Proof of Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences

Each result cited is universally quantified over the data in its own statement.

We work in the setting of The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness. By its clause The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals, $\mathbb{R}$ is an ordered field, so the notation of Commutative Rings, Fields and Ordered Fields: Standard Notation applies to it and the rules of Rules of Arithmetic and Order in an Ordered Field are in force by Commutative Rings, Fields and Ordered Fields: Standard Notation §ordered-fields. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §identification, $\mathbb{N}\subseteq\mathbb{R}$: a natural number $n$ standing where a real number is required is identified with the element of $\mathbb{R}$ it denotes, and by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement sums, products and the order agree whether formed in $\mathbb{N}$ or in $\mathbb{R}$. In particular every $n\in\mathbb{N}$ satisfies $0<n$ in $\mathbb{R}$, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals; so $n\neq0$, and $1/n=n^{-1}$ is defined. Elementary rearrangements in the field $\mathbb{R}$, such as $x-0=x$ or $L-x=-(x-L)$ (the latter by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §signs), are used without comment. In particular, $0+0=0$ and $1\cdot1=1$ by Commutative Rings §ring, so $-0=0$ by Negatives, Differences, Reciprocals and Quotients §negative, whence $x-0=x+0=x$ for $x\in\mathbb{R}$, and $1^{-1}=1$ by Negatives, Differences, Reciprocals and Quotients §reciprocal.

Clause reciprocals. We show that the sequence $(1/n)_{n\in\mathbb{N}}$ converges to $0$ in the sense of Convergent Sequences of Real Numbers §converges.

Let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$. By Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, $0<\varepsilon^{-1}$. By the Archimedean property The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean, applied to $x=\varepsilon^{-1}$, there is $N\in\mathbb{N}$ with $\varepsilon^{-1}<N$. Let $n\in\mathbb{N}$ with $n\ge N$. Then $N\le n$ also as real numbers, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement, so $\varepsilon^{-1}<n$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. Now Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, applied to $0<\varepsilon^{-1}<n$, gives $n^{-1}<(\varepsilon^{-1})^{-1}$, and applied to $0<n$ it gives $0<n^{-1}$; moreover $(\varepsilon^{-1})^{-1}=\varepsilon$ by Rules of Arithmetic and Order in an Ordered Field §reciprocals. Since $-\varepsilon<0$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-negative, we have $-\varepsilon<1/n-0<\varepsilon$, and so $|1/n-0|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §absolute-strict. Thus $N$, chosen after $\varepsilon$, satisfies the definition, and $1/n\to0$.

Clause geometric. Let $r\in\mathbb{R}$ with $|r|<1$, and let powers be as in Powers with Exponents in the Natural Numbers with Zero §power.

If $r=0$, then $r^{n}=0$ for every $n\in\mathbb{N}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §product, since $n\ge1$ by Arithmetic and Order of the Natural Numbers §least. So $(r^{n})$ is the constant sequence $(0)_{n\in\mathbb{N}}$, which converges to $0$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §constant.

Let now $r\neq0$. Then $0<|r|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value. From $0<|r|<1$, Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal gives $1=1^{-1}<|r|^{-1}$, with $1^{-1}=1$ as noted at the start. Put $h=|r|^{-1}-1$. Adding $-1$ to both sides of $1<|r|^{-1}$ gives $0<h$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, and $1+h=|r|^{-1}$. Since $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-negative gives $-1<-0=0$; so $h>0>-1$, hence $-1<h$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive and $h\ge-1$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict.

