[proofplan] We replace the given modulus by an increasing one, so that later accuracy requirements correspond to later indices. Then we sample the original sequence along this modulus, defining $b_k=a_{\mu(k+1)}$. The Cauchy estimate for $a$ immediately gives the regular estimate for $b$, and the same tail estimate shows that the sampled sequence is equivalent to the original representative. [/proofplan] [step:Replace the modulus by an increasing modulus] Define the function \begin{align*} \mu:\mathbb{N}\to\mathbb{N} \end{align*} by \begin{align*} \mu(k)=\max\{\nu(1),\nu(2),\dots,\nu(k)\}. \end{align*} Then $\mu$ is increasing: if $k\leq \ell$, then the finite set defining $\mu(k)$ is contained in the finite set defining $\mu(\ell)$, hence $\mu(k)\leq \mu(\ell)$. Also, $\mu$ is still a Cauchy modulus for $a$. Indeed, for every $k\in\mathbb{N}$ we have $\mu(k)\geq \nu(k)$. Therefore, if $i,j\in\mathbb{N}$ satisfy $i,j\geq \mu(k)$, then $i,j\geq \nu(k)$, and the defining Cauchy estimate for $(a,\nu)$ gives \begin{align*} |a_i-a_j|\leq 2^{-k}. \end{align*} [guided] The original modulus $\nu$ tells us how far along the sequence $a$ we must go to obtain a prescribed accuracy. The proof will choose terms of $a$ at indices depending on the required accuracy, so it is useful to make sure that higher accuracy never asks us to move backward in the sequence. Define \begin{align*} \mu:\mathbb{N}\to\mathbb{N} \end{align*} by \begin{align*} \mu(k)=\max\{\nu(1),\nu(2),\dots,\nu(k)\}. \end{align*} This definition uses only the first $k$ values of $\nu$, so the maximum is taken over a finite non-empty set of natural numbers. If $k\leq \ell$, then every term appearing in the maximum for $\mu(k)$ also appears in the maximum for $\mu(\ell)$. Hence \begin{align*} \mu(k)\leq \mu(\ell). \end{align*} Thus $\mu$ is increasing. We must also check that replacing $\nu$ by $\mu$ does not destroy the Cauchy estimate. Fix $k\in\mathbb{N}$, and suppose $i,j\in\mathbb{N}$ satisfy $i,j\geq \mu(k)$. Since $\mu(k)\geq \nu(k)$, we also have $i,j\geq \nu(k)$. The defining property of the Cauchy representative $(a,\nu)$ then gives \begin{align*} |a_i-a_j|\leq 2^{-k}. \end{align*} So $\mu$ is an increasing Cauchy modulus for the same sequence $a$. [/guided] [/step] [step:Sample the original sequence along the increasing modulus] Define the sequence \begin{align*} b:\mathbb{N}\to\mathbb{Q} \end{align*} by \begin{align*} b_k=a_{\mu(k+1)}. \end{align*} This is a well-defined rational sequence because $a:\mathbb{N}\to\mathbb{Q}$ and $\mu(k+1)\in\mathbb{N}$ for every $k\in\mathbb{N}$. [/step] [step:Prove that the sampled sequence is regular and has the identity modulus] Let $m,n\in\mathbb{N}$. By symmetry of the absolute value, it suffices to treat the case $m\leq n$. Since $\mu$ is increasing and $m+1\leq n+1$, we have \begin{align*} \mu(m+1)\leq \mu(n+1). \end{align*} Thus both selected indices $\mu(m+1)$ and $\mu(n+1)$ are at least $\mu(m+1)$. Applying the Cauchy estimate for $a$ with modulus $\mu$ at accuracy level $m+1$ gives \begin{align*} |b_m-b_n|=|a_{\mu(m+1)}-a_{\mu(n+1)}|\leq 2^{-(m+1)}. \end{align*} Since $2^{-(m+1)}\leq 2^{-m}$ and $2^{-n}>0$, we obtain \begin{align*} |b_m-b_n|\leq 2^{-m}+2^{-n}. \end{align*} Therefore $b$ is a regular [Cauchy sequence](/page/Cauchy%20Sequence). It remains in this step to verify that $\operatorname{id}_{\mathbb{N}}:\mathbb{N}\to\mathbb{N}$ is a Cauchy modulus for $b$. Fix $k\in\mathbb{N}$ and let $m,n\in\mathbb{N}$ satisfy $m,n\geq k$. If $m\leq n$, then the previous estimate at accuracy level $m+1$ gives \begin{align*} |b_m-b_n|\leq 2^{-(m+1)}\leq 2^{-k}. \end{align*} The case $n\leq m$ is identical after exchanging $m$ and $n$. Hence $\operatorname{id}_{\mathbb{N}}$ is a Cauchy modulus for $b$. [/step] [step:Verify the tail equivalence with the original representative] We prove that the Cauchy representative $(b,\operatorname{id}_{\mathbb{N}})$ is equivalent to $(a,\nu)$ in the sense stated in the theorem. Fix $k\in\mathbb{N}$ and define the tail threshold \begin{align*} N_k=\max\{\mu(k),k\}. \end{align*} If $i,j\in\mathbb{N}$ satisfy $i\geq N_k$ and $j\geq N_k$, then $i\geq \mu(k)$ and, because $\mathbb{N}=\{1,2,3,\dots\}$, also $j+1\geq k$. Since $\mu$ is increasing, $j+1\geq k$ implies \begin{align*} \mu(j+1)\geq \mu(k). \end{align*} Thus both $i$ and $\mu(j+1)$ lie in the tail of $a$ controlled by $\mu(k)$. Applying the Cauchy estimate for $a$ with modulus $\mu$ at accuracy level $k$ gives \begin{align*} |a_i-b_j|=|a_i-a_{\mu(j+1)}|\leq 2^{-k}. \end{align*} This verifies the defining tail-equivalence condition between $(a,\nu)$ and $(b,\operatorname{id}_{\mathbb{N}})$. Hence $b$ represents the same Cauchy real $x$. Combining this equivalence with the regular estimate and identity-modulus verification proved above, $x$ has a representative given by the regular Cauchy sequence $b$. [guided] We now check that the sampled representative really lies in the same equivalence class as the original representative. The [equivalence relation](/page/Equivalence%20Relation) in the theorem asks us to prove the following: for each requested accuracy $2^{-k}$, we must find one tail threshold $N_k$ such that every sufficiently late original term $a_i$ is within $2^{-k}$ of every sufficiently late sampled term $b_j$. Fix $k\in\mathbb{N}$ and define \begin{align*} N_k=\max\{\mu(k),k\}. \end{align*} Now take arbitrary $i,j\in\mathbb{N}$ with $i\geq N_k$ and $j\geq N_k$. From $i\geq N_k$ we get $i\geq \mu(k)$. From $j\geq N_k$ we get $j\geq k$, and since $\mathbb{N}=\{1,2,3,\dots\}$ this implies $j+1\geq k$. The monotonicity of $\mu$ gives \begin{align*} \mu(j+1)\geq \mu(k). \end{align*} The definition of $b$ says $b_j=a_{\mu(j+1)}$, so both indices appearing in $|a_i-b_j|$ are now indices of $a$ at least $\mu(k)$: the first is $i$, and the second is $\mu(j+1)$. Therefore the Cauchy estimate for $a$ with the modulus $\mu$ applies at accuracy level $k$, and it yields \begin{align*} |a_i-b_j|=|a_i-a_{\mu(j+1)}|\leq 2^{-k}. \end{align*} This is exactly the equivalence condition stated in the theorem for the two representatives $(a,\nu)$ and $(b,\operatorname{id}_{\mathbb{N}})$. Therefore $(b,\operatorname{id}_{\mathbb{N}})$ represents the same Cauchy real $x$ as $(a,\nu)$. Since the previous step also proved that $b$ is regular and that $\operatorname{id}_{\mathbb{N}}$ is a Cauchy modulus for $b$, the desired regular representative has been constructed. [/guided] [/step]