Consider the Complex projective $n$-space $\mathbb{C}P^n$ and the sets $U_i=\{ [z]\in \mathbb{C}P^n : z_i\ne0\}$ and $V_i=\mathbb{C}P^n\backslash\{[0:...:0:\underset{i}{\underbrace{1}}:0:...:0]\}$. I read somewhere that $$U_{n}\cap V_{n}\cong\mathbb{C}^{n}\backslash\{0\}$$ Where $\cong$ is the homeomorphism relation. But I have no idea why this is true and how to prove it.
I know that $U_{n}\cap V_{n}=\{ [z\in\mathbb{C}P^n: z_i\ne0 \wedge (\exists k\ne i\text{ such that } z_k\ne 0) \}$, but I do not know what to do with this.
$U_n$ is an open which verifies $U_n \cong \mathbb C^n$, the isomorphism being $$ (x_0, x_1, \dots, x_n) \mapsto (\frac{x_0}{x_n}, \dots, \frac{x_{n-1}}{x_n}) $$ In particular, $(0,0, \dots, 0,1) \mapsto (0, \dots, 0)$.
Now, $V_n$ is simply the complement of $(0, \dots, 0, 1)$ in $\mathbb P^n$, so $U_n \cap V_n \cong \mathbb C^n \backslash \{(0, \dots, 0\}$.