Let $n\in\mathbb{N}$. Bernoulli's inequality Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §bernoulli, applied with $x=h$, gives $(1+h)^{n}\ge1+nh$. Since $0<n$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals and $0<h$, we have $0<nh$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product. Since $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, $0\le1$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict, and Rules of Arithmetic and Order in an Ordered Field §order-sum gives $nh\le1+nh$. Hence $0<nh\le(1+h)^{n}$, and by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §reciprocal-order

$$\big((1+h)^{n}\big)^{-1}\le(nh)^{-1}.$$

By Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §reciprocal, applied to $|r|\neq0$, we have $|r|^{n}\neq0$ and $(1+h)^{n}=(|r|^{-1})^{n}=(|r|^{n})^{-1}$. Hence $\big((1+h)^{n}\big)^{-1}=|r|^{n}$ by Rules of Arithmetic and Order in an Ordered Field §reciprocals. Also $(nh)^{-1}=h^{-1}\,n^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §reciprocals. Finally $|r^{n}|=|r|^{n}$ by Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality §sign. Together:

$$|r^{n}-0|=|r|^{n}\le h^{-1}\cdot\frac{1}{n}\qquad\text{for every }n\in\mathbb{N}.$$

By the clause reciprocals proved above, $1/n\to0$. By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §arithmetic, with $\lambda=h^{-1}$, the sequence $(h^{-1}\cdot1/n)_{n\in\mathbb{N}}$ converges to $h^{-1}\cdot0$, which is $0$ by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field §zero; so it is a null sequence. By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §domination, applied with $c_{n}=r^{n}$, $L=0$, $b_{n}=h^{-1}\cdot1/n$ and $K=1$, we conclude $r^{n}\to0$.

Clause monotone. Let $(a_{n})$ be a sequence in $\mathbb{R}$ and $S=\{a_{n}:n\in\mathbb{N}\}$ the set of its terms. $S$ is nonempty, since $a_{1}\in S$.

First let $(a_{n})$ be nondecreasing and bounded above. By Bounded Sequences of Real Numbers §bounded, $S$ is bounded above in the sense of Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum, the nonempty set $S$, being bounded above, has a supremum $s=\sup S$, which is an upper bound of $S$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum.

Let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$. By Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers §epsilon-above, applied to the nonempty set $S$, which is bounded above, there is an element of $S$ greater than $s-\varepsilon$. That element is a term $a_{N}$ with $N\in\mathbb{N}$, so $s-\varepsilon<a_{N}$. Let $n\in\mathbb{N}$ with $n\ge N$. Then $a_{N}\le a_{n}$ by Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §monotone, and $a_{n}\le s$ since $s$ is an upper bound of $S$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, $s-\varepsilon<a_{n}$, so $-\varepsilon<a_{n}-s$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Also, adding $-s$ to $a_{n}\le s$ gives $a_{n}-s\le s-s=0$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, and $0<\varepsilon$, so $a_{n}-s<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. Hence $|a_{n}-s|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §absolute-strict. Since $N$ was chosen after $\varepsilon$, this shows $a_{n}\to\sup S$ in the sense of Convergent Sequences of Real Numbers §converges.

Now let $(a_{n})$ be nonincreasing and bounded below. By Bounded Sequences of Real Numbers §bounded, $S$ is bounded below. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §infimum, the nonempty set $S$, being bounded below, has an infimum $t=\inf S$, which is a lower bound of $S$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum.

Let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$. By Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers §epsilon-below, applied to the nonempty set $S$, which is bounded below, there is an element of $S$ less than $t+\varepsilon$. That element is a term $a_{N}$ with $N\in\mathbb{N}$, so $a_{N}<t+\varepsilon$. Let $n\in\mathbb{N}$ with $n\ge N$. Then $t\le a_{n}$, and $a_{n}\le a_{N}$ by Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §monotone, so $a_{n}<t+\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. By Rules of Arithmetic and Order in an Ordered Field §order-sum, $t\le a_{n}$ gives $0\le a_{n}-t$, and adding $-t$ to $a_{n}<t+\varepsilon$ gives $a_{n}-t<\varepsilon$. Since $0<\varepsilon$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-negative gives $-\varepsilon<-0=0$. Thus $-\varepsilon<0\le a_{n}-t<\varepsilon$, so $-\varepsilon<a_{n}-t$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, and $|a_{n}-t|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §absolute-strict. Hence $a_{n}\to\inf S$.

