My Proof: Every convergent sequence is a Cauchy sequence.

Question

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

Answer 1

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.

Answer 2

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.