# Proof Verification: Regarding equivalent sequences and Cauchy sequences, Tao Analysis I: Exercise 5.2.1.

Proposition: If $$(a_n)_{n=1}^{\infty}$$ and $$(b_n)_{n=1}^{\infty}$$ are equivalent sequences of rationals, then $$(a_n)_{n}^{\infty}$$ is a Cauchy sequence iff $$(b_n)_{n}^{\infty}$$ is a Cauchy sequence.

Reason for duplicating: I did it somewhat differently. So I'm not certain if I did it correctly. Would be grateful if someone could point out erroneous areas of the proof and how it could be better.

Definition $$\bf 5.2.6$$ (Equivalent sequences). Two sequences $$(a_n)_{n=0}^\infty$$ and $$(b_n)_{n=0}^\infty$$ are equivalent iff for each rational $$\varepsilon>0$$, the sequences $$(a_n)_{n=0}^\infty$$ and $$(b_n)_{n=0}^\infty$$ are eventually $$\epsilon$$-close. In other words, $$a_0,a_1,a_2,\ldots$$ and $$b_0,b_1,b_2,\ldots$$ are equivalent iff for every rational $$\varepsilon>0$$, there exists an $$N\geqslant0$$ such that $$|a_n-b_n|\leqslant\varepsilon$$ for all $$n\geqslant N$$.

Proof: Let $$(a_n)_{n=1}^{\infty}$$ and $$(b_n)_{n=1}^{\infty}$$ be equivalent sequences and $$(a_n)_{n=1}^{\infty}$$ be a Cauchy sequence. Then, by definition of equivalent sequences \begin{align}\forall\epsilon\in \Bbb{Q}^{+}\exists n\in\Bbb{N}:( n\geqslant 1 \land \vert a_n-b_n\vert\leqslant\epsilon).\end{align} By definition of a Cauchy sequence, \begin{align}\forall\epsilon\in\Bbb{Q}^+\exists n\in\Bbb{N}: (j,k\geqslant n \Rightarrow \vert a_j=a_k\vert\leqslant\epsilon).\end{align} Note that \begin{align}\vert b_j-b_k\vert=\vert(a_k-b_k)-(a_k-b_j)\vert\ \quad and \quad \vert(a_k-b_k)-(a_k-b_j)\vert\leqslant\vert a_k-b_k\vert +\vert a_k-b_j\vert\end{align} Thus, $$\vert b_j-b_k\vert\leqslant\vert a_k-b_k\vert +\vert a_k-b_j\vert\leqslant\epsilon'+\vert a_k-b_j\vert$$ such that $$\epsilon'\in\Bbb{Q}^+$$ and $$\forall x\geqslant k$$, $$\vert a_k-b_k\vert\leqslant\epsilon'$$. Also, $$\vert a_k-b_j\vert=\vert(a_j-b_j)-(a_j-a_k)\vert\leqslant\vert a_j-b_j\vert +\vert a_j-a_k\vert$$ such that $$\vert a_j-b_j\vert\leqslant\overline{\epsilon}$$ and $$\vert a_j-a_k\vert\leqslant\epsilon^*$$, since $$(a_n)_{n=1}^{\infty}$$ and $$(b_n)_{n=1}^{\infty}$$ are equivalent sequences and $$(a_n)_{n=1}^{\infty}$$ is a Cauchy sequence.

Thus, $$\vert a_k-bj\vert\leqslant\vert a_j-b_j\vert + \vert a_j-a_k\vert\leqslant \overline{\epsilon}+\epsilon^*$$, and $$\vert b_j-b_k\vert\leqslant \vert a_k-b_k\vert +\vert a_k-b_j\vert\leqslant \epsilon'+\vert a_k-b_j\vert\leqslant\epsilon'+\overline{\epsilon}+\epsilon^*$$. Since, $$\epsilon',\overline{\epsilon},\epsilon^*$$ are arbitrary elements in $$\Bbb{Q}^+$$, then $$\epsilon' +\overline{\epsilon}+ \epsilon^*=\hat{\epsilon}\in\Bbb{Q}^+$$ is also arbitrary. Therefore, $$\forall \hat{\epsilon}\in\Bbb{Q}^+\exists n\in\Bbb{N}$$ such that if $$j,k\geqslant n$$, then $$\vert b_j-b_k\vert\leqslant\hat{\epsilon}$$, which implies that $$(b_n)_{n=1}^{\infty}$$ is a Cauchy sequence. The converse of the proposition can be argued backwards.

• putting spaces somewhere wont kill you :) – Masacroso Apr 2 '18 at 10:57