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In this problem, we will investigate a new property which some sequences may have. Here is a new definition.

Definition: A sequence $\{a_n\}_{n=0}^∞$ is said to be Cauchy iff

$$\forall \varepsilon,\exists N\in\mathbb{N}s.t.\forall n\in\mathbb{N},\forall m\in\mathbb{N}(n,m\geq N\rightarrow|a_n-a_m|<\varepsilon)$$

Let $\{a_n\}_{n=0}^∞$ be a sequence of real numbers that converges to a number L. Show that $\{a_n\}_{n=0}^∞$ must be Cauchy


My attempts:

Proof.

Let $\{a_n\}_{n=0}^∞$ be a sequence of real numbers

Assume $\{a_n\}_{n=0}^∞$ converges to a number L where $L\in \mathbb{R}$

Show $\{a_n\}_{n=0}^∞$ must be Cauchy

By assumption we have

1.$\forall \varepsilon>0,\exists n_0\in\mathbb{N}s.t.\forall n\in\mathbb{N},(n\geq n_0\rightarrow L-\varepsilon<a_n<L+\varepsilon)$

WTS

$\forall \varepsilon>0,\exists N\in \mathbb{N}s.t.\forall n,m\in \mathbb{N},n,m\geq N\rightarrow a_m-\varepsilon<a_n<a_m+\varepsilon$

By 1. we have

$\forall \varepsilon>0,\exists N\in\mathbb{N}s.t.\forall n,m\in\mathbb{N},$

$n\geq N\rightarrow L-\varepsilon<a_n<L+\varepsilon $

$\wedge m\geq N\rightarrow L-\varepsilon<a_m<L+\varepsilon$

Since $2\varepsilon>0$, this also hold for $2\varepsilon$

Implies the following:

$\forall \varepsilon>0,\exists N\in\mathbb{N}s.t.\forall n,m\in\mathbb{N},$

$n\geq N\rightarrow\underbrace{L-2\varepsilon<a_n<L+2\varepsilon}_\alpha $

$\wedge m\geq N\rightarrow\underbrace{L-\varepsilon<a_m<L+\varepsilon}_\beta$

$\Leftrightarrow \forall \varepsilon>0,\exists N\in\mathbb{N}s.t.\forall n,m\in\mathbb{N},\underbrace{n,m\geq N}_p$

$\rightarrow \underbrace{L-\varepsilon<a_n<L+\varepsilon\wedge L-\varepsilon<a_m<L+\varepsilon}_q$

Since $\alpha -\beta$ have $-\varepsilon<a_n-a_m<\varepsilon$

That $a_m-\varepsilon<a_n<a_m+\varepsilon$

$\forall \varepsilon>0,\exists N\in\mathbb{N}s.t.\forall n,m\in\mathbb{N},\underbrace{n,m\geq N}_p$

$\rightarrow ((\underbrace{L-\varepsilon<a_n<L+\varepsilon\wedge L-\varepsilon<a_m<L+\varepsilon}_q)\rightarrow \underbrace{a_m-\varepsilon<a_n<a_m+\varepsilon}_r)$

Consider $((p\rightarrow (q\rightarrow r))\rightarrow(p\rightarrow r))$

It's only False when p is true, q is false and r is false.

However we had show $p\rightarrow q$ is true,

but when p is true and q is false that $p\rightarrow q$ is false

Therefore this False case can never happen by contradiction

In another word $((p\rightarrow q)\wedge(p\rightarrow (q\rightarrow r)))\rightarrow(p\rightarrow r)$ is a tautology

Implies $\forall \varepsilon>0,\exists N\in\mathbb{N}s.t.\forall n,m\in\mathbb{N},\underbrace{n,m\geq N}_p \rightarrow \underbrace{\vert a_n-a_m\vert<\varepsilon}_r$

Therefore $\{a_n\}_{n=0}^∞$ must be Cauchy


Is my proof correct? Any suggestion would be appreciated.

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Although the proof seems correct from a logical point of view, those references to first-order logic make it quite hard to read and there is no reason for them. It is much simpler to write that, given $\varepsilon>0$, if you take $N\in\mathbb N$ such that $n\geqslant N\implies\lvert a_n-L\rvert<\frac\varepsilon2$, then, if $m,n\geqslant N$,$$\lvert a_m-a_n\rvert=\lvert a_m-L+L-a_n\rvert\leqslant\lvert a_m-L\rvert+\lvert a_n-L\rvert<\varepsilon.$$

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