compact subsets of $ \mathbb{R^2}$ Which of the following subsets of $ \mathbb{R^2}$ are compact?

*

*$\displaystyle \{(x, y) : xy = 1\}$


*$\displaystyle \{(x, y) : x^{\large\frac{2}{3}} + y^{\large\frac{2}{3}} = 1\}$


*$\displaystyle \{(x, y) : x^2 + y^2 < 1\}$
I am stuck on this problem. Can anyone help me please?
 A: For $1)$ use the fact that $|(x,y)| \ge \max \{|x|, |y| \}$.
For $2)$, recall that a continuous mapping of a compact set is compact.  This set looks awfully similar to a very well-known compact set.  Can you identify it, and can you think of a continuous surjection from that set onto yours?
For $3)$ recall that $(0,1) \subset \mathbb{R}$ is not closed because we may find a convergent sequence (say $\{\frac{1}{n}\}_{n=1}^{\infty}$) which converges but not to an element of $(0,1)$.  See if this can be adapted.
A: As mentioned in the comments, it helps us help you if you tell us what you've tried and what tools are at your disposal so we can best direct our help.
For example, do you know that in $\mathbb{R}^2$, that a set $S$ is compact if and only if it is closed and bounded? If so, use this characterization of compactness (in $\mathbb{R}^2$) to reason as follows (each time $S$ represents the set you gave):
(1) This is the graph of a hyperbola. $S$ is certainly unbounded so by what I cited above, $S$ cannot be compact.
(2) Here's a plot of $S$:

$S$ here is closed and bounded (be sure you understand why), so $S$ is compact. 
(3) This is the interior of the unit disk, but with its boundary $x^2+y^2=1$ excluded. 

$S$ here is bounded but not closed, so $S$ is not compact.
Hope that helps.
