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I wonder if my idea of proof is correct. Prove that $\{s_n\}_n\subset\mathbb{R}$ is convergent if it is cauchy sequence. idea of pf: By definition, for every $\epsilon> 0\exists N$ s.t.$|s_m-s_n|<\epsilon \forall m,n >N$ so $|s_n|<|s_m|+\epsilon$ and by completeness we have a supremum $s_0=\sup\{s_n:n\in\mathbb{N}\}$ and then proving $s_n$ converge to $s_0$. |
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A hint: It seems that you are allowed to use sup and inf. Assume for a moment that the theorem is true. How could you express $\lim_{n\to\infty} s_n$ in terms of sup's and inf's of certain sets connected with the $s_n$? Maybe you also have a theorem about bounded monotone sequences at your disposal. |
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$(x_n)\in\mathbb{R}^\mathbb{N}$ $(x_n)$ converges to $x$ $\Leftrightarrow \forall \epsilon > 0, \exists n_\epsilon\in\mathbb{N}, \forall n \ge n_\epsilon, |x-x_n|<\epsilon$ $(x_n)$ is Cauchy $\Leftrightarrow \forall \alpha > 0, \exists n_\alpha\in\mathbb{N}, \forall n \ge n_\alpha,\forall m \ge n_\alpha, |x_m-x_n|<\alpha$ The idea behind this proof is that you can take two sequences, the sup and the inf of the terms after some $n$ and then proving that these two sequences converge because they are adjacent sequences (after proving they exists by proving the original sequence is bounded). And since the original sequence is between the sup sequence and the inf sequence for all $n$, you get the convergence of your original sequence at the same time. Suppose $(x_n)$ converges to $x$ Take some $\alpha>0$ Use the convergence property with $\epsilon=\cfrac{\alpha}{2}$ Take some $n_\epsilon$ such as $\forall n \ge n_\epsilon, |x-x_n|<\epsilon$ Then $\forall n \ge n_\epsilon,\forall m \ge n_\epsilon, |x_m-x_n|\le|x-x_n|+|x-x_m|=\cfrac{\alpha}{2}+\cfrac{\alpha}{2} = \alpha$ So $(x_n)$ is Cauchy Suppose So $(x_n)$ is Cauchy Use the fact it is Cauchy with $\alpha=1$ to take a $n_1\in\mathbb{N}$ such as $\forall n \ge n_1,\forall m \ge n_1, |x_m-x_n|<1$ Now let $m=n_1$ $\forall n \ge n_1,|x_{n_1}-x_n|<1$ This means that $\forall n \ge n_1, x_n \in ]x_{n_1}-1, x_{n_1}+1[$ so $\{x_n, n\ge n_1\}$ is bounded $\{x_n, n< n_1\}$ is finite and hence bounded So $\{x_n, n\in\mathbb{N}\}=\{x_n, n< n_1\}\cup\{x_n, n\ge n_1\}$ is bounded $\forall n \in\mathbb{N},$ let $S_n=\{x_m, m\ge n\}$ It is a subset of $\{x_n, n\in\mathbb{N}\}$ so it is bounded Let $(i_n)=(\inf S_n)$ and $(s_n)=(\sup S_n)$ $(i_n)$ is increasing and majored by $s_0$ so it converges $(s_n)$ is decreasing and minored by $i_0$ so it converges Suppose $d=\lim\limits_{n\to\infty}s_n-i_n > 0$ Then $\forall \epsilon>0, \exists n_0 \in \mathbb{N}, \forall n\ge n_0, |(s_n-i_n)-d|<\epsilon$ Take $\epsilon=\cfrac{d}{2}$ and chose $n_0 \in \mathbb{N}$ such as $\forall n\ge n_0, |(s_n-i_n)-d|<\cfrac{d}{2}$ $\forall n\ge n_0, d-\cfrac{d}{2}=\cfrac{d}{2}<(s_n-i_n)$ But $\forall n\ge n_0, \exists p \ge n, \exists q \ge n, i_n=x_p, s_n=x_q$ So you get $\forall n\ge n_0, \exists p \ge n, \exists q \ge n, \cfrac{d}{2}<(x_p-x_q)$ But since your sequence is Cauchy, with $\alpha = \cfrac{d}{2}$ $\exists n_0\in\mathbb{N}, \forall p \ge n_0,\forall q \ge n_0, |x_p-x_q|<\cfrac{d}{2}$ And these two statements contradict each other So you get $d=\lim\limits_{n\to\infty}s_n-i_n = 0$ Now you have
You must have a theorem that proves that both converge and $\lim\limits_{n\to\infty}i_n=\lim\limits_{n\to\infty}s_n$ And since you have $\forall n \in \mathbb{N}, i_n \ge x_n \ge \s_n$, you get that $(x_n)$ converges and $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}i_n=\lim\limits_{n\to\infty}s_n$ |
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