Clause bolzano-weierstrass. Let $(a_{n})$ be bounded, and fix $M\in\mathbb{R}$ with $|a_{n}|\le M$ for every $n\in\mathbb{N}$. Then every sequence of the form $a\circ\sigma$, with $\sigma:\mathbb{N}\to\mathbb{N}$, is bounded with the same $M$, since its terms are terms of $(a_{n})$. By Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences §two-sided, every such sequence is therefore both bounded above and bounded below. We let $\mathbb{N}$ carry its order, a total order with strict relation $<$ by Arithmetic and Order of the Natural Numbers §partial-order and Arithmetic and Order of the Natural Numbers §trichotomy, as in the preamble of the definition of subsequences.

Call $m\in\mathbb{N}$ a peak if $a_{n}\le a_{m}$ for every $n\in\mathbb{N}$ with $m\le n$, and let $P$ be the set of peaks, a subset of $\mathbb{N}$. Either for every $N\in\mathbb{N}$ there is a peak $m$ with $N\le m$, or there is $N_{0}\in\mathbb{N}$ such that no $m\in\mathbb{N}$ with $N_{0}\le m$ is a peak. We treat the two cases separately. In both, a least element of a subset of $\mathbb{N}$ is unique and is written $\min$, as in Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §least.

Case 1: for every $N\in\mathbb{N}$ there is a peak $m\ge N$. For $u\in\mathbb{N}$, the set $P_{u}=\{m\in P:u\le m\}$ is nonempty by the case hypothesis. So it has a least element by Arithmetic and Order of the Natural Numbers §well-order, and $\min P_{u}\in\mathbb{N}$ is defined, $\min$ being used properly at the nonempty subset $P_{u}$; hence $p:\mathbb{N}\to\mathbb{N}$, $u\mapsto\min P_{u}$, is a map by Sets and Maps: Ordinary Notation §maps. Apply Recursion on the Natural Numbers Starting at One §recursion with the set $\mathbb{N}$, the element $c=p(1)$, and the map $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$, $(k,u)\mapsto p(u+1)$, given by Sets and Maps: Ordinary Notation §maps. It yields a map $\sigma:\mathbb{N}\to\mathbb{N}$ with

$$\sigma(1)=p(1),\qquad\sigma(k+1)=p(\sigma(k)+1)\quad\text{for every }k\in\mathbb{N}.$$

Every $j\in\mathbb{N}$ is $1$ or of the form $k+1$ with $k\in\mathbb{N}$, by Arithmetic and Order of the Natural Numbers §predecessor. So every value $\sigma(j)$ is a value of $p$, and hence a peak.

Let $k\in\mathbb{N}$. Since $\sigma(k+1)\in P_{\sigma(k)+1}$, we have $\sigma(k)+1\le\sigma(k+1)$. Also $\sigma(k)<\sigma(k)+1$ by Arithmetic and Order of the Natural Numbers §successor. So $\sigma(k)<\sigma(k+1)$ by Arithmetic and Order of the Natural Numbers §partial-order. Thus $\sigma$ is strictly increasing, and $a\circ\sigma$ is a subsequence of $(a_{n})$. Moreover $\sigma(k)$ is a peak, and $\sigma(k)\le\sigma(k+1)$ because $\sigma(k)<\sigma(k+1)$, by Arithmetic and Order of the Natural Numbers §partial-order; so $a_{\sigma(k+1)}\le a_{\sigma(k)}$. Hence $a\circ\sigma$ is nonincreasing. It is bounded below, as noted above, so by the clause monotone proved above it converges, to $\inf\{a_{\sigma(k)}:k\in\mathbb{N}\}$, which exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §infimum, this set of terms being nonempty and, by Bounded Sequences of Real Numbers §bounded, bounded below.

