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I must prove that for every real number $x$ there exists a sequence of rational numbers $(q_n)_{n=1}^\infty$ with $$\lim_{n \rightarrow \infty} q_{n}=x$$

What I have is by the Archimedean property, for every $x\in \mathbb{R}$, there exists some $y \in \mathbb{R}$ and $z \in \mathbb{R}$ such that $z<x<y$. So by the density of $\mathbb{Q}$ in $\mathbb{R}$, a sequence $(q_n)_{n=1}^\infty$ exists for each $x \in \mathbb{R}$ such that $\lim_{n \rightarrow \infty} q_{n}=x$

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    $\begingroup$ You shouldn't assume $\mathbb Q$ density because that's what you're asked, right? $\endgroup$ – Oussama Boussif May 28 '17 at 16:17
  • $\begingroup$ @claire we need more context in this question, what you knows about the density of $\Bbb Q$ in $\Bbb R$? What is the original definition of density where you start? $\endgroup$ – Masacroso May 28 '17 at 16:21
  • $\begingroup$ Also: Theorem. (The Archimedean Property of R) The set N of natural numbers is unbounded above in R. $\endgroup$ – user261263 May 28 '17 at 16:32
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Hint: Consider the sequences $x + \frac{1}{n}$ and $x - \frac{1}{n}$ in combination with the density of $\mathbb{Q}$ (it is unclear from your question whether you are allowed to use this or not...?)

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Theorem ($\mathbb{Q}$ is dense in $\mathbb{R}$). For every $x, y \in \mathbb{R}$ such that $x < y$, there exists a rational number $r$ such that $x < r < y$.


Let $q_n \in \mathbb{Q}$ defined: $q_{n}$ is a rational from $(x-\frac 1 n, x)$ interval, whose existence is quaranteed by the density theorem. Then $\lim_{n \rightarrow \infty} q_{n}=x$ from squeeze theorem.

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  • $\begingroup$ It's a shame to give the answer away rather than provide clues to allow op to work out the answer themselves - thus learning more in the process. $\endgroup$ – lux May 29 '17 at 2:20

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