Well-Define of Summation of Real Number : Cauchy/Quotient Set Approach

Question

Definition of $\Bbb R$

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

Define $\alpha \sim \beta$ when $\forall e\;\;$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.

Definition of Summation among Real Numbers

Define $[\alpha]+[\beta] = [\alpha+\beta]$

$\alpha + \beta$ is defined as $i \mapsto \alpha(i) + \beta(i)$

Question

my exercise book requires me to prove whether the $[\alpha]+[\beta] = [\alpha+\beta]$ is well-defined.

I understand the "Well-define" is try to find out another equivalence class to $[\alpha]$ and $[\beta]$ such as $[\alpha] \sim [\alpha]'$ and $[\beta] \sim [\beta]'$ then whether the definition still works, but frankly my understanding doesn't reach to 'what is the meaning of definition still works?' or 'what exactly in which verification' my textbook requires me.

Any guidance would be grateful.

Well-defined means: if $\alpha \sim \alpha'$ and $\beta\sim\beta'$, then $[\alpha+\beta] = [\alpha'+\beta']$. — Rigel, May 19 '17 at 11:19
@Rigel How to check "=" of $[\alpha+\beta']=[\alpha+\beta']$? — Beverlie, May 19 '17 at 11:21

score 3 · Accepted Answer · answered May 19 '17 at 11:55

3

$\alpha, \alpha', \beta, \beta' : \Bbb N \rightarrow \Bbb Q$

Let $\alpha\sim\alpha'$ and $\beta\sim\beta'$ then

$\forall e \in \Bbb Q>0, \;\; \exists N_1\; s.t. \; \forall i\ge N_1\; \rvert\alpha(i) - \alpha'(i)\lvert \lt {e \over 2}\;$ and $\forall e\in \Bbb Q>0, \;\; \exists N_2\; s.t. \; \forall i\ge N_2\; \rvert\beta(i) - \beta'(i)\lvert \lt {e \over 2}\;$

Now let

$\;\gamma(i) := \alpha(i) + \beta(i) $ $\;\gamma'(i) := \alpha'(i) + \beta'(i)$ $\;N = sup(N_1, N_2)$

then

$\; \forall e \in \Bbb Q>0, \;\; \exists N\; s.t.\; |\gamma(i) - \gamma'(i)|= |\alpha(i) - \alpha'(i)+ \beta(i) - \beta'(i)| \le\; \rvert\alpha(i) - \alpha'(i)\lvert + \rvert\beta(i) - \beta'(i)\lvert \lt e$

Thus $[\gamma] = [\gamma']$ since $\gamma \sim \gamma'$

answered May 19 '17 at 11:55

Beverlie

2,645

@Rigel How's this proof? – Beverlie May 19 '17 at 11:56
It's ok. . . . . – Rigel May 19 '17 at 12:19
@Rigel one more additional question: If I want to check whether to "Associativity" works for the above operator, is it sufficient to check $([\alpha]+[\beta])+[\gamma] = [\alpha + \beta + \gamma] and [\alpha]+([\beta]+[\gamma]) = [\alpha + \beta + \gamma]$? – Beverlie May 19 '17 at 12:25
I think it's enough. – Rigel May 19 '17 at 12:27

score 1 · Answer 2 · answered May 19 '17 at 11:27

Hint: let $\alpha\sim\alpha'$ and $\beta\sim\beta'$, and define $$ \gamma(i) := \alpha(i) + \beta(i), \qquad \gamma'(i) := \alpha'(i) + \beta'(i). $$ You have to show that $[\gamma] = [\gamma']$, i.e., for every $\epsilon > 0$ there exists $N\in\mathbb{N}$ such that $|\gamma(i) - \gamma'(i)| < \epsilon$ for every $i\geq N$.

Got it. Will add my answer soon – Beverlie May 19 '17 at 11:28 — Beverlie, May 19 '17 at 11:28

Well-Define of Summation of Real Number : Cauchy/Quotient Set Approach

2 Answers2