Proof of Convergence and the Cauchy Condition for Real Sequences Agree with Those in the Real Line as a Metric Space

Throughout, $\mathbb{R}$, $d_{\mathbb{R}}$, $\mathbb{N}$, $(a_m)_{m\in\mathbb{N}}$ and $L$ are as in the statement, and the reading of $(a_n)_{n=1}^{\infty}$ and of index inequalities stipulated there is in force; $\iota:\mathbb{N}\to\mathbb{R}$ is the canonical map of that stipulation. Recall that $s<t$ means $s\le t$ and $s\ne t$, so that $s<s$ holds for no $s\in\mathbb{R}$; the order-arithmetic steps in $\mathbb{R}$ below use claim 2 of Elementary Order Arithmetic in an Ordered Field.

Step 0 (the index conditions agree). Let $m,N\in\mathbb{N}$. We first check that $\iota(N)\le\iota(m)$ if and only if $N\le m$, and then that the integers $z$ with $\iota(N)\le z$ are exactly the values $\iota(m)$ with $m\in\mathbb{N}$ and $N\le m$.

Suppose $N\le m$. Then either $N=m$, in which case $\iota(N)=\iota(m)$, or $N<m$, in which case $\iota(N)<\iota(m)$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field; either way $\iota(N)\le\iota(m)$. Conversely suppose $\iota(N)\le\iota(m)$. By trichotomy, claim 3 of Properties of the Order on the Natural Numbers, exactly one of $N<m$, $N=m$ and $m<N$ holds. If $m<N$ then claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field gives $\iota(m)<\iota(N)$, and then claim 2 of Elementary Order Arithmetic in an Ordered Field, applied to $\iota(N)\le\iota(m)$ and $\iota(m)<\iota(N)$, gives $\iota(N)<\iota(N)$, which is impossible. Hence $N<m$ or $N=m$, so $N\le m$ by claim 1 of Properties of the Order on the Natural Numbers.

Now let $z$ be an integer with $\iota(N)\le z$. By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field we have $0<\iota(N)$, so claim 2 of Elementary Order Arithmetic in an Ordered Field, applied to $0<\iota(N)$ and $\iota(N)\le z$, gives $0<z$, and claim 1 of Arithmetic, Order and Discreteness of the Integers provides $m\in\mathbb{N}$ with $z=\iota(m)$, and the previous paragraph gives $N\le m$. Conversely, if $m\in\mathbb{N}$ and $N\le m$ then $\iota(m)$ is an integer by The Integers as a Subset of the Real Numbers and $\iota(N)\le\iota(m)$.

Consequently, for $N\in\mathbb{N}$, the index condition "for all integers $n\ge N$" occurring in Limit of a Sequence of Real Numbers and in Cauchy Sequence of Real Numbers, read as stipulated in the statement, constrains exactly those terms $a_n$ with $n$ an integer and $\iota(N)\le n$; by what has just been shown, and by the reading of $a_n$ stipulated in the statement, these are exactly the terms $a_m$ with $m\in\mathbb{N}$ and $N\le m$ — which is precisely the range of indices constrained in Convergent Sequence in a Metric Space and in Cauchy Sequence in a Metric Space.

Claim 1. By the definition of $d_{\mathbb{R}}$ given in the statement, $d_{\mathbb{R}}(a_m,L)=|a_m-L|$ for every $m\in\mathbb{N}$. Consequently, for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and every $N\in\mathbb{N}$, the condition

which by Step 0 is what Limit of a Sequence of Real Numbers requires of $\varepsilon$ and $N$, and the condition

which is what Convergent Sequence in a Metric Space requires of $\varepsilon$ and $N$ in the metric space $(\mathbb{R},d_{\mathbb{R}})$, are one and the same condition. The two definitions quantify over $\varepsilon$ and $N$ in the same way, namely for every such $\varepsilon$ there exists such an $N$, so each of them holds exactly when the other does.

Claim 2. Likewise $d_{\mathbb{R}}(a_m,a_{\ell})=|a_m-a_{\ell}|$ for all $m,\ell\in\mathbb{N}$. Consequently, for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and every $N\in\mathbb{N}$, the condition

which by Step 0 is what Cauchy Sequence of Real Numbers requires of $\varepsilon$ and $N$, and the corresponding condition with $|a_m-a_{\ell}|$ replaced by $d_{\mathbb{R}}(a_m,a_{\ell})$, which is what Cauchy Sequence in a Metric Space requires of $\varepsilon$ and $N$ in $(\mathbb{R},d_{\mathbb{R}})$, are one and the same condition. Since the two definitions quantify over $\varepsilon$ and $N$ in the same way, each holds exactly when the other does.