please how to prove these two properties:

$(1)\quad\forall x\in\mathbb{R},\ \forall \varepsilon>0,\ \exists r\in\mathbb{Q},\ |x-r|\leq\varepsilon$

$(2)\quad \forall x\in\mathbb{R},\ \exists (r_n)\in\mathbb{Q},\ \displaystyle\lim_{n\to\infty} r_n=x$

  • 1
    $\begingroup$ It refers to density. $\endgroup$ – Wuestenfux Jan 11 at 9:37
  • 2
    $\begingroup$ What are your definitions of $\Bbb Q$ and of $\Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another. $\endgroup$ – Arthur Jan 11 at 9:37
  • $\begingroup$ The second property is how my professor in mathematics defined $\mathbb{R}$. And the first property is a consequence of the second one $\endgroup$ – Damien Jan 11 at 10:27

We can show $(1)$ as follows. Let $N > \frac{1}{\varepsilon}$ be an integer. Then we observe that $$Nx - \left \lfloor{Nx}\right \rfloor < 1 \Leftrightarrow x - \frac{\left \lfloor{Nx}\right \rfloor}{N} < \frac{1}{N} < \varepsilon$$ So taking $r = \frac{\left \lfloor{Nx}\right \rfloor}{N}$, we are done.

$(2)$ follows from $(1)$ by considering a sequence $\varepsilon_1,\varepsilon_2,...$ with $\varepsilon_n > 0$ and $\lim_{n \rightarrow \infty}\varepsilon_n = 0$ (for instance $\varepsilon_n = \frac{1}{n}$). Then we can find a sequence $r_n \in \mathbb{Q}$ such that $|x - r_n| < \varepsilon_n$, from which $\lim_{n \rightarrow \infty} r_n = x$ follows since $\forall \varepsilon > 0$ there is an $N$ such that $n > N$ implies $$|x - r_n| < \varepsilon_n < \varepsilon$$ as required.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.