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

Throughout, $\mathbb{R}$ is an ordered field by The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum §ordered-field, and, as Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §numbers and Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §fields provide, a natural number $n$ standing where a real number is required denotes $\kappa_{\mathbb{R}}(n)$; this reading preserves sums, products and the order. In particular every $n\in\mathbb{N}$ is positive as a real number: $1\le n$ by Arithmetic and Order of the Natural Numbers §least, and $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares; so $0<1\le n$, hence $0<n$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, so $n\neq0$, and $1/n=n^{-1}$ is defined. Elementary rearrangements in the field $\mathbb{R}$, such as $x-0=x$ or $L-x=-(x-L)$, are used without comment.

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. The ordered field $\mathbb{R}$ satisfies the hypothesis of Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema, namely that every nonempty subset bounded above has a supremum, by The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum §supremum. The natural number that Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §archimedean provides for $\mathbb{R}$, read through $\kappa_{\mathbb{R}}$, is that natural number as a real number in the sense of Ordinary Mathematical Language for Analysis: Sets, Maps, Numbers and Ordered Fields §fields.

Let $\varepsilon\in\mathbb{R}$ be positive. By Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, $0<\varepsilon^{-1}$. By the Archimedean property Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §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, 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 $n\ge1$. 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 §tails.

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 $1=1^{-1}<|r|^{-1}$. 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 $h>0>-1$, the latter by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-negative, we have $h\ge-1$.

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 $n$ and $h$ are positive, $0<nh$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §positive-product. Since $0\le1$, 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 and Order in an Ordered 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 Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum §supremum, $S$ 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 $\varepsilon\in\mathbb{R}$ be positive. By Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers §epsilon-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 $a_{n}-s\le0<\varepsilon$. 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. Since $\mathbb{R}$ satisfies the hypothesis of Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema by The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum §supremum, Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §infimum shows that $S$ 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 $\varepsilon\in\mathbb{R}$ be positive. By Arbitrary Positive Slack, and Approximation of Suprema and Infima, in the Real Numbers §epsilon-below there is $N\in\mathbb{N}$ with $a_{N}<t+\varepsilon$. 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. Thus $-\varepsilon<0\le a_{n}-t<\varepsilon$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, 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 Monotone Sequences and Subsequences: Comparison of All Terms, Growth of the Indices, and Subsequences of Subsequences §bounded, every such sequence is therefore both bounded above and bounded below. We let $\mathbb{N}$ carry its order, as in Subsequences §subsequence.

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 $p:\mathbb{N}\to\mathbb{N}$, $u\mapsto\min P_{u}$, is a map. Apply Recursion on the Natural Numbers Starting at One §recursion with the set $\mathbb{N}$, the element $c=p(1)$, and the map $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ given by $g(k,u)=p(u+1)$. 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 $\sigma(k)\le\sigma(k+1)$, 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 $\inf\{a_{\sigma(k)}:k\in\mathbb{N}\}$.

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 $a_{u}\le a_{n}$. Since $a_{u}\le a_{u}$, we have $a_{n}\neq a_{u}$, 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, and $q:\mathbb{N}\to\mathbb{N}$, $u\mapsto\min T_{u}$, is a map 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 $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ given by $g(k,u)=q(u)$. 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 $\sup\{a_{\sigma(k)}:k\in\mathbb{N}\}$.

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 $\varepsilon\in\mathbb{R}$ be positive. 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 $n\mapsto n+N_{1}$, 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 $\varepsilon\in\mathbb{R}$ be positive; 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 $n\ge N_{2}$. 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.