Proof of Limits of Sequences in an Ordered Field: Uniqueness, Boundedness, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences

Throughout, convergence is that of Convergent Sequences of Real Numbers §converges, and $\mathbb{N}$ carries its order, a total order with strict relation $<$, as recorded in Subsequences §subsequence. Three facts are used repeatedly.

First, for $r\in\{2,3\}$ (the only cases used below) and $N_{1},\dots,N_{r}\in\mathbb{N}$ the set $\{N_{1},\dots,N_{r}\}$ is a nonempty finite subset of $\mathbb{N}$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §small and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §union, so it has a greatest element $\max\{N_{1},\dots,N_{r}\}$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §extremes; every $n\ge\max\{N_{1},\dots,N_{r}\}$ satisfies $n\ge N_{i}$ for each $i$, by transitivity of the order.

Second, for $x,y\in F$ we have $y-x=-(x-y)$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs and $|-z|=|z|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, so $|x-y|=|y-x|$.

Third, for $x,y,\varepsilon\in F$, $|x-y|<\varepsilon$ holds if and only if $-\varepsilon<x-y<\varepsilon$, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §absolute-strict, that is, adding $y$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, if and only if $y-\varepsilon<x<y+\varepsilon$.

Clause unique. Suppose $c_{n}\to L$, $c_{n}\to L'$ and $L\neq L'$. Then $L-L'\neq0$, so $0<|L-L'|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value, and $\varepsilon=|L-L'|/2$ is positive with $\varepsilon+\varepsilon=|L-L'|$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. Choose $N_{1}$ with $|c_{n}-L|<\varepsilon$ for $n\ge N_{1}$ and $N_{2}$ with $|c_{n}-L'|<\varepsilon$ for $n\ge N_{2}$, and put $N=\max\{N_{1},N_{2}\}$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §triangle-three-points, the second fact above and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum,

$$|L-L'|\le|L-c_{N}|+|c_{N}-L'|=|c_{N}-L|+|c_{N}-L'|<\varepsilon+\varepsilon=|L-L'|,$$

so $|L-L'|<|L-L'|$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, which Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive excludes. Hence $L=L'$.

Clause bounded. Let $a_{n}\to A$. Since $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, there is $N\in\mathbb{N}$ with $|a_{n}-A|<1$ for $n\ge N$; for such $n$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §triangle-three-points with $x=a_{n}$, $y=0$, $z=A$ and Rules of Arithmetic and Order in an Ordered Field §order-sum give $|a_{n}|\le|a_{n}-A|+|A|\le1+|A|=|A|+1$. The set $[N]=\{1,\dots,N\}$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §naturals, so its image $\{|a_{k}|:k\in[N]\}$ under $k\mapsto|a_{k}|$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §image, and

$$S=\{|a_{k}|:k\in[N]\}\cup\{|A|+1\}$$

is a nonempty finite subset of $F$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §small and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §union. The order of the ordered field $F$ is total, so $S$ has a greatest element $M=\max S$ by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice §extremes. Let $n\in\mathbb{N}$. By Arithmetic and Order of the Natural Numbers §trichotomy, $n\le N$ or $N\le n$. If $n\le N$, then $n\in[N]$, as $1\le n$ by Arithmetic and Order of the Natural Numbers §least, so $|a_{n}|\le M$; if $N\le n$, then $|a_{n}|\le|A|+1\le M$. Thus $|a_{n}|\le M$ for every $n$, and $(a_{n})$ is bounded.

Clause tails. For the constant sequence, $|\lambda-\lambda|=|0|=0<\varepsilon$ for every positive $\varepsilon$ and every $n\in\mathbb{N}$, by Rules of Arithmetic and Order in an Ordered Field §absolute-value; so $N=1$ serves. Next let $c_{n}=a_{n}$ for $n\ge K$. Given a positive $\varepsilon$, choose $N_{1}$ with $|a_{n}-A|<\varepsilon$ for $n\ge N_{1}$ and put $N=\max\{N_{1},K\}$; for $n\ge N$, $|c_{n}-A|=|a_{n}-A|<\varepsilon$. Finally let $p\in\mathbb{N}$. Given a positive $\varepsilon$, choose $N$ with $|a_{n}-A|<\varepsilon$ for $n\ge N$. For $n\ge N$ we have $n<n+p$ by Arithmetic and Order of the Natural Numbers §difference, hence $n+p\ge N$ and $|a_{n+p}-A|<\varepsilon$. So $(a_{n+p})_{n\in\mathbb{N}}$ converges to $A$.

