Show if $Q ∩[0,1]$ is compact?

Question

Show if $Q ∩[0,1]$ is compact.

It is bounded. It contains it limit point $1$, therefore it is closed = > compact.

What is wrong with this conclusion?

Answer 1

It is not compact. Take any irrational point $p$ in $(0,1)$. Then $p\in\overline{\mathbb{Q}\cap(0,1)}$, but $p\notin\mathbb{Q}\cap(0,1)$. Therefore, $\mathbb{Q}\cap(0,1)$ is not a closed subset of $\mathbb R$ and so it cannot be compact.

Answer 2

For each $\alpha\in\mathbb Q\cap (0,1)$, let $$I_\alpha=\begin{cases}[0,\alpha)&\alpha^2<\frac{1}{2}\\(\alpha,1]&\alpha^2>\frac{1}{2}\end{cases}$$

Show that the $I_\alpha$ are open in $X=\mathbb Q\cap [0,1]$, and that they cover $X,$ but there is no finite sub-cover.

Answer 3

The set is not compact in $\Bbb R$. Take the following subsequence $$a_n=\text{the first n digits in the decimal representation of } \pi -3\text{ after decimal point}$$for example$$a_1=0.1\\a_2=0.14\\a_3=0.151\\.\\.\\.$$then any subsequence of $a_n$ tends to $\pi -3\notin Q\cap[0,1]$.