In Smith's An Invitation to Algebraic Geometry, following the definition of the projective closure of an affine variety, it was remarked that "the closure may be computed in either the Zariski topology on $\mathbb{P}^n$, or in the Euclidean topology on $\mathbb{P}^n$; the result is the same, and both correspond to our intuitive idea of a closure.'' (Varieties in this book are taken to be over $\mathbb{C}$.)
I was wondering why this is true, since the Zariski topology is coarser than the Euclidean topology. Can someone sketch a proof of this fact? Smith offers no explanation for this.
Partly I think I'm confused about the notion of "Euclidean topology" on projective space. There are at least two topologies that could be considered the "Euclidean topology", and I hope they're the same:
The standard affine cover of $\mathbb{P}^n$ gives rise to charts where the open sets are affine $n$-space $\mathbb{C}^n$. If $\mathbb{C}^n$ is equipped with the Euclidean topology, this makes $\mathbb{P}^n$ a complex manifold.
There is a surjective map from $\pi: \mathbb{C}^{n+1} \setminus \{0\} \to \mathbb{P}^n$ that identifies lines given by $\pi(z_0,\ldots,z_n) = [z_0:\cdots:z_n]$. If $\mathbb{C}^{n+1}$ is given the Euclidean topology, then $\mathbb{P}^n$ can be given the quotient topology. This should be the same as declaring that a set $V$ in $\mathbb{P}^n$ is closed iff its affine cone $\pi^{-1}(V) \cup \{0\}$ is closed in $\mathbb{C}^{n+1}$ with the Euclidean topology. (A related question: If $\mathbb{C}^{n+1}$ is given the Zariski topology instead, is the quotient topology the Zariski topology on $\mathbb{P}^n$?)