Which of the following sets are compact?
$1.\{({1\over n},{1\over m})\in \mathbb R^2:m,n\in \mathbb Z\}\bigcup \{(0,0)\}\bigcup \{({1\over n},0):n\in \mathbb Z\}\bigcup \{(0,{1\over n}): n\in \mathbb Z\}$
$2.\{(x,y,z)\in \mathbb R^3: |x|+|2y|+|3z|\le 1\}$
$3.\{(x,y,z)\in \mathbb R^3:x^2+2y^2-3z^2=1\}$
$4.\{(x_1,x_2,....,x_n)\in \mathbb R^n:x_1+x_2+x_3+.....+x_n=0\}$
Option $4$ is clearly unbounded so not compact.
For option $3$, $x^2+2y^2=1+3z^2$ so the set , is unbounded for arbitrarily large values of $z$ will take up arbitrarily large values of $x$ and $y$ so unbounded.
Option $2$ is bounded since the equation is basically sum of three positive numbers less than $1$.How do I show that it is closed as well?
For Option $1$,the set is bounded and I can see the set $ \{(0,0)\}\bigcup \{({1\over n},0):n\in \mathbb Z\}\bigcup \{(0,{1\over n}): n\in \mathbb Z\}$ is closed itself and is a set of limit points of the set $\{({1\over n},{1\over m})\in \mathbb R^2:m,n\in \mathbb Z\}$. Now if only I could show that those are the only limit points the set $\{({1\over n},{1\over m})\in \mathbb R^2:m,n\in \mathbb Z\}$ can have then I would be sure that it was compact. How do I prove that? Thank you.