In my self-study of Algebraic geometry, I came across the following argument;
The Projective plane $\mathbb{P}^2$ comes with a standard Affine cover consisting of the three open subsets; $\mathcal{U}_j = \{ [z_0 : z_1 : z_2] : z_j \neq 0 \} $ for j = 0, 1, 2.
I wanted to play with this statement with a concrete example of the open subset and Here is how I tried to concretise the form for j = 0;
$\mathcal{U}_0 = \{ [1 : 0 : 0] , [1 : 1 : 0] , [1 : 0 : 1] , [1 : 1 : 1] \}$.
Could someone correct me if I'm mistaken?
Here is (in my view) the correct way to understand open sets of $\Bbb{P}^2$ (or more generally $\Bbb{P}^n$ with appropriate modification). Don't forget what $\Bbb{P}^2$ is. It is a moduli space of lines in $\Bbb{A}^3$ passing through the origin. This is a fancy way of saying that it is the set of lines in $\Bbb{A}^3$ through the origin equipped with a topology and a variety structure (i.e. sheaf of rings). So, picture $\Bbb{A}^3$ as the $3$-space that we live in and consider that this space comes equipped with $3$ coordinate planes: $x=0, y=0,z=0$.
Now, the open sets in $\Bbb{P}^2$ that you are considering are defined by incidence relations. Namely, a line $\ell \in \Bbb{P}^2$ is in $\mathcal{U}_0$ if and only if for any nonzero $p\in \ell \subset \Bbb{A}^3$, we have $x(p)\ne 0$. That is to say, $\ell \in \mathcal{U}_0$ if and only if $\ell$ is not contained in the plane $x\ne 0$. Similarly, $\ell \in \mathcal{U}_1$ if and only if $\ell$ is not contained in the plane $y=0$, and finally, $\ell \in \mathcal{U}_2$ if and only if $\ell$ is not contained in the plane $z=0$.
Since no line in $\Bbb{A}^3$ through the origin is contained in all three coordinate planes, the $\mathcal{U}_i$ form a cover. Moreover, since a "generic" line is not contained in a given plane, the condition of being in $\mathcal{U}_i$ is open. This is intuitively why the cover is open.