Monotone Convergence Property $\iff$ Order Completeness in an arbitrary ordered field.

Question

I have found an answer here but having some doubts about the first answer(highest rep.) posted by Jonas Meyer.

There in Claim 1. He states for all $\epsilon>0$, $U-\epsilon:=\{u- \epsilon: u \in U\}$ is not contained in $U$.
He proved by method of contradiction, he writes "If $U-\epsilon \subseteq U$, then $U-n\epsilon \subseteq U$" for all $n$.

My question is how can we conclude this statement?

Edit: I have a few more questions:

1. After proving claim 2, he moves on to construct an increasing sequence. He writes "There exist $n_2>n_1$ such that $x_2 \notin U-\frac{1}{n_1}$" How
   can we obtain that $n_2>n_1$? Can it not be smaller than $n_1$?

2. After creating the monotone increasing sequence $(x_k)$ it is written "(x_k)" is bounded above by each element of U. How can we conclude this?

3. In the second last line it is written "$\bigcap_{j=1}^\infty U-\frac{1}{n_j}=\bigcap_{n=1}^\infty U-\frac{1}{n}$". How can we conclude this?

Answer 1

By induction on $n$. Suppose that $U-\epsilon\subseteq U$. If $U-n\epsilon\subseteq U$ for some $n\in\Bbb N$, then

$$U-(n+1)\epsilon=(U-\epsilon)-n\epsilon\subseteq U-n\epsilon\subseteq U\,.$$

That’s the induction step, and the claim follows immediately.

Added to answer the additional questions.

For the first question, note that $U-\frac1n\subseteq U-\frac1m$ whenever $n\ge m$.¹ Thus, if $x\notin U-\frac1m$, then $x\notin U-\frac1n$ for $n\ge m$. There is certainly some $m$ such that $x_2\notin U-\frac1m$, and we’ve just shown that in that case $x_2\notin U-\frac1n$ for all $n\ge m$, so we can simply take $$n_2=\max\{m,n_1+1\}$$ to ensure that $n_2>n_1$.

¹ To see this, suppose that $n\ge m$, and let $x\in U-\frac1n$. Then $$x+\frac1m\ge x+\frac1n\in U\,,$$ so $x+\frac1m\in U$, and hence $x\in U-\frac1m$.

For your second question, note that if $u\in U$, and $v>u$, then $v\in U$. If the sequence is not bounded above by each element of $U$, there are a $k\in\Bbb Z^+$ and a $u\in U$, such that $u<x_k$. But then $x_k\in U$, contradicting the fact that every point of the sequence was chosen not to be in $U$.

For your third question, the fact that

$$\bigcap_{j=1}^\infty\left(U-\frac1{n_j}\right)=\bigcap_{n=1}^\infty\left(U-\frac1n\right)$$

is an immediate consequence of the fact, proved above, that $$U-\frac1n\subseteq U-\frac1m$$ whenever $n\ge m$:

$$U-1\supseteq U-\frac12\supseteq U-\frac13\supseteq\ldots\;.$$

If a point is not in some $U-\frac1n$, then it isn’t in $U-\frac1{n_j}$ for any $n_j\ge n$, and there is certainly an $n_j\ge n$. (In fact the intersection of any infinite subsequence of a decreasing sequence of sets is the same as the intersection of the entire sequence.)