Proof of Order Properties of Limits of Real Sequences

Throughout, $|x|$ is the absolute value of the real number $x$ as described in the statement; by claim 8 of Properties of Complex Conjugation and Modulus it is the modulus of $x$ as a complex number, so $0\le|x|$, $|x+y|\le|x|+|y|$ by claim 7, and $x\le|x|$ and $-x\le|x|$ by claim 6 (the real part of a real number being the number itself, by Real and Imaginary Parts of a Complex Number). In particular, for real $x$ and $\varepsilon$, if $-\varepsilon<x<\varepsilon$ then $|x|<\varepsilon$, because $|x|$ equals $x$ or $-x$. Convergence is as in Limit of a Sequence of Real Numbers, and we use the ordered field properties of $\mathbb{R}$, including that $\le$ is a total order.

Claim 1. Suppose $A\le B$ fails. By totality $B\le A$ and $B\neq A$, so $0<A-B$ and $\varepsilon=(A-B)/2$ is positive. Choose $N_{1}$ with $|a_{n}-A|<\varepsilon$ for $n\ge N_{1}$ and $N_{2}$ with $|b_{n}-B|<\varepsilon$ for $n\ge N_{2}$, and let $n$ be at least as large as both. From $A-a_{n}\le|a_{n}-A|<\varepsilon$ we get $A-\varepsilon<a_{n}$, and from $b_{n}-B\le|b_{n}-B|<\varepsilon$ we get $b_{n}<B+\varepsilon$. Since $A-\varepsilon=(A+B)/2=B+\varepsilon$, this gives $b_{n}<a_{n}$, contradicting the hypothesis $a_{n}\le b_{n}$. Hence $A\le B$.

Claim 2. Let $\varepsilon$ be a positive real number. Choose $N_{1}$ with $|a_{n}-L|<\varepsilon$ for $n\ge N_{1}$ and $N_{2}$ with $|b_{n}-L|<\varepsilon$ for $n\ge N_{2}$, and let $N$ be the larger of the two. For $n\ge N$ we have $L-\varepsilon<a_{n}$ and $b_{n}<L+\varepsilon$ as in claim 1, so

whence $-\varepsilon<c_{n}-L<\varepsilon$ and therefore $|c_{n}-L|<\varepsilon$. Thus $(c_{n})$ converges to $L$.

Claim 3. Let $\varepsilon$ be a positive real number and choose $N$ with $|b_{n}-0|<\varepsilon$ for all $n\ge N$. For such $n$ we have $0\le|c_{n}-L|\le b_{n}$, so $0\le b_{n}$ and hence $|b_{n}|=b_{n}$; therefore

Thus $(c_{n})$ converges to $L$.

Claim 4. For every $n$, the triangle inequality gives $|a_{n}|\le|a_{n}-A|+|A|$ and $|A|\le|A-a_{n}|+|a_{n}|=|a_{n}-A|+|a_{n}|$, using $|-x|=|x|$, which holds because $|-x|=|-1||x|$ and $|-1|=1$ by claims 4 and 8. Hence both $|a_{n}|-|A|$ and $-(|a_{n}|-|A|)$ are at most $|a_{n}-A|$, and since $\bigl||a_{n}|-|A|\bigr|$ is one of these two numbers,

Given a positive real $\varepsilon$, choose $N$ with $|a_{n}-A|<\varepsilon$ for $n\ge N$; then $\bigl||a_{n}|-|A|\bigr|<\varepsilon$ for such $n$. Thus $(|a_{n}|)$ converges to $|A|$.