Case 2: there is $N_{0}\in\mathbb{N}$ such that no $m\ge N_{0}$ is a peak. We first show the following. If $u\in\mathbb{N}$ and $N_{0}\le u$, then there is $n\in\mathbb{N}$ with $u<n$ and $a_{u}<a_{n}$. Indeed, $u$ is not a peak, so there is $n\in\mathbb{N}$ with $u\le n$ and not $a_{n}\le a_{u}$. The order of $\mathbb{R}$ is total by Ordered Fields §ordered-field and Partial and Total Orders on a Set and the Associated Strict Relation §total, so $a_{u}\le a_{n}$. If $a_{n}=a_{u}$, then $a_{n}\le a_{u}$ by reflexivity of the order of $\mathbb{R}$ (Partial and Total Orders on a Set and the Associated Strict Relation §partial), contrary to the choice of $n$; so $a_{n}\neq a_{u}$, and thus $n\neq u$. Hence $a_{u}<a_{n}$ by Partial and Total Orders on a Set and the Associated Strict Relation §strict, and $u<n$ by Arithmetic and Order of the Natural Numbers §partial-order.

For $u\in\mathbb{N}$ let

$$T_{u}=\{n\in\mathbb{N}:u<n\ \text{and}\ (u<N_{0}\ \text{or}\ a_{u}<a_{n})\}.$$

If $u<N_{0}$, then $u+1\in T_{u}$ by Arithmetic and Order of the Natural Numbers §successor. Otherwise $N_{0}\le u$ by Arithmetic and Order of the Natural Numbers §trichotomy, and $T_{u}$ is nonempty by the preceding paragraph. In either case $T_{u}$ has a least element by Arithmetic and Order of the Natural Numbers §well-order, so $\min T_{u}\in\mathbb{N}$ is defined, $\min$ being used properly at the nonempty subset $T_{u}$, and $q:\mathbb{N}\to\mathbb{N}$, $u\mapsto\min T_{u}$, is a map by Sets and Maps: Ordinary Notation §maps, with $u<q(u)$ for every $u\in\mathbb{N}$. Moreover $a_{u}<a_{q(u)}$ whenever $N_{0}\le u$, since then $u<N_{0}$ fails by Arithmetic and Order of the Natural Numbers §trichotomy.

Apply Recursion on the Natural Numbers Starting at One §recursion with the set $\mathbb{N}$, the element $c=N_{0}$, and the map $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$, $(k,u)\mapsto q(u)$, given by Sets and Maps: Ordinary Notation §maps. It yields a map $\sigma:\mathbb{N}\to\mathbb{N}$ with $\sigma(1)=N_{0}$ and $\sigma(k+1)=q(\sigma(k))$ for every $k\in\mathbb{N}$. Then $\sigma(k)<q(\sigma(k))=\sigma(k+1)$ for every $k$, so $\sigma$ is strictly increasing, and $a\circ\sigma$ is a subsequence of $(a_{n})$.

For $k\in\mathbb{N}$ we have $1\le k$ by Arithmetic and Order of the Natural Numbers §least. So either $k=1$, or $1<k$ and then $\sigma(1)<\sigma(k)$ by Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §monotone. In both cases $N_{0}\le\sigma(k)$, and therefore $a_{\sigma(k)}<a_{q(\sigma(k))}=a_{\sigma(k+1)}$. Hence $a\circ\sigma$ is nondecreasing, by Partial and Total Orders on a Set and the Associated Strict Relation §strict. It is bounded above, as noted above, so by the clause monotone it converges, to $\sup\{a_{\sigma(k)}:k\in\mathbb{N}\}$, which exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §supremum, this set of terms being nonempty and, by Bounded Sequences of Real Numbers §bounded, bounded above.

In both cases $(a_{n})$ has a convergent subsequence.

Clause cauchy. Let $(a_{n})$ be a sequence in $\mathbb{R}$, with Cauchy sequences as in Cauchy Sequences of Real Numbers §cauchy.

Convergent implies Cauchy. Let $a_{n}\to L$ with $L\in\mathbb{R}$, and let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$. Then $\varepsilon/2$ is positive by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. Choose $N\in\mathbb{N}$, after $\varepsilon$, with $|a_{n}-L|<\varepsilon/2$ for every $n\ge N$. Let $m,n\in\mathbb{N}$ with $m,n\ge N$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §triangle-three-points,

$$|a_{m}-a_{n}|\le|a_{m}-L|+|L-a_{n}|,$$

and $|L-a_{n}|=|-(a_{n}-L)|=|a_{n}-L|$ by Rules of Arithmetic and Order in an Ordered Field §absolute-value. Then Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum gives $|a_{m}-L|+|a_{n}-L|<\varepsilon/2+\varepsilon/2$, which is $\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. So $|a_{m}-a_{n}|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, and $(a_{n})$ is a Cauchy sequence.