Clause arithmetic. Sums. Given a positive $\varepsilon$, the element $\varepsilon/2$ is positive with $\varepsilon/2+\varepsilon/2=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. 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 put $N=\max\{N_{1},N_{2}\}$. For $n\ge N$, since $(a_{n}+b_{n})-(A+B)=(a_{n}-A)+(b_{n}-B)$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs and the commutative ring laws, the triangle inequality Rules of Arithmetic and Order in an Ordered Field §triangle, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed give

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

Scalar multiples. If $\lambda=0$, then $\lambda a_{n}=0=\lambda A$ for every $n$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §zero, and the claim is clause tails for the constant sequence $0$. Let $\lambda\neq0$, so $0<|\lambda|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value. Given a positive $\varepsilon$, the element $\delta=\varepsilon\,|\lambda|^{-1}$ is positive by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product, and $|\lambda|\,\delta=\varepsilon$. Choose $N$ with $|a_{n}-A|<\delta$ for $n\ge N$. For $n\ge N$, as $\lambda a_{n}-\lambda A=\lambda(a_{n}-A)$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs and the commutative ring laws, Rules of Arithmetic and Order in an Ordered Field §absolute-value and Rules of Arithmetic and Order in an Ordered Field §order-product give $|\lambda a_{n}-\lambda A|=|\lambda|\,|a_{n}-A|<|\lambda|\,\delta=\varepsilon$.

Negatives and differences. By Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs, $-a_{n}=(-1)a_{n}$ and $-A=(-1)A$, so $-a_{n}\to-A$ is the case $\lambda=-1$. Then $a_{n}-b_{n}=a_{n}+(-b_{n})$ by Negatives, Differences, Reciprocals and Quotients §negative, and the case of sums, applied to $(a_{n})$ and $(-b_{n})$, gives $a_{n}-b_{n}\to A+(-B)=A-B$.

Products. By clause bounded there is $M\in F$ with $|a_{n}|\le M$ for every $n$; then $0\le|a_{1}|\le M$. Put $C=M+|B|+1$. Since $0\le M$ and $0\le|B|$, we have $0\le M+|B|$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, hence $0<C$ by $0<1$ and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum; and Rules of Arithmetic and Order in an Ordered Field §order-sum gives $|a_{n}|\le C$ for every $n$ and $|B|\le C$. For every $n$, the identity $a_{n}b_{n}-AB=a_{n}(b_{n}-B)+B(a_{n}-A)$, from Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs and the commutative ring laws, together with Rules of Arithmetic and Order in an Ordered Field §triangle, Rules of Arithmetic and Order in an Ordered Field §absolute-value, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling and Rules of Arithmetic and Order in an Ordered Field §order-sum, gives

$$|a_{n}b_{n}-AB|\le|a_{n}|\,|b_{n}-B|+|B|\,|a_{n}-A|\le C\,|b_{n}-B|+C\,|a_{n}-A|.$$

Given a positive $\varepsilon$, put $\delta=(\varepsilon/2)\,C^{-1}$, which is positive by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving, Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product, and satisfies $C\delta=\varepsilon/2$. Choose $N_{1}$ with $|a_{n}-A|<\delta$ for $n\ge N_{1}$ and $N_{2}$ with $|b_{n}-B|<\delta$ for $n\ge N_{2}$, and put $N=\max\{N_{1},N_{2}\}$. For $n\ge N$, Rules of Arithmetic and Order in an Ordered Field §order-product gives $C\,|b_{n}-B|<\varepsilon/2$ and $C\,|a_{n}-A|<\varepsilon/2$, so $|a_{n}b_{n}-AB|<\varepsilon/2+\varepsilon/2=\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving.

