compactness of $\{(x,y):x^p+y^p=1\}$ I would like to know whether there is a rule other than drawing the curve to find out the set $\{(x,y):x^p+y^p=1\}$ is compact or non compact subset in $\mathbb{R}^2$
The set is closed for any values of $p\in\mathbb{R}^{+}$ right?
when $p$ is even positive integer, clearly the set is bounded and unbounded when $p$ is odd.
what about when $p$ is positive rational number and irrational number? anything can be concluded? Thank you.
 A: Taking a more general view, you want to know when a set $\{x\in\mathbb R^n: f(x)=c\}$ is compact, where $f:\mathbb R^n\to \mathbb R$ is continuous. The relevant term is proper map, which has the more general property that preimage of every compact set is compact.  
Fact.  A continuous function $f:\mathbb R^n\to \mathbb R$ is proper if and only if $\lim_{|x|\to\infty} |f(x)|=\infty$. 
The necessity is obvious because for every $M$ the set $\{x:|f(x)|\le M\}$ is compact. The sufficiency is best handled by passing to one-point compactifications of the spaces involved. The condition $\lim_{|x|\to\infty} |f(x)|=\infty$ says precisely that $f$ extends continuously to a map between compactified spaces. Its properness immediately follows, since every continuous map from a compact space to a Hausdorff space is  proper. $\quad \Box$
Returning to your situation: when $p$ is a positive even integer, the above applies. When $p$ is a positive odd integer, the map is not proper (consider $f(x,-x)=0$) which suggests that the compactness of other level sets might fail as well - and it does.
