Proof of Cauchy Sequences in a Metric Space: Constant Sequences, Convergence of Termwise Distances, the Null Relation, and Independence of Representatives

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

Conventions. The four conditions of Metric Space are referred to as conditions 1 to 4. For a real number $\varepsilon>0$ write $\varepsilon/2$ for $\varepsilon\cdot 2^{-1}$; by clause 8 of Elementary Order Arithmetic in an Ordered Field, $0<\varepsilon/2$ and $\varepsilon/2+\varepsilon/2=\varepsilon$. A sum of two inequalities of which one is strict is strict by clause 3 of Elementary Order Arithmetic in an Ordered Field, and a chain mixing $\le$ and $<$ is strict by clause 2 there. The order $\le$ of $\mathbb{R}$ is a total order, as required in Ordered Field; in particular it is antisymmetric. Given $N_{1},N_{2}\in\mathbb{N}$, one of $N_{1}<N_{2}$, $N_{1}=N_{2}$, $N_{2}<N_{1}$ holds by clause 3 of Properties of the Order on the Natural Numbers; letting $N$ be the larger of the two, every $k\ge N$ satisfies $k\ge N_{1}$ and $k\ge N_{2}$ by clause 1 there.

A real sequence $(a_{k})_{k\in\mathbb{N}}$ converges to $L$ in the sense of Limit of a Sequence of Real Numbers if and only if it converges to $L$ in the real line $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Line, because both conditions require $|a_{k}-L|<\varepsilon$ for all large $k$ and $d_{\mathbb{R}}(a_{k},L)=|a_{k}-L|$. Hence, by Uniqueness of Limits in a Metric Space applied in the real line, a convergent real sequence has exactly one limit; this makes the phrase "its limit" in clause 2 meaningful. A constant real sequence with value $c$ converges to $c$, since $c-c=0$ and $|0|=0<\varepsilon$ for every $\varepsilon>0$ by clause 1 of Properties of the Absolute Value in an Ordered Field.

Proof of clause 1 (Constant sequences). Let $a\in X$ and let $\varepsilon>0$. By condition 2, $d(a,a)=0<\varepsilon$. Take $N=1$. For all $m,l\ge N$ the $m$-th and $l$-th terms of $\bar{a}$ are both $a$, so their distance is $d(a,a)<\varepsilon$. By Cauchy Sequence in a Metric Space, $\bar{a}$ is a Cauchy sequence, that is $\bar{a}\in\mathcal{C}(X)$.

Proof of clause 2 (Distances). Step 1: a two-sided estimate. We show that for all $p,q,p',q'\in X$,

Put $c=d(p,p')+d(q,q')$. Applying condition 4 twice and condition 3 once gives $d(p,q)\le d(p,p')+d(p',q)\le d(p,p')+d(p',q')+d(q',q)=d(p',q')+c$. By clause 3 of Elementary Arithmetic in an Ordered Field (translation), this says $d(p,q)-d(p',q')\le c$. Exchanging the roles of $(p,q)$ and $(p',q')$, and using condition 3 to rewrite $d(p',p)+d(q',q)$ as $c$, gives $d(p',q')-d(p,q)\le c$. By clause 4 of Elementary Order Arithmetic in an Ordered Field (sign reversal), this is $-c\le d(p,q)-d(p',q')$. Clause 6 of Properties of the Absolute Value in an Ordered Field now gives $(\ast)$.

Step 2: convergence. Let $x,y\in\mathcal{C}(X)$ and put $a_{k}=d(x_{k},y_{k})$ for $k\in\mathbb{N}$. Let $\varepsilon>0$. By Cauchy Sequence in a Metric Space, first choose $N_{1}\in\mathbb{N}$ with $d(x_{k},x_{l})<\varepsilon/2$ for all $k,l\ge N_{1}$, then $N_{2}\in\mathbb{N}$ with $d(y_{k},y_{l})<\varepsilon/2$ for all $k,l\ge N_{2}$, and let $N$ be the larger of the two. For $k,l\ge N$, the estimate $(\ast)$ with $p=x_{k}$, $q=y_{k}$, $p'=x_{l}$, $q'=y_{l}$ gives

Thus $(a_{k})$ is a Cauchy sequence of real numbers, and it converges by Every Cauchy Sequence of Real Numbers Converges. Its limit is unique by the conventions above, so $\delta(x,y)$ is well defined.

Step 3: the four properties. Let $x,y,z\in\mathcal{C}(X)$ and $a,b\in X$.

Nonnegativity: $0\le d(x_{k},y_{k})$ for every $k$ by condition 1, and the constant sequence $0$ converges to $0$; clause 1 (comparison) of Order Properties of Limits of Real Sequences gives $0\le\delta(x,y)$.

Symmetry: by condition 3, $d(x_{k},y_{k})=d(y_{k},x_{k})$ for every $k$, so the two real sequences defining $\delta(x,y)$ and $\delta(y,x)$ are the same sequence, and by uniqueness of limits $\delta(x,y)=\delta(y,x)$.

Triangle inequality: by condition 4, $d(x_{k},z_{k})\le d(x_{k},y_{k})+d(y_{k},z_{k})$ for every $k$. The right-hand side converges to $\delta(x,y)+\delta(y,z)$ by clause 1 (sums) of Arithmetic of Limits of Real Sequences, and the left-hand side converges to $\delta(x,z)$. Clause 1 (comparison) of Order Properties of Limits of Real Sequences gives $\delta(x,z)\le\delta(x,y)+\delta(y,z)$.

Constant sequences: $\bar{a},\bar{b}\in\mathcal{C}(X)$ by clause 1, and the $k$-th term of the sequence defining $\delta(\bar{a},\bar{b})$ is $d(a,b)$ for every $k$. This constant sequence converges to $d(a,b)$, so $\delta(\bar{a},\bar{b})=d(a,b)$ by uniqueness of limits.

Proof of clause 3 (Null relation). Let $x,y,z\in\mathcal{C}(X)$.

Reflexivity: $d(x_{k},x_{k})=0$ for every $k$ by condition 2, and the constant sequence $0$ converges to $0$, so $\delta(x,x)=0$, that is $x\approx x$.

Symmetry: if $\delta(x,y)=0$, then $\delta(y,x)=\delta(x,y)=0$ by the symmetry proved in clause 2.

Transitivity: if $\delta(x,y)=0$ and $\delta(y,z)=0$, then clause 2 gives $0\le\delta(x,z)\le\delta(x,y)+\delta(y,z)=0$, and antisymmetry of $\le$ gives $\delta(x,z)=0$.

Classes: suppose $x\approx y$. If $w\in[y]$, then $y\approx w$, hence $x\approx w$ by transitivity, so $w\in[x]$. If $w\in[x]$, then $x\approx w$; also $y\approx x$ by symmetry, so $y\approx w$ by transitivity, and $w\in[y]$. Hence $[x]=[y]$. Conversely suppose $[x]=[y]$. Since $y\approx y$ by reflexivity, $y\in[y]=[x]$, which by definition of $[x]$ means $x\approx y$.

Proof of clause 4 (Representatives). Let $x\approx x'$ and $y\approx y'$, so $\delta(x,x')=0$ and $\delta(y,y')=0$; by the symmetry in clause 2 also $\delta(x',x)=0$ and $\delta(y',y)=0$. Applying the triangle inequality of clause 2 twice,

By antisymmetry of $\le$, $\delta(x,y)=\delta(x',y')$. $\blacksquare$