Let $(x_n)_{n\in\Bbb N}$ be a real sequence.

$\textbf{Definition 1.}$ $(x_n)$ is $\textit{convergent}$ iff there is an $x\in\Bbb R$ such that, for every $\varepsilon\in\Bbb R$ with $\varepsilon>0$, there is an $N\in\Bbb N$ such that, for every $n\in\Bbb N$ with $n>N$, we have $|x_n-x|<\varepsilon$.

$\textbf{Definition 2.}$ $(x_n)$ is a $\textit{Cauchy sequence}$ iff, for every $\varepsilon \in\Bbb R$ with $\varepsilon > 0$, there is an $N\in\Bbb N$ such that, for every $m,n\in\Bbb N$ with $m,n > N$, we have $|x_m - x_n| < \varepsilon$.

$\textbf{Theorem.}$ If $(x_n)$ is convergent, then it is a Cauchy sequence.

Proof: Since $(x_n)\to x$ we have the following for for some $\varepsilon_1, \varepsilon_2 > 0$ there exists $N_1, N_2 \in \Bbb N$ such for all $n_1>N_1$ and $n_2>N_2$ following holds $$|x_{n_1}-x|<\varepsilon_1\\ |x_{n_2}-x|<\varepsilon_2$$ So both will hold for all $n_1, n_2 >\max(N_1, N_2)=N$, say $\varepsilon = \max(\varepsilon_1, \varepsilon_2)$ then $$|x_{n_1}-x-(x_{n_2}-x)|<\varepsilon\\\implies |x_{n_1}-x_{n_2}|<\varepsilon$$ Hence all convergent sequences are Cauchy.

Is this proof correct? I also saw this question and copied some of the content(definition and theorem) from there.https://math.stackexchange.com/q/1105255

  • $\begingroup$ No. The proof has a fatal error. Neither of the definitions say the an epsilon exist that does what you want. They both say ALL epsilons do what you want. So you certainly can't say let epsilon = max. You must define epsilon first. $\endgroup$ – fleablood Mar 19 '18 at 6:44
  • $\begingroup$ Yes, I see that now. $\endgroup$ – Sonal_sqrt Mar 19 '18 at 6:47
  • $\begingroup$ Perhaps I was too harsh. You proof is flawed in that looks for a specific rather than starting with the general. But the mechanics for the most part is good. So the proof is salvageable if you redo it. $\endgroup$ – fleablood Mar 19 '18 at 18:25

Idea is right, but the execution misses out on a couple of points.

So both will hold for all $n_1, n_2 > max(N_1, N_2)=N$, say $\epsilon = max(\epsilon_1, \epsilon_2)$

Technically $\,\epsilon\,$ is a given, you don't get to choose it.

then $\quad|x_{n_1}-x-(x_{n_2}-x)|<\epsilon \quad\implies\quad |x_{n_1}-x_{n_2}|<\epsilon$

The RHS does not follow from the stated premise that $\,|x_{n_1}-x| \lt \epsilon_1\,$ and $\,|x_{n_2}-x| \lt \epsilon_2$. At best, from the triangle inequality:

$$ |x_{n_1} - x_{n_2}| = |(x_{n_1}-x)-(x_{n_2}-x)| \le |x_{n_1}-x| + |x_{n_2}-x| \lt \epsilon_1 + \epsilon_2 $$

To fix it, just assume $\,\epsilon\,$ is given, choose $\,\epsilon_1=\epsilon_2=\epsilon / 2\,$, then proceed along the same line.

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    $\begingroup$ I think it's worth pointing out that the implication written does follow, but the implicit implication $$|x_{n_1} - x| < \epsilon_1 \text{ and } |x_{n_2} - x| < \epsilon_2 \implies |x_{n_1} - x - (x_{n_2} - x)| < \epsilon$$ is not valid. $\endgroup$ – Theo Bendit Mar 19 '18 at 6:18
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    $\begingroup$ Alright I got it, thanks to all you guys. $\endgroup$ – Sonal_sqrt Mar 19 '18 at 6:25
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    $\begingroup$ @PiyushDivyanakar I know you just got it, but here's the counterexample I was just about to post: Take $\epsilon_1 = \epsilon_2 = 1$ (hence $\epsilon = 1$), $x = 0$, $x_{n_1} = 0.75$, and $x_{n_2} = -0.75$. $\endgroup$ – Theo Bendit Mar 19 '18 at 6:26
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    $\begingroup$ @PiyushDivyanakar Or, if you really wanted to annoy someone, you could take $\epsilon_1 = \epsilon / \pi$ and $\epsilon_2 = (1 - 1/ \pi)\epsilon\,$ ;-) Point being that there is not a unique choice of $\epsilon_{1,2}$ in such cases, but you'd usually go with what's the easiest, most obvious one. $\endgroup$ – dxiv Mar 19 '18 at 6:39
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    $\begingroup$ LOL. I really love this site. $\endgroup$ – Sonal_sqrt Mar 19 '18 at 6:43

It should not be that for some $\epsilon_{1},\epsilon_{2}>0$. Rather, one fixes an arbitrary $\epsilon>0$, and we find $N_{1},N_{2}$ such that $|x_{n_{1}}-x|<\epsilon/2$ and $|x_{n_{2}}-x|<\epsilon/2$ for all $n_{1}>N_{1}$, $n_{2}>N_{2}$.

For all $n_{1},n_{2}>\max(N_{1},N_{2})$, then $|x_{n_{1}}-x_{n_{2}}|=|x_{n_{1}}-x-(x_{n_{2}}-x)|\leq|x_{n_{1}}-x|+|x_{n_{2}}-x|<\epsilon/2+\epsilon/2=\epsilon$.

Actually just one $N$ for which $|x_{n}-x|<\epsilon/2$, $n\geq N$ is enough.

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    $\begingroup$ If I am not mistaken, then you actually only need one $N$ such that $|x_n - x| < \varepsilon/2$ for all $n > N$, right? There is no need for $N_1$ and $N_2$ and taking the max. $\endgroup$ – Matthias Klupsch Mar 19 '18 at 8:54
  • $\begingroup$ Yes, true, I just followed what OP wrote. $\endgroup$ – user284331 Mar 19 '18 at 13:54

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