# A proof related to the convergence of Cauchy sequences

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

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

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

$$\wedge m\geq N\rightarrow 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

$$\wedge m\geq N\rightarrow\underbrace{L-\varepsilon

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

$$\rightarrow \underbrace{L-\varepsilon

Since $$\alpha -\beta$$ have $$-\varepsilon

That $$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

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.

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 $$m,n\geqslant N$$,$$\lvert a_m-a_n\rvert=\lvert a_m-L+L-a_n\rvert\leqslant\lvert a_m-L\rvert+\lvert a_n-L\rvert<\varepsilon.$$