How to prove that a set, by the open cover definition, is not compact For an exam I have to be able to prove whether certain sets are open, closed or neither and, by extension, (ab)using the Heine-Borel theorem to prove if these sets are compact or not.
Because I prefer not to use the Heine-Borel theorem I would like to use the definition of a compact set to prove whether the following sets are compact or not (spoiler: they are not):
$$ V = \lbrace x \in \mathbb{R}^2 : -1 \leq x_1 \lt 2, -3 \lt x_2 \leq 4 \rbrace $$ 
$$ W = \lbrace x \in \mathbb{R}^2 : x_1^2 + x_2^2 \lt 1 \rbrace $$
Obviously, $W = B(0;1)$ and by definition, open balls are open and therefore it is not compact. But how would you prove this with the open cover definition
Could anyone give me some hints how to do these kind of problems?
Thanks in advance!
 A: The basic idea for a metric space is (usually) to find a set of open sets that cover more and more of a sequence of points that lie within the set but have limit outside. In the case of $W$, others have already pointed out that choosing balls of increasing radius but never reaching $1$ will work, and can be seen as covering more and more of a sequence that heads towards the boundary of the open set $W$. For $V$ there are a number of options, and we can choose for example the sequence $( (2-\frac{1}{n},0) : n \in (1,2,...) )$ with limit $(2,0)$. A suitable open cover is $\{ (-2,2-\frac{1}{n}) \times (-4,5)) \}$ which gradually covers more and more of the sequence but not all at any time.
A: For $W$, take sets $A_n$ defined as $A_n=B(0, 1-1/n)$. You can see that 
$$\bigcup_{n\in\mathbb N}A_n = W.$$ Can you find a finite subcovering?
When you understand why such a covering does not exist, you can try and prove $V$ is not compact in the same way.
A: Use an infinite cover of $W$ by a family of balls of radius $r$ for all $r<1$.  Here it does not matter whether the balls are open or closed: one obviously can't choose a finite subcover.
