In reading these notes, I stumbled upon the statement in the title. More precisely, denoting with $\mathbb{RP}^2$ the projective plane and with $\mathbb{RP}^1$ the projective line, they show that $$\mathbb{RP}^2=\mathbb A^2\sqcup\mathbb{RP}^1,$$ and how this statement generalises to $\mathbb{RP}^n$.
What I don't quite understand is why we write this using $\mathbb A^2$ as opposed to, say, the normal plane $\mathbb R^2$. As far as I understand, what they are saying is that the set of equivalence classes (points of the projective plane) of the form $[X:Y:1]$ is invariant up to scaling, and is therefore an affine plane. But the same applies to $\mathbb R^2$, being it a linear space.
So could I equivalently write $\mathbb{RP}^2\simeq\mathbb R^2\sqcup\mathbb{RP}^1$, or am I missing something?
The set of equivalence classes of the form $[x:y:1]$ is not literally the same set as the set of pairs of the form $(x,y)$. There's a nice bijection between the two sets, but they aren't the same.
One nice property of the $\mathbb A^2$ formalism is that you won't be tempted to treat it as a vector space, whereas $\mathbb R^2$ is just daring you to add, subtract, and scale its elements.
By the way, the author of that PDF is not very careful with their notation or language in Propositions 2.2, 2.3, and 2.4, and at one point they even mistake $n+1$ for $n$, so... feel free to write your own notes in a way that makes more sense!