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 $\mathbb{P}^n$?)
This has little to do with the projective space.
Let $X$ be an algebraic variety over $\mathbb C$. It can be endowed with the natural topology induced by the complex absolute value. This topology is usually called the complex topology.
Lemma Let $Z$ be a complex algebraic variety. Let $T$ be a Zariski closed subset of $Z$, Zariski nowhere dense in $Z$. Then $T$ is complex nowhere dense in $Z$.
Let $Z^0$ be an open subset of a closed subset of $X$ (all in the sense of Zariski topology). We want to show
Claim: the Zariski closure of $Z^0$ coincides with its complex closure.
In your question, $X$ is a projective space, and $Z^0$ is an affine subvariety of $X$.
Proof: Let $Z$ be the Zariski closure of $Z^0$ and let $Z^c$ be the complex closure of $Z^0$. As the Zariski topology is coarse than the complex one, we have $Z^c\subseteq Z$. As $Z\setminus Z^0$ is Zariski closed and Zariski nowhere dense in $Z$, by the previous lemma, $Z\setminus Z^0$ is complex nowhere dense in $Z$, hence $Z^0$ is complex dens in $Z$. This implies that $Z^c=Z$.
It remains to prove the lemma. It is well-known, but I don’t have a reference, so let me give a proof here. Shrinking $Z$ if necessary, we can suppose $Z$ is affine and Zariski closed in some $\mathbb C^n$. Let $I, J$ be the respective definining (radical) ideals in $\mathbb C[z_1,\dots, z_n]$ of $Z$ and $T$. By hypothesis, there exists
a complex open subset $U$ of $\mathbb C^n$ such that
$U\cap Z=U\cap T\ne\emptyset$.
We can suppose wlog that $p:=(0,..,0)\in U\cap T$. In the local ring $\mathcal O_{(\mathbb C^n)^{an},p}$ of germs of holomorphic functions, by analytic Nullstellensatz, we have $I \mathcal O_{(\mathbb C^n)^{an},p}=J\mathcal O_{(\mathbb C^n)^{an},p}$ because $U\cap Z=U\cap T$. Passing to the formal completion, we obtain
$$I\mathbb C[[z_1,\dots, z_n]]=J\mathbb C[[z_1,\dots, z_n]].$$
By the faithfull flatness of the formal completion $\mathbb C[z_1,\dots, z_n]_{\mathfrak m} \to \mathbb C[[z_1,\dots, z_n]]$ (where $\mathfrak m$ is the maximal ideal corresponding to $p$), we have
$$I\mathbb C[z_1,\dots, z_n]_{\mathfrak m}=J\mathbb C[z_1,\dots, z_n]_{\mathfrak m}.$$
This means that $T$ and $Z$ coincide in a Zