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$\mathbb{Q}\cap [0,1]$ is not closed since any irrational numbers in $(0,1)$ is a limit point but not in $\mathbb{Q}\cap [0,1]$.

By Heine-Borel theorem, $\mathbb{Q}\cap [0,1]$ is not compact.

Now, instead of using Heine-Borel theorem, i want to use the definition of compact set to prove $\mathbb{Q}\cap [0,1]$ is not compat, i.e. find a sequence that converges to an irrational number $\in [0,1]$, example $\frac{1}{\sqrt{2}}$ and hence every subsquence converges to that irrational number as well.

I will appreciate if you prove the sequence converges to its limit instead of just stating it.

DonAntonio
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Little Rookie
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    Is your probem that you cannot find such a sequence of rational numbers? – Arthur Mar 21 '17 at 12:56
  • its a typing mistake, i have edited it. Yes my problem is i cant think of such convergent sequence. – Little Rookie Mar 21 '17 at 12:58
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    Well, ${.4,.41,.414,.4142,.41421,\cdots }$ converges to $\sqrt 2-1\approx .414213562373$. – lulu Mar 21 '17 at 12:59
  • Hint: http://math.stackexchange.com/questions/226793/show-that-mathbbq-is-dense-in-the-real-numbers-using-supremum using that condtruct the sequence. – Santropedro Mar 21 '17 at 13:01
  • Similarly, $\frac 1{\sqrt 2}\approx 0.707106781186547$ so take ${.7,.707,.7071,.707106, \cdots}$. – lulu Mar 21 '17 at 13:01
  • You can prove that compact implies closed without Heine-Borel (in fact it's quite easy), and that's all you need for a contradiction here since $\Bbb Q\cap [0,1]$ is dense in $[0,1]$. – Adam Hughes Mar 21 '17 at 13:05
  • @lulu how do i construct a sequence using that? – Little Rookie Mar 21 '17 at 13:09
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    $a_n=\lfloor \frac 1{\sqrt 2}\times 10^n\rfloor\times 10^{-n}$. – lulu Mar 21 '17 at 13:23
  • @lulu How do i prove that it converges to $\frac{1}{\sqrt{2}}$? – Little Rookie Mar 21 '17 at 13:53
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    $|\frac 1{\sqrt 2}-a_n|<10^{-n+1}$ – lulu Mar 21 '17 at 13:56
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    This is just how you get a decimal expansion for any real number. Just use the greedy algorithm...at each stage, take the greatest decimal number, truncated to $n$ places, which is less than your given number. – lulu Mar 21 '17 at 14:01

2 Answers2

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It think it is almost trivial to show $\;\left(1+\cfrac1n\right)^n\;$ is a rational number$\;>1\;$, for any $\;n\in\Bbb N\;$. Well, now just take

$$x_n:=\left(\frac n{n+1}\right)^n=\frac1{\left(1+\cfrac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e\notin\Bbb Q\cap[0,1]$$

Or perhaps a simpler one:

$$y_n:=\left(1-\frac1n\right)^n\xrightarrow[n\to\infty]{}\frac1e$$

DonAntonio
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Alternatively: Let be two monotonic sequences of rational numbers $0<r_n<r_{n+1}<s_{n+1}<s_n<1$ with the same limit $l\not\in\Bbb Q$. The family $$\{[0,r_n)\cap\Bbb Q\}_{n\in\Bbb N}\cup\{(s_n,1]\cap\Bbb Q\}_{n\in\Bbb N}$$ is an open cover of the set without finite subcover.

Martín-Blas Pérez Pinilla
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