Cauchy implies convergent. Let $(a_{n})$ be a Cauchy sequence.

A bounded tail. Applying the Cauchy property with $\varepsilon=1$, which is positive by Rules of Arithmetic and Order in an Ordered Field §squares, gives $N_{1}\in\mathbb{N}$ with $|a_{m}-a_{n}|<1$ for all $m,n\ge N_{1}$. Let $\rho:\mathbb{N}\to\mathbb{N}$ be the map $n\mapsto n+N_{1}$, given by Sets and Maps: Ordinary Notation §maps as $n+N_{1}\in\mathbb{N}$, and let $b=a\circ\rho$, that is, $b_{n}=a_{n+N_{1}}$.

The map $\rho$ is strictly increasing. Indeed, $n<n+1$ by Arithmetic and Order of the Natural Numbers §successor, hence $n+N_{1}<(n+1)+N_{1}$ by Arithmetic and Order of the Natural Numbers §order.

Moreover, for every $n\in\mathbb{N}$ we have $N_{1}<N_{1}+n=n+N_{1}$ by Arithmetic and Order of the Natural Numbers §difference and Arithmetic and Order of the Natural Numbers §commutative. So $|b_{n}-a_{N_{1}}|<1$. By Rules of Arithmetic and Order in an Ordered Field §triangle and Rules of Arithmetic and Order in an Ordered Field §order-sum,

$$|b_{n}|=|(b_{n}-a_{N_{1}})+a_{N_{1}}|\le|b_{n}-a_{N_{1}}|+|a_{N_{1}}|\le1+|a_{N_{1}}|.$$

Thus $(b_{n})$ is bounded.

A convergent subsequence. By the clause bolzano-weierstrass proved above, $(b_{n})$ has a convergent subsequence $b\circ\tau$, with $\tau:\mathbb{N}\to\mathbb{N}$ strictly increasing and $b\circ\tau$ converging to some $L\in\mathbb{R}$. Put $\sigma=\rho\circ\tau$. By Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §composition, $\sigma$ is strictly increasing and $b\circ\tau=(a\circ\rho)\circ\tau=a\circ\sigma$. Hence $a_{\sigma(k)}\to L$.

The whole sequence converges to $L$. Let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$; then $\varepsilon/2$ is positive by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving. We make three choices, in this order, each after $\varepsilon$.

First, by the Cauchy property, choose $N_{2}\in\mathbb{N}$ with $|a_{m}-a_{n}|<\varepsilon/2$ for all $m,n\ge N_{2}$.

Second, by $a_{\sigma(k)}\to L$, choose $K\in\mathbb{N}$ with $|a_{\sigma(k)}-L|<\varepsilon/2$ for every $k\ge K$.

Third, by Arithmetic and Order of the Natural Numbers §trichotomy, let $k$ be the larger of $K$ and $N_{2}$, so that $K\le k$ and $N_{2}\le k$. Then $k\le\sigma(k)$ by Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §index, hence $N_{2}\le\sigma(k)$ by Arithmetic and Order of the Natural Numbers §partial-order.

Now let $n\in\mathbb{N}$ with $n\ge N_{2}$. Then $|a_{n}-a_{\sigma(k)}|<\varepsilon/2$ by the first choice, as $n\ge N_{2}$ and $\sigma(k)\ge N_{2}$, and $|a_{\sigma(k)}-L|<\varepsilon/2$ by the second choice, as $k\ge K$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §triangle-three-points, 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 §halving,

$$|a_{n}-L|\le|a_{n}-a_{\sigma(k)}|+|a_{\sigma(k)}-L|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$

and so $|a_{n}-L|<\varepsilon$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed. Hence $N_{2}$ satisfies Convergent Sequences of Real Numbers §converges for $\varepsilon$, and $a_{n}\to L$. So $(a_{n})$ is convergent.