Clause quotient. Let $B\neq0$ and $b_{n}\neq0$ for every $n$. For $x\in F$ with $x\neq0$ we have $|x|\,|x^{-1}|=|x\,x^{-1}|=|1|=1$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value and Rules of Arithmetic and Order in an Ordered Field §squares, so $|x^{-1}|=|x|^{-1}$ by Negatives, Differences, Reciprocals and Quotients §reciprocal. We first show $b_{n}^{-1}\to B^{-1}$. As $0<|B|$, the element $\beta=|B|/2$ is positive with $\beta+\beta=|B|$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. Choose $N_{1}$ with $|b_{n}-B|<\beta$ for $n\ge N_{1}$. For $n\ge N_{1}$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §reverse-triangle and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed give $\big||b_{n}|-|B|\big|<\beta$, so by the third fact above $|B|-\beta<|b_{n}|$, that is $0<\beta<|b_{n}|$, and therefore $|b_{n}|^{-1}<\beta^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. Put $\Gamma=\beta^{-1}|B|^{-1}$, which is positive by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product. For every $n$, $b_{n}^{-1}-B^{-1}=(B-b_{n})\,b_{n}^{-1}B^{-1}$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs and Negatives, Differences, Reciprocals and Quotients §reciprocal; hence for $n\ge N_{1}$, by Rules of Arithmetic and Order in an Ordered Field §absolute-value, the second fact above and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §nonnegative-scaling,

$$|b_{n}^{-1}-B^{-1}|=|b_{n}-B|\,|b_{n}|^{-1}\,|B|^{-1}\le \Gamma\,|b_{n}-B|.$$

Given a positive $\varepsilon$, the element $\varepsilon \Gamma^{-1}$ is positive by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product. Choose $N_{2}$ with $|b_{n}-B|<\varepsilon \Gamma^{-1}$ for $n\ge N_{2}$, and put $N=\max\{N_{1},N_{2}\}$. For $n\ge N$, Rules of Arithmetic and Order in an Ordered Field §order-product and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed give $|b_{n}^{-1}-B^{-1}|<\Gamma\,\varepsilon \Gamma^{-1}=\varepsilon$. So $b_{n}^{-1}\to B^{-1}$, and the product case of clause arithmetic, applied to $(a_{n})$ and $(b_{n}^{-1})$, gives $a_{n}/b_{n}=a_{n}b_{n}^{-1}\to AB^{-1}=A/B$, quotients being as in Negatives, Differences, Reciprocals and Quotients §reciprocal.

Clause absolute. Given a positive $\varepsilon$, choose $N$ with $|a_{n}-A|<\varepsilon$ for $n\ge N$. For $n\ge N$, $\big||a_{n}|-|A|\big|\le|a_{n}-A|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §reverse-triangle and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed.

Clause finite-sums. The addition of $F$ is associative and commutative with neutral element $0$, so the sums are those of Sums and Products over a Finite Set and over an Interval §operation and Sums and Products over a Finite Set and over an Interval §empty. If $I=\emptyset$, then $\sum_{i\in I}d(i,n)=0$ for every $n$ and $\sum_{i\in I}D_{i}=0$ by Sums and Products over a Finite Set and over an Interval §empty, and the claim is clause tails for the constant sequence $0$.

Let $I\neq\emptyset$, let $q=\#I\in\mathbb{N}$, and let $\varphi$ be a bijection from $[q]$ onto $I$, as in Sums and Products over a Finite Set and over an Interval §operation. By that clause and Sums and Products over a Finite Set and over an Interval §intervals, for every $n\in\mathbb{N}$

$$\sum_{i\in I}d(i,n)=\sum_{j=1}^{q}d(\varphi(j),n),\qquad\sum_{i\in I}D_{i}=\sum_{j=1}^{q}D_{\varphi(j)}.$$

