Connected components defined by a given curve in the affine plane, in the projective plane and in the complex affine plane 
Let $E$ be the plane affine curve defined by $y^2=x^3-1$. Prove that:
(i) removing the points of $E(\mathbb R)$ from the affine plane
  $\mathbb R^2$ we get a disconnected set with two connected components.
(ii) removing the closure of $E$ in the projective real plane we get a
  connected set.
(iii) removing the set of complex points $E(\mathbb C)$ from the
  affine plane we obtain a connected space.

Now, I see intuitively point (i), and after writing the equation for the projective closure, point (ii) too. But intuition fails in order to imagine point (iii). Furthermore, I would like to know how to compose a formal proof in the three cases, since intuition is not a proof. An example of proof would be then interesting to me.
Thank you in advance.
 A: Well, I think for point (i) you simply draw a picture of the curve defined by $y^2=x^3-1$ in the real plane. We can write the curve as $x^3 = y^2 + 1$ so for real values of $y$ the real variable $x$ is uniquely defined (the other solutions are complex, actually complex conjugate to each other). Thus, in the real plane the curve looks like  $x = x(y) = \sqrt[3]{y^2-1}$ so it is a graph of a function defined for all $y \in \mathbb{R}$. Hence it splits the real plane into two components -- "below" the graph and "above" the graph. Notice ti is even, i.e. symmetric with respect to the change $x \mapsto -x$ which is the elliptic involution of the curve. 
(ii) To see what happens at infinity, perform a projective change of variables $[x:y:1] = \left[\frac{x}{y} : 1 : \frac{1}{y}\right] = [z: 1 : w]$ which is the same as
\begin{align*}
z&= \frac{x}{y}\\
w&= \frac{1}{y}
\end{align*} 
with inverse
\begin{align*}
x&= \frac{z}{w}\\
y&= \frac{1}{w}
\end{align*}
and follow the computation 
$$0 = x^3 - y^2-1 = \left(\frac{z}{w}\right)^3 - \left(\frac{1}{w}\right)^2 - 1 = \frac{z^3 - w - w^3}{w^3}$$ so the curve in $z,w$ coordinates looks like 
$$z^3 - w - w^3 = 0.$$ Therefore the projectivisation of your curve extends to the point $[0:1:0]$ at infinity. It's only one point and the curve is smoothly embedded there. Thus in the projective plane it is a smoothly (actually real analytically) embedded one dimensional submanifold without self intersections. It intersects the line at infinity at exactly one point. Hence it has only one component, as the real projective plane is a non-orientable surface. You can simply go from the component "above" the graph of $x(y)$ to the component "below" the graph by walking along the $x$ axis and passing the "infinity border" at the point $[1:0:0]$ which is not on the curve.  
(iii) The affine complex algebraic curve in $\mathbb{C}^2 = E(\mathbb{C})$ is a two dimensional real submanifold, while the ambient space is a real four dimensional manifold. The real codimension of the algebraic curve is two, so the complement of the curve is a connected space. It is like having a smooth curve in three space. Its complement is always connected.      
