Is this the basis for the topology of the complex projective plane? I am doing this as a thought exercise to test my understanding of the quotient topology. 
Anyway, given an open set $U \subset \mathbb{CP}^2$, if $\pi: \mathbb{C}^3\setminus\{(0,0,0)\} \to \mathbb{CP}^2$ is the defining quotient map, then $\pi^{-1}(U)$ is open. Moreover, since each point of $\mathbb{CP}^2$ is an equivalence class equal to a line through the origin $(0,0,0)$ minus the origin, the fibers of $\pi$ must be lines through the origin minus the origin. Thus the preimage of any set in $\mathbb{CP}^2$ must be the union of lines through the origin minus the origin, including the pre-image of any open set in $\mathbb{CP}^2$, which thus also must be open in $\mathbb{C}^3$.
The only open sets of this form which I could think of would be "double cones without boundary", for example the union of the first and third quadrant minus the $x$ and $y$ axes and the origin.
Question: Do these "double cones" form a basis for the topology of $\mathbb{CP}^2$? This would be a nice way to visualize the topology if it were true. I cannot think of a simple way to verify it however.
Also, if I wanted to write this basis for the topology in homogeneous coordinates, would it consist of sets of the form (for $(x,y) \in \mathbb{C^2}$ and $(a:b:0)$ in the line at infinity): $$( B_{\varepsilon}(x) :B_{\varepsilon}(y) : 1 ) \quad and \quad (a:b:B_{\varepsilon}(0)) ? $$ $B_{\varepsilon}(z)$ is the ball of radius $\varepsilon$ centered at the point $z \in \mathbb{C}$.
 A: Open sets
I have never worked with topological bases before, so I'm working with what Wikipedia tells me. The double cones you describe are indeed reasonable preimages of open sets in $\mathbb{CP}^2$. They are of course not the only ones, since you can build unions of these, but that's probably what you meant anyway.
Your basis
Now to show that what you'd write down is a basis, according to Wikipedia you have to show two things.


*

*These sets cover $\mathbb{CP}^2$. Sure, for any finite point you have a neighborhood of the first form, and for every infinite point you have one of the second form.

*For two such sets $B_1,B_2$, any point $x$ in the intersection $I=B_1\cap B_2$ will belong to a basis set $B_3\subseteq I$. You'd choose your basis as a neighborhood of $x$. To show that there exists a suitable $\varepsilon$, observe that $B_1$ and $B_2$ are open sets, so any point outside these sets has to be a positive “distance” away from $x$.
My suggestion for a basis
But this definition of “distance” might be a bit hard to make rigorous here in this context. What metric would you use? In order to address that problem, I'd suggest a different basis, namely one which does not depend on a distinguished line at infinity or a specific coordinate system for $\mathbb{C}^3$. Since you want to describe neighborhoods, I'd start with a “neighborhood” of radius zero and ask you this: when do two vectors $v$ and $w$ describe the same point in $\mathbb{CP}^2$? That is the case if they point in the same (or opposite) direction. Which depends on the angle. Which in $\mathbb{RP}^2$ you can check using the scalar product (dot product). The two vectors are equal if and only if
$$\langle v,w\rangle^2 = \langle v,v\rangle\langle w,w\rangle$$
since in that case, both sides of the equation equal the product of the squared lengths of the vectors. Otherwise the left hand side falls short by a factor of $\cos^2\alpha$ with $\alpha$ the angle between the vectors. So when are two vectors almost equal? If
$$\langle v,w\rangle^2 > \langle v,v\rangle\langle w,w\rangle
\cdot(1-\varepsilon)$$
So given a vector $v$ and a “radius” $\varepsilon$, this will define the set of vectors $w$ in the open neighborhood of $v$.
How well does this translate to $\mathbb C$? Well, for one thing you will have to take the standard sesquilinear product (i.e. $\langle v,w\rangle=\bar v_1w_1+\bar v_2w_2+\bar v_3w_3$) to ensure that the right hand side really represents something like real lengths. And for the left hand side you should use a sesquilinear product and its conjugate, which is equal to the argument-swapped version.
$$\langle v,w\rangle\langle w,v\rangle > \langle v,v\rangle\langle w,w\rangle
\cdot(1-\varepsilon)$$
Then you can show that you get equality only of $w=zv$ for some $z\in\mathbb C$, and almost-equality for small $\varepsilon$.
Now yo can again check coverage and the intersection property, as above.


*

*If you choose $\varepsilon>1$ then any $v$ will give you a basis element which contains all of $\mathbb{CP}^2$. Or you observe that around any point $v$ you have a whole non-empty bunch of neighborhoods in your basis.

*This boils down to the same “distance” argument as above, only this time you may have a better notion of what a “distance” actually is.
