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.

Identical questions: (1)Equivalent Cauchy sequences. (2)Two sequences are equivalent. Prove that one is Cauchy iff the other is Cauchy.

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.

  • 3
    $\begingroup$ putting spaces somewhere wont kill you :) $\endgroup$ – Masacroso Apr 2 '18 at 10:57

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