# Prove $\mathbb{Q}\cap [0,1]$ is not compact without using Heine-Borel theorem. [duplicate]

$\mathbb{Q}\cap [0,1]$ 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.

• Is your probem that you cannot find such a sequence of rational numbers? – Arthur Mar 21 '17 at 12:56
• Well, $\{.4,.41,.414,.4142,.41421,\cdots \}$ converges to $\sqrt 2-1\approx .414213562373$. – lulu Mar 21 '17 at 12:59
• $a_n=\lfloor \frac 1{\sqrt 2}\times 10^n\rfloor\times 10^{-n}$. – lulu Mar 21 '17 at 13:23
• $|\frac 1{\sqrt 2}-a_n|<10^{-n+1}$ – lulu Mar 21 '17 at 13:56
• 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

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
$$x_n:=\left(\frac n{n+1}\right)^n=\frac1{\left(1+\cfrac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e\notin\Bbb Q\cap[0,1]$$
$$y_n:=\left(1-\frac1n\right)^n\xrightarrow[n\to\infty]{}\frac1e$$
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.