A confusion about the use of triangular inequality and abolute value in a proof

Question

In the book of Mathematical Analysis I by V.A Zorich, at page 85, it is given that

Let ${x_n}$ and ${y_n}$ be two convergent sequences with $lim_{n→∞} x_n = A$ and $lim_{n→∞} y_n = B$. If $A<B$, then there exists an index $N ∈ \mathbb{N}$ such that $x_n < y_n$ for all $n \geq N.$

Proof: Choose a number $C$ such that $A<C<B$. By definition of limit, we can find numbers $N'$and $N''$such that $|x_n − A| < C − A$ for all $n>N'$and $|y_n − B| < B −C$ for all $n>N''$. Then for $n>N = max{N',N''}$ we shall have $x_n < A+C − A = C = B − (B − C)<y_n$.

However, I couldn't understand how the author get the last inequality, namely $$x_n < A+C − A = C = B − (B − C)<y_n.$$

I mean from the triangular inequality we can have $$|x_n - A| \leq |x_n| + |A| < C-A,$$ but after that how can we get rid of the absolute value of $A$ ? And similarly for the upper left of the inequality ?

Answer 1

If $|x_{n}-A|<C-A$, as $x_{n}-A\leq|x_{n}-A|$, so $x_{n}-A<C-A$, so $x_{n}<A+C-A$. If $|y_{n}-B|<B-C$, as $-(y_{n}-B)\leq|y_{n}-B|$, so $-(y_{n}-B)<B-C$, so $y_{n}-B>C-B$, so $y_{n}>C-B+B=C$.

Answer 2

$|x_n-A|\lt C-A$ implies $x_n-A\lt C-A$ because $x_n-A\le|x_n-A|$. (In general, $x\le|x|$.) It is not anything more profound than that.

Answer 3

What do you mean by $|x|<a$? Just $-a<x<a$. Now put $x_{n} - A$ in place of $x$ and $C-A$ in place of $a$ to get $$-(C-A) <x_{n} - A<C-A$$ the second inequality above gives $$x_{n} <A+(C-A) =C$$ Similarly we have $$-(B-C) <y_{n} - B<B-C$$ and the first inequality gives $$C=B-(B-C) <y_{n} $$

---

Most of the proofs (the easy ones at least) in analysis need just a basic understanding of inequalities (and not logical quantifiers as most people believe). The problem is that most of high school curriculum is focused on development of deep skills in algebraic manipulation using symbols like $+, -, \times, /$ and the symbols $<, >$ are left out.