Convergence in $\mathbb{Q}$ How will I prove that a sequence in $\mathbb{Q}$ which is bounded below and decreasing is Cauchy, without using the knowledge of reals?
 A: HINT: Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb Q$ that is decreasing and not Cauchy. Then there is an $\epsilon>0$ such that for each $m\in\Bbb N$ there are $k_m,\ell_m\in\Bbb N$ such that $k_m,\ell_m\ge m$ and $|x_{k_m}-x_{\ell_m}|\ge\epsilon$.


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*Show that $\sigma$ has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ such that $|x_{n_{2k+1}}-x_{n_{2k}}|\ge\epsilon$ for each $k\in\Bbb N$.  

*Show that this subsequence is not bounded below, and conclude that $\sigma$ is not bounded below, either.

A: We do it for sequences $(r_n)$ that are bounded above and increasing, because the intuition is clearer. Modification for decreasing and bounded below is straightforward.
Suppose our sequence is not Cauchy. Then there is an $\epsilon\gt 0$ such that no matter what $k$ we pick, there is an $l\gt k$ such that  $r_l-r_k\ge \epsilon$. We may choose $\epsilon$ to be rational.
Pick $m_1=1$. There is an $m_2\gt m_1$ such that $r_{m_2}-r_{m_1}\ge \epsilon$. 
But then there is an $m_3\gt m_2$ such that $r_{m_3}-r_{m_2}\ge \epsilon$.
But then there is an $m_4\gt m_3$ such that $r_{m_4}-r_{m_3}\ge \epsilon$.
And so on. Note that it follows that $r_{m_t}\ge r_{m_1}+(t-1)\epsilon$. But $t$, and hence $t\epsilon$, can be made arbitrarily large. This contradicts the fact that our sequence is bounded above.
A: Suppose WLOG that the sequence is not eventually constant (since such a sequence would be Cauchy). Let $\epsilon > 0$ be fixed. Since the sequence is bounded below, choose a lower bound $l$ such that $l+\epsilon$ is not a lower bound. If no such $l$ exists, then every element of the sequence is larger than $l+n\epsilon$ for all $n$, but there is no such rational number let alone a sequence of them. Hence such a lower bound exists. Then choose an element of the sequence that is less than $l+\epsilon$, if it has index $N$, then for all $n,m >N$, (WLOG $n\le m$) $|x_n-x_m|=x_n-x_m<l+\epsilon -l =\epsilon$.
A: A sequence $(x_n)$ is Cauchy if
$$\forall \epsilon >0, \, \exists n_0 \in \mathbb{N}, \, \forall m,n \geq n_0, \, |x_m-x_n|< \epsilon$$
Hence, if $(x_n)$ is not Cauchy then
$$\exists \epsilon >0, \, \forall n_0 \in \mathbb{N}, \, \exists m,n \geq n_0, \, |x_m-x_n| \geq \epsilon$$
In other words, there is some positive $\epsilon$ with the property that one can always find arbitrarily large indices $m$ and $n$ such that $x_m$ and $x_n$ are at least $\epsilon$ apart.
Can you use this and the fact that $(x_n)$ is decreasing to produce a contradiction with the fact that $(x_n)$ is bounded from below?
