For what values of $\alpha$ is this set compact? I was given the set $M=\{(x_1,x_2)\in \mathbb{R}^2 | |x_1|^\alpha + |x_2|^\alpha \leq 1 \}$.
For which values of $\alpha$ is this set:
a) closed
b) compact
c) convex.
What I know: if a set is closed and bounded, then it's compact. And a closed set has to contain all limit points. So whatever works for b) will have to work for a). But I can't really figure out how to apply that theory. 
 A: If $\alpha<0$, then $M$ is unbounded (in particular, it is not compact) and disconneted (in particular, it is not convex).
If $\alpha=0$, then $M=\emptyset$, which is both compact and convex.
If $\alpha>0$, then $M$ is closed ($M=f^{-1}\bigl([0,1]\bigr)$ with $f(x_1,x_2)=\lvert x_1\rvert^\alpha+\lvert x_2\rvert^\alpha$, and $f$ is continuous) and bounded; therefore it is compact. But it is convex if and only if $\alpha\geqslant1$ (if $\alpha\in(0,1)$, then $(1,0),(0,1)\in M$, but no point of the line segment joining them belongs to $M$).
A: Note that if $\alpha < 0$, then the set is not compact for it is not bounded.
To show this, note that $(x_0, x_0) \in M$ for every $x_0 \ge \left(\frac{1}{2}\right)^{1/\alpha}$. (Simply check.)
It is not convex either. Note that $M$ isn't empty (by above exercise). So, pick any point $(x_0, y_0) \in M$. We also have that $(-x_0, -y_0) \in M.$ Now, $(0, 0)$ lies on the line segment joining these two points (it's the midpoint) but $(0, 0) \notin M$. We will later show that $M$ is closed.
If $\alpha = 0$, then the set is empty and all the properties are (vacuously) true.
Let us now assume that $\alpha > 0.$
One can straightaway note that $|x| \le 1$ and $|y| \le 1$. Thus, $M$ is bounded. This means that it is compact iff it is closed. We will later see that this is indeed the case. Let us first deal with convexity.  
Claim 1. If $0 < \alpha < 1$, then $M$ is not convex.
Proof. Consider the points $(0, 1)$ and $(1, 0)$. It is clear that these points belong to $M$.
Now, check that $\left(\frac{1}{2}, \frac{1}{2}\right)$ is the midpoint of these points. However, it does not belong to $M$.
To check this, note that $\left(\frac{1}{2}\right)^\alpha + \left(\frac{1}{2}\right)^\alpha = 2^{1-\alpha} > 2^0  =1.$
Claim 2. If $\alpha \ge 1$, then $M$ is convex.
Proof. Use inequalities carefully.  
Now, we show that $M$ is always closed irrespective of $\alpha$.  
Note that the function $(x_1, x_2) \mapsto |x_1|^\alpha + |x_2|^\alpha$ is continuous and the set $M$ is simply the inverse image of the closed set $[0, 1]$. As inverse image of closed sets are closed under continuous maps, we are done. 

To sum it up:
$\alpha < 0 : $ only closed
$\alpha = 0 : $ all
$0 < \alpha < 1:$ only closed and compact
$1 \le \alpha:$ all
