Show that E is not compact. Given the set $E = \{x\in\mathbb{Q}: 2 < x^2 < 3\}$, considered as a subset of $\mathbb{Q}$.
Exercise: Show that $E$ is not compact.
What I've tried:
take any finite $n$ number of points $x_1,...,x_n \in \mathbb{Q}$ and take $\epsilon < \frac{\sqrt3-\sqrt2}{2n}$. The union $\bigcup\limits_{i = 1}^{n}B_\epsilon(x_i)$ does not cover $E$. This means that $E$ is not totally bounded and hence not compact.
Question: I think I'm going in the right direction, however, my solution feels a bit too intuitive. Is my answer sufficient, or should I be more careful/elaborate when concluding that $\bigcup\limits_{i = 1}^{n}B_\epsilon(x_i)$ does not cover $E$?
 A: Consider the sequence
$$
x_n=\frac{\lfloor n\sqrt{3}\rfloor}{n}, \quad n\ge 2
$$
It is easy to check that 
$$
\sqrt{2}<x_n<\sqrt{3}, \quad\text{when}\,\,\, n\ge 2
$$
and hence $\{x_n\}_{n\ge 2}\subset E$, while 
$$3-x_n^2\to 0,$$
and hence $\{x_n\}_{n\ge 2}$ does not possess a converging subsequence in $E$.
A: It can be proved that the set $A = \left\{ {p:{p^2} < 3,\, p \in {\Bbb Q^+}} \right\}$ has no greatest element (see this). From this it is easy to conclude that the set $E = \{x\in\mathbb{Q}: 2 < x^2 < 3\}$ also does not contain a greatest element.
Recall that closed subsets of compact spaces are also compact.
We use the above in the following two solutions. Note that these demonstrations do not require the construction of the real numbers or the fact that compactness in a metric space is equivalent to being sequentially compact.

Solution 1:
Enumerate all the elements of the countable set $E$ with a sequence $(p_k)_{\, k \in \mathbb N}$. Define closed sets in $\mathbb{Q}$ as follows:
$\tag 1 I_n = [max(p_1,p_2,\dots,p_n), +\infty) \; \bigcap \; E$
The sequence of nonempty closed sets $I_n$ is a decreasing chain $A_{n+1} \subset A_n$. If we assume that $E$ is compact, then we can apply Cantor's intersection theorem and conclude that the intersection of all the $I_n$ must be nonempty. But by design no $p_j$ can be in this intersection (no greatest element), a contradiction.

Solution 2:
We define $r_1 = 1.7$ so that $r_1 \in E$. We recursively define
$\tag 2  r_{n+1} = r_n + d \, 10^{-(n+1)} \text{, } d \in \{0,1,\dots,9\} \text{ largest digit | } r_{n+1} \in E$
Define closed sets in $\mathbb{Q}$ as follows:
$\tag 3 I_n = [r_n, +\infty) \; \bigcap \; E$
The sequence of nonempty closed sets $I_n$ is a decreasing chain $A_{n+1} \subset A_n$. If we assume that $E$ is compact, then we can apply Cantor's intersection theorem and conclude that the intersection of all the $I_n$ must be nonempty. But if $q$ is any point in $E$ it is easy to see that some $r_m$ must be greater than $q$, leading to a contradiction.
A: Let $G_n = \left( 2+ \dfrac 1n, 3 \right)=\{x \in \mathbb{Q}|2+ \dfrac 1n < x < 3 \}.$ Then   $\displaystyle \{\cup G_n \}_{n \in \mathbb{N}}$ is an open cover of $(2, 3)$ but we can show (by contradiction) that any finite subcover of it will not cover $(2, 3)$.
A: Let $G_n = \left( \sqrt2 + \dfrac 1n, \sqrt3 \right)=\{x \in \mathbb{Q}\,| \, \sqrt2+ \dfrac 1n < x < \sqrt 3 \}.$ Then   $\displaystyle \bigcup_{n \in \mathbb{N}} \, G_n $ is an open cover of the closed set $E = [\sqrt2, \sqrt3]$  in $\mathbb{Q}$. Note that for any $k$, $\, G_k \subset G_{k+1}$.
It is easy to see that for any $k$, $\,G_k \ne E$. But this means that no finite subcover of $(G_n)$ can be found, so $E$ is not compact.

$E$ is a closed and totally bounded subset of the metric space $\mathbb Q$. See
Why does Totally bounded need Complete in order to imply Compact?