We induct along the enumeration $\varphi$: by Arithmetic and Order of the Natural Numbers §induction, applied to the set of those $m\in\mathbb{N}$ for which $m\le q$ implies that $\big(\sum_{j=1}^{m}d(\varphi(j),n)\big)_{n\in\mathbb{N}}$ converges to $\sum_{j=1}^{m}D_{\varphi(j)}$, it suffices to treat $m=1$ and the step from $m$ to $m+1$. For $m=1$, the interval $\{1,\dots,1\}$ is $\{1\}$ by antisymmetry of the order, so by Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §singleton the claim reads $d(\varphi(1),n)\to D_{\varphi(1)}$, which holds by hypothesis as $\varphi(1)\in I$. Suppose the claim for $m$, and let $m+1\le q$; then $m<m+1\le q$ by Arithmetic and Order of the Natural Numbers §successor, so $m\le q$ and the claim for $m$ applies. As $1\le m+1\le q$, $m+1\in[q]$, so $\varphi(m+1)\in I$. By Iterated Operations over Finite Sets: Singletons, Disjoint Unions, Reindexing, Products of Sets, Termwise Combination, Homomorphisms and Intervals §interval-recursion, with the interval sums of Sums and Products over a Finite Set and over an Interval §sums,

$$\sum_{j=1}^{m+1}d(\varphi(j),n)=\sum_{j=1}^{m}d(\varphi(j),n)+d(\varphi(m+1),n),\qquad\sum_{j=1}^{m+1}D_{\varphi(j)}=\sum_{j=1}^{m}D_{\varphi(j)}+D_{\varphi(m+1)},$$

and $d(\varphi(m+1),n)\to D_{\varphi(m+1)}$ by hypothesis; so the case of sums in clause arithmetic gives the claim for $m+1$. Taking $m=q$ proves the clause.

Clause order. Let $a_{n}\le b_{n}$ for every $n\ge K$, and suppose that $A\le B$ fails. Then $B<A$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, so $0<A-B$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, and $\varepsilon=(A-B)/2$ is positive with $\varepsilon+\varepsilon=A-B$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving; hence $B+\varepsilon=A-\varepsilon$. 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 put $N=\max\{N_{1},N_{2},K\}$. By the third fact above,

$$b_{N}<B+\varepsilon=A-\varepsilon<a_{N},$$

so $b_{N}<a_{N}$, while $a_{N}\le b_{N}$ as $N\ge K$; this contradicts Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Hence $A\le B$. The constant sequence $\lambda$ converges to $\lambda$ by clause tails. If $a_{n}\le\lambda$ for $n\ge K$, the case just proved, with the constant sequence $\lambda$ in place of $(b_{n})$, gives $A\le\lambda$; if $\lambda\le a_{n}$ for $n\ge K$, it gives $\lambda\le A$ when applied with the constant sequence $\lambda$ in place of the first sequence and $(a_{n})$ in place of the second.

Clause squeeze. Let $A=B$ and $a_{n}\le c_{n}\le b_{n}$ for every $n\ge K$. Given a positive $\varepsilon$, choose $N_{1}$ with $|a_{n}-A|<\varepsilon$ for $n\ge N_{1}$ and $N_{2}$ with $|b_{n}-A|<\varepsilon$ for $n\ge N_{2}$, and put $N=\max\{N_{1},N_{2},K\}$. For $n\ge N$, the third fact above and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed give

$$A-\varepsilon<a_{n}\le c_{n}\le b_{n}<A+\varepsilon,$$

so $A-\varepsilon<c_{n}<A+\varepsilon$, that is $|c_{n}-A|<\varepsilon$ by the third fact again.

Clause domination. Let $(b_{n})$ be a null sequence and $|c_{n}-L|\le b_{n}$ for every $n\ge K$. Given a positive $\varepsilon$, choose $N_{1}$ with $|b_{n}-0|<\varepsilon$ for $n\ge N_{1}$, and put $N=\max\{N_{1},K\}$. For $n\ge N$, using $b_{n}\le|b_{n}|$ from Rules of Arithmetic and Order in an Ordered Field §absolute-value and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed,

$$|c_{n}-L|\le b_{n}\le|b_{n}|=|b_{n}-0|<\varepsilon .$$

Clause subsequence. Let $a\circ\sigma$ be a subsequence of $(a_{n})$, with $\sigma:\mathbb{N}\to\mathbb{N}$ strictly increasing. Given a positive $\varepsilon$, choose $N$ with $|a_{n}-A|<\varepsilon$ for $n\ge N$. For $k\ge N$ we have $\sigma(k)\ge k\ge N$ by Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §index, so $|a_{\sigma(k)}-A|<\varepsilon$. Hence $a_{\sigma(k)}\to A$.