Proof of Arithmetic of Limits of Real Sequences

For a real number $x$, $|x|$ denotes its absolute value in the ordered field $\mathbb{R}$. By Properties of the Absolute Value in an Ordered Field §nonnegative, $0\le|x|$ and $|x|=0$ only for $x=0$; by Properties of the Absolute Value in an Ordered Field §multiplicative, $|xy|=|x|\,|y|$; by Properties of the Absolute Value in an Ordered Field §triangle, $|x+y|\le|x|+|y|$; and by Properties of the Absolute Value in an Ordered Field §symmetry, $|-x|=|x|$. For $x\neq0$ one has $|x^{-1}|=1/|x|$: indeed $|x|\,|x^{-1}|=|x\,x^{-1}|=|1|=1$ by multiplicativity, where $|1|=1$ because $0\le1$ (Elementary Arithmetic in an Ordered Field §unit-nonnegative), and $|x|\neq0$. Convergence is as in Limit of a Sequence of Real Numbers. We use the ordered field properties of $\mathbb{R}$, in particular the order reversal of reciprocals: if $0<p\le q$ then $1/q\le1/p$, and if $0<p<q$ then $1/q<1/p$ (Elementary Order Arithmetic in an Ordered Field §reciprocal). For all real $y,z$ we have $|y-z|=|z-y|$, by Properties of the Absolute Value in an Ordered Field §symmetry applied to $y-z=-(z-y)$.

Claim 1. Let $\varepsilon$ be a positive real number. Choose $N_{1}$ with $|a_{n}-A|<\varepsilon/2$ for $n\ge N_{1}$ and $N_{2}$ with $|b_{n}-B|<\varepsilon/2$ for $n\ge N_{2}$, and let $N$ be the larger. For $n\ge N$,

$$\bigl|(a_{n}+b_{n})-(A+B)\bigr|=\bigl|(a_{n}-A)+(b_{n}-B)\bigr|\le|a_{n}-A|+|b_{n}-B|<\varepsilon/2+\varepsilon/2=\varepsilon .$$

Claim 2. By claim 2 of Uniqueness of Limits and Boundedness of Convergent Real Sequences there is a positive real $M$ with $|a_{n}|\le M$ for every $n$. Put $K=M+|B|+1$, so that $K$ is positive and $M+|B|\le K$. Let $\varepsilon$ be positive; then $\varepsilon/(2K)$ is positive. Choose $N_{1}$ with $|a_{n}-A|<\varepsilon/(2K)$ for $n\ge N_{1}$, and $N_{2}$ with $|b_{n}-B|<\varepsilon/(2K)$ for $n\ge N_{2}$; let $N$ be the larger. For $n\ge N$, using $a_{n}b_{n}-AB=a_{n}(b_{n}-B)+(a_{n}-A)B$,

$$|a_{n}b_{n}-AB|\le|a_{n}|\,|b_{n}-B|+|a_{n}-A|\,|B|\le M|b_{n}-B|+|B|\,|a_{n}-A|<(M+|B|)\frac{\varepsilon}{2K}\le\frac{\varepsilon}{2}<\varepsilon ,$$

the strict inequality in the middle because $M$ is positive. Hence $(a_{n}b_{n})$ converges to $AB$.

Claim 3. The constant sequence with every term equal to $c$ converges to $c$, since $|c-c|=0<\varepsilon$ for every positive $\varepsilon$. Applying claim 2 to $(a_{n})$ and this constant sequence gives that $(a_{n}c)=(ca_{n})$ converges to $Ac=cA$. Taking $c=-1$ shows $(-a_{n})$ converges to $-A$, and then claim 1 applied to $(a_{n})$ and $(-b_{n})$ shows that $(a_{n}-b_{n})$ converges to $A-B$.

Claim 4. We first show that $(1/b_{n})$ converges to $1/B$. Since $B\neq0$ we have $0<|B|$, so $|B|/2$ is positive. Choose $N_{0}$ with $|b_{n}-B|<|B|/2$ for $n\ge N_{0}$. For such $n$,

$$|B|=\bigl|(B-b_{n})+b_{n}\bigr|\le|b_{n}-B|+|b_{n}|<|B|/2+|b_{n}| ,$$

so $|B|/2<|b_{n}|$, and in particular $1/|b_{n}|<2/|B|$ by the strict reciprocal property. Since $\frac{1}{b_{n}}-\frac{1}{B}=(B-b_{n})\,b_{n}^{-1}B^{-1}$, multiplicativity, $|x^{-1}|=1/|x|$ and $|B-b_{n}|=|b_{n}-B|$ give, for $n\ge N_{0}$,

$$\Bigl|\frac{1}{b_{n}}-\frac{1}{B}\Bigr|=\frac{|b_{n}-B|}{|b_{n}|\,|B|}\le\frac{2\,|b_{n}-B|}{|B|^{2}} ,$$

the inequality by multiplying $1/|b_{n}|<2/|B|$ by the nonnegative number $|b_{n}-B|/|B|$ (Elementary Arithmetic in an Ordered Field §scaling). Let $\varepsilon$ be positive; then $\varepsilon|B|^{2}/2$ is positive, so there is $N_{1}$ with $|b_{n}-B|<\varepsilon|B|^{2}/2$ for $n\ge N_{1}$. For $n$ at least as large as both $N_{0}$ and $N_{1}$, multiplying $|b_{n}-B|<\varepsilon|B|^{2}/2$ by the positive number $2/|B|^{2}$ (Elementary Order Arithmetic in an Ordered Field §strict-scaling) and combining with the display above (Elementary Order Arithmetic in an Ordered Field §mixed-transitivity) gives $|1/b_{n}-1/B|<\varepsilon$. Thus $(1/b_{n})$ converges to $1/B$, and applying claim 2 to $(a_{n})$ and $(1/b_{n})$ shows that $(a_{n}/b_{n})=(a_{n}\cdot(1/b_{n}))$ converges to $A\cdot(1/B)=A/B$.