt [\beta]$ Then

$\exists d>0 \in \Bbb Q $ and $ N \in \Bbb N$

s.t. $\forall i \ge N$, $\lvert \alpha(i)-\beta(i)\rvert \lt d$

one can take $N_1, N_2$ s.t. $\forall i \ge N_1$, $\lvert \alpha(i)-\alpha'(i)\rvert \lt e$ and $\forall i \ge N_2$, $\lvert \beta(i)-\beta(i)'\rvert \lt e$

since $\alpha'\sim\alpha,\; \beta' \sim \beta$

Now let $N_{sup} = sup\{N_1, N_2\}$ then

$\exists d>0 \in \Bbb Q $ and $ N_{sup} \in \Bbb N$

s.t. $\forall i \ge N_{sup}$, $\lvert \alpha'(i)-\beta'(i)\rvert \lt d$

since for $N_{sup}$,

$\lvert \alpha(i)\rvert-\lvert\alpha'(i)\rvert \le\lvert \alpha(i)-\alpha'(i)\rvert \lt e$ (1)

$\lvert \beta(i)\rvert-\lvert\beta(i)'\rvert \le \lvert \beta(i)-\beta'(i)\rvert \lt e$ (2)

$\lvert \alpha(i)\rvert-\lvert\beta(i)\rvert \le \lvert \alpha(i)-\beta(i)\rvert \lt d$ (3)

(2)-(1)+(3) derives $\lvert \alpha'(i)\rvert-\lvert\beta'(i)\rvert \le\lvert \alpha'(i)-\beta'(i)\rvert \lt d$