Is $(x^2-y^2-1)^2+4x^2y^2 <1$ an open set? I must answer, for some sets, if they are: i) open, ii)closed, iii)bounded, iv) compact
I've got the following set:
$$\{z\in \mathbb{C}; |z^2-1|<1\}$$
if I suppose $z=x+iy$ I get that this set if the set of reals $x,y$ that satisfy:
$$(x^2-y^2-1)^2+4x^2y^2 <1$$
Which is this, according to Wolfram Alpha:

Well, it looks like open, but what about that point in the middle? I also think it's not closed, so not compact. It's, however, bounded because I can fit it entirely inside an open ball
 A: It is perhaps interesting to get the general result. Let $n$ be a natural, $a$ a real and $P$ a polynomial of $n$ variables. Then the set $$S_{a,P}=\{(x_1,...,x_n) : P(x_1,...,x_n)<a\}$$ is open in $\mathbb{R}^n$ with the usual topology. To notice that, just remark that $$S_{a,P} = P^{-1}((-\infty,a))$$ and that $P$ is continuous since it is a polynomial. Of course you can can replace $P$ by any continuous function.
A: The function $f:\mathbb{R}^2\to\mathbb{R}$ defined by
$$f(x,y)=(x^2-y^2-1)^2+4x^2y^2$$
is continuous.  The set $\{x\in\mathbb{R}:x<1\}$ is open.  Now using the technical definition of a continuous function (that the preimage of an open set is open), we see that 
$$f^{-1}((-\infty,1))=\{(x,y)\in\mathbb{R}^2:(x^2-y^2-1)^2+4x^2y^2<1\}$$
is an open set.
A: To show that $\{z\in\mathbb{C}: |z^2-1|<1\}$ is open, let $z\in\mathbb{C}$ satisfy $|z^2-1|<1$. We want to show that
$$\exists\delta>0: (|h|<\delta\implies |(z+h)^2-1|<1). $$
Note that
$$ ((z+h)^2-1)-(z^2-1) = 2zh+h^2 $$
so by the triangle inequality
$$ |(z+h)^2-1| \le |z^2-1| + |2zh+h^2|. $$
If we write $|z^2-1| = 1-\epsilon$, then $\epsilon>0$, and so it suffices to find $\delta>0$ such that
$$ |h|<\delta \implies |2zh+h^2| < \epsilon. $$
Now let $\delta>0$ satisfy
$$ y < \delta \implies 2|z|y+y^2 < \epsilon. $$
(It should be clear that such a $\delta$ exists. For example, try to show that $\delta = \min\left(1,\frac{\epsilon}{2|z|+1}\right)$ works.)
Then
$$ |h| < \delta\implies |2zh+h^2|\le 2|z||h|+|h|^2 < \epsilon $$
as desired.
