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Defn 1): A set $\textbf{A}$ is dense in $\textbf{B}$ if $\textbf{B}$ is contained in $\overline{A}$

Defn 2) The closure of A, ia the set $\overline{A}$ consisting of all limit pints of A.

$\textbf{i)}$ Show that the set of irrational numbers = $\mathbb{Q}^{c}$ is dense in $\mathbb{R}$

$\textbf{ii)}$ Hence show that $\mathbb{Q}$ has empty interior

Attempt at i)

I want to show that $$\mathbb{R}\subset \overline{\mathbb{Q}^{c}}$$ This means I want to show that all the limit points of $\mathbb{R} \subset \overline{\mathbb{Q}^{c}}$. Now since I am in $\mathbb{Q}^{c}$, I don't have to worry about showing the limit points of the irrationals. What is left to show though is that all the rationals $\mathbb{Q}$ are limit points of $\mathbb{Q}^{c}$.

Take the sequence $$\frac{1}{\sqrt{2}n}$$ in irrationals. This is where I get stuck. My professor that I have to show that this sequence converges to 0 which would show $\mathbb{Q}^{c}$ is irrational, but I don't know how and I don't fully understand how it satisfies the condition for all other rationals that are not $0$

Attempt at ii)

Here I am clueless. WHat I do know is that the inerior of a set is the largest open set of a set. Also going on previous knowledge if this interior was not empty it would mean that for every point, $a$ in the set there exists a ball $B_{r}(a)$ such that the ball is contained within the set. I don't see how to use these facts.

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  • $\begingroup$ I suggest you show that Q has empty interior and use that to show (i) $\endgroup$ – Mariano Suárez-Álvarez Feb 8 '18 at 3:28
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You want to show that a rational $q$ is a limit of irrationals. For this, you can use that "rational plus irrational" is irrational. So you could form $$ q+\frac{\sqrt2}n,\ \ \ \ n\in\mathbb N. $$ All numbers in this sequence are irrational, and the sequence converges to $q$.

That $\mathbb Q$ has empty interior follows directly from the above. Given $q\in\mathbb Q$ and any interval $I=(q-\varepsilon,q+\varepsilon)$, by the above there are irrationals in $I$. So $I\not\subset\mathbb Q$, which implies that $\mathbb Q$ has empty interior.

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  • $\begingroup$ Maybe I'm being pedantic, because I see that taking the limit of that sequence would converge to a rational. But wouldn't I have to use the formal $\epsilon$ - argument to show convergence? $\endgroup$ – dc3rd Feb 8 '18 at 3:55
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    $\begingroup$ Yes. You have $$\left|q+\frac{\sqrt2}n-q\right|=\frac{\sqrt2}n.$$ Then, given any $\varepsilon>0$, you can use the Archimedean property to find $n>\sqrt2/\varepsilon$. $\endgroup$ – Martin Argerami Feb 8 '18 at 4:19
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Hint: $a_n=\frac1{\sqrt2 n}$ is a sequence in $\mathbb Q^c$ converging to $0$. For any other $r\in \mathbb Q$, just shift the sequence by $r$... that is, $b_n=a_n+r$ is a sequence in $\mathbb Q^c$ converging to $r$...

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See you can prove the convergence of the sequence $\frac{1}{\sqrt2n}$ by Archimedean property. Once you showed that take any rational $r \in \mathbb{Q}$ and consider the sequence $r+\frac{1}{\sqrt2n}$ which converges to $r$.Therfore any nbd of a rational contains a irrational which proves the second statement too.The Archimedean property of $\mathbb{R}$ says for any positive $x,y \in \mathbb{R} $, there exists a natural number $n$ such that $ny>x$. Now take $y=\epsilon >0$ and $x=\frac{1}{\sqrt2}$. The convergence follows.

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  • $\begingroup$ How would the Archimeidan property apply? $\endgroup$ – dc3rd Feb 8 '18 at 3:58
  • $\begingroup$ You both managed to misspell the word Archimedean. $\endgroup$ – Mariano Suárez-Álvarez Feb 8 '18 at 4:01
  • $\begingroup$ I edited . Thanks @MarianoSuárez-Álvarez. $\endgroup$ – user250285 Feb 8 '18 at 4:06
  • $\begingroup$ so then what you're saying is that $\frac{1}{\sqrt{2}n} < \frac{1}{\sqrt{2}N} < \frac{\epsilon}{\sqrt{2}} < \epsilon$? $\endgroup$ – dc3rd Feb 8 '18 at 4:11
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    $\begingroup$ I think you wanted to say $\frac{1}{\sqrt2 n} <\frac{1}{\sqrt2 N}< \epsilon $ for all $n > N$. Then you are right. $\endgroup$ – user250285 Feb 8 '18 at 4:15
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Here are two alternate proofs.

For any not empty interval I, I is uncountable.
As there are only countably many rationals, there has
to be irrationals in I. Thus the irrationals are dense.

$\overline{R-Q} = R - Q^o = R.$

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