Are $A=[0,1]\times[0,1]$ and $B=([0,1]\cap\mathbb{Q})\times([0,1]\cap\mathbb{Q})$ compact in $\mathbb{R}^2$? $ A= \{ (x_1,x_2): x_1,x_2 \in [0,1] \} $ in $\mathbb{R}^2$
$ B= \{ (x_1,x_2): x_1,x_2 \in [0,1] \cap \mathbb{Q} \} $ in $\mathbb{R}^2$
I am using the usual Euclidean metric in $\mathbb{R}^2$. So, a set is compact iff its closed and bounded.
I have some issues to understand if those sets are compact. Do you have a good explanation? 
 A: HINT(S): $A$ can be rewritten as $[0,1]\times[0,1]$. Is this set bounded? Is it true that $[0,1]\times[0,1]\subset B_{10}(0)?$ Here $B_r(p)$ is the ball of radius $r$ about $p\in \mathbf{R}^2$. 
Is $A$ it closed? Is $A^C=\mathbf{R}^2\setminus [0,1]\times[0,1]$ open? Around any point in $A^C$ can you find a ball of radius $r>0$ contained in $A^C$? If so, $A^C$ is open and $A$ is closed. 
$B$ can be rewritten as $([0,1]\cap \mathbf{Q})\times ([0,1]\cap \mathbf{Q}).$ Is $\mathbf{Q}$ closed in $\mathbf{R}?$ Is the limit of a convergent sequence of rationals rational? By the same reasoning, we can determine whether or not $[0,1]\cap \mathbf{Q}$ is closed in $[0,1]$. 
To finish the problem, if we fix a rational point $p\in [0,1]\cap \mathbf{Q}$ and consider sequences 
$$ (p,q_n)_{n\in \mathbf{N}}\in B$$
then we are back to the $1-$dimensional case.
A: $[0,1]^2$ is compact because you can prove that the Cartesian product of compact sets is compact if and only if their components are compact. We know $[0,1]$ is compact. (I am assuming the underlying topology is the standard one). 
Note that  $[0,1]\cap \mathbb{Q}$  is not compact, since $\mathbb{Q}$ isn't even closed. To see this, take the sequence $\{x_n\}=\{\frac{1}{4}(1+\frac{1}{n})^n\}$. This sequence certainly contains only rationals in the unit interval, but $x_n \rightarrow \frac{e}{4}$, which is irrational, so we have shown $[0,1]\cap \mathbb{Q}$ is not even closed, so it cannot be compact (again, under standard topology). Since this set is not compact,
$B=([0,1]\cap \mathbb{Q})^2$ is not compact. 
A: Clearly $[0,1]\times[0,1]$ is sequentially compact since $[0,1]$ is compact and for any sequence  $\{(x_n,y_n)\}_{n\in\mathbb{N}}$ on $\mathbb{R^2}$ converges to $(K,L)$ iff  $\{x_n\}_{n\in\mathbb{N}}$ converges to $K$ and $\{y_n\}_{n\in\mathbb{N}}$ converges to $L$.
It's clear that $\mathbb{Q}\cap [0,1]$ is not sequentially compact since there exists irrationals $q\in\mathbb{Q}\cap [0,1]$ and we can construct a sequence on $\mathbb{Q}\cap [0,1]$ that converges to $q$.

Example : Consider the sequence $\{\frac{F_n}{F_{n+1}}\}_{n\in\mathbb{N}}$ where $F_n$ is the $n$ th
Fibonacci number. This is a sequence on $\mathbb{Q}\cap [0,1]$ that
converges to $\frac1\varphi$ where $\varphi\in\mathbb{R-Q}$ is the
golden ratio.

Therefore, if $\{x_n\}_{n\in\mathbb{N}}$ is a sequence on $\mathbb{Q}\cap [0,1]$ that converges to a irrational, then $\{(0,x_n)\}_{n\in\mathbb{N}}$ is a sequence on $\mathbb{Q}\cap [0,1]\times\mathbb{Q}\cap [0,1]$ that converges to a point in $\mathbb{R^2-Q^2}$. Hence $\mathbb{Q}\cap [0,1]\times\mathbb{Q}\cap [0,1]$ is not compact.
