Doubts regarding a set exercise and proof verification

59 Views Asked by At

I took this exercise from textbook A Course in Mathematical Analysis by Prof D. J. H. Garling and have two questions regarding how I should infer the stimulus.

Thank you for your help!

enter image description here

My questions:

  1. Are $A_1,A_2,A_3,...$ all ordered by the same kind of order, or every $A_i$ is ordered by a different order?

  2. What is the order (or orders) that the author refers to?

  3. Please check my below proof.

For any $x,y \in A$, we have two possible cases.

  1. $x=y$

$\implies x \leq y$

  1. $x \neq y$

$\implies \exists t \in \mathbb{N}, x_t \neq y_t \implies X= \{n \in \mathbb{N} \mid x_n \neq y_n\} \neq \varnothing$.

By Well-ordering Principle, $X$ has the least element and this least element is $\inf X$.

$\implies k(x,y)=\inf X \in X \implies x_{k(x,y)} \neq y_{k(x,y)}$.

Since $A_i$ is TOTALLY ordered for all $i \in \mathbb{N}$ and $x_{k(x,y)} \neq y_{k(x,y)}$, then either $x_{k(x,y)} < y_{k(x,y)}$ or $y_{k(x,y)} < x_{k(x,y)}$.

If $x_{k(x,y)} < y_{k(x,y)}$, then $x \leq y$.

If $y_{k(x,y)} < x_{k(x,y)}$, then $y \leq x$.

To sum up, we have $x=y$ or $x \leq y$ or $y \leq x$.

Thus $A$ is totally ordered by $\leq$.