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I want to show that Cauchy sequences are bounded without using the fact that Cauchy sequences are convergent sequences and convergent sequences are bounded.

proof: Suppose $a_m$ is a Cauchy sequence, that is $\forall\varepsilon\gt0, \exists N\in\mathbb{R}$ s.t.

$$m,n\gt N\implies |a_m-a_n|\lt \varepsilon$$

So we have $$-\varepsilon\lt a_m-a_n\lt\varepsilon$$ and $$|a_n|-|\varepsilon|\le a_n-\varepsilon\lt a_m\lt a_n+\varepsilon \lt |a_n|+|\varepsilon|$$ $$|a_m|\lt |a_n|+\varepsilon$$

Am I on the right track at all? Can I then say that $M$ will be the bound for $a_m$ as: $$M=\max\{a_1,...,|a_{n-1}|,|a_n|+\varepsilon\}$$Or am I making some incorrect assumption or something else?

Cotton Headed Ninnymuggins
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1 Answers1

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I would suggest that you break the sequence up into a finite sequence and the remaining infinite sequence. So, fill in the details in the following proof.

Proof: Given that $(a_n)_{n=1}^\infty$ is a Cauchy sequence, for $\varepsilon =1$, there exists an $N$ such that for all $j,k>N$, we have that $|a_j-a_k|<1$

Now split the sequence into a finite sequence $(a_n)_{n=1}^N$ and the remaining infinite sequence $(a_n)_{n=N+1}^\infty$. Now use the theorem(use induction to prove it) that every finite sequence is bounded. So there exists a real number $M$ such that the sequence $(a_n)_{n=1}^N$ is bounded by $M$.

Now for the sequence $(a_n)_{n=N+1}^\infty$, we have the inequality that for every $j>N$ and fixed $k>N$ we have that $|a_j-a_k|<1$.

Take it from here to conclude that $|a_j|<1+|a_k|$. Then use your previous result that $(a_n)_{n=1}^N$ is bounded by $M$ to conclude that the entire sequences $(a_n)_{n=1}^\infty$ is bounded by $1+|a_k|+M$.

Seeker
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  • Do you mean "So there exists a real number $M$ such that the sequence $(a_n)_{n=1}^{\color{red}{N}}$ is bounded by $M$."? – Cotton Headed Ninnymuggins Nov 22 '22 at 03:27
  • @CottonHeadedNinnymuggins Yes. Sorry for the typo – Seeker Nov 22 '22 at 03:29
  • Is there anything wrong with my initial approach? How would I use induction to prove every finite sequence is bounded? – Cotton Headed Ninnymuggins Nov 22 '22 at 03:36
  • @CottonHeadedNinnymuggins I was suggesting a shorter and cleaner proof I knew. As for proving that every finite sequence is bounded, here is a proof: Let $a_1,\dots,a_n$ be a finite sequence. Clearly $a_1$ is bounded by $|a_1|$. Now suppose inductively that the sequence $a_1,\dots,a_{n-1}$ is bounded by some real number $M$. We have to show that $a_1,\dots,a_n$ is bounded. Using our induction hypothesis we know that $a_1,\dots,a_{n-1}$ is bounded by $M$. We also know that $a_n$ is bounded by $|a_n|$. Thus, the entire sequence $a_1,\dots,a_n$ is bounded by $M+|a_n|$. – Seeker Nov 22 '22 at 03:41
  • That is pretty simple, thank you – Cotton Headed Ninnymuggins Nov 22 '22 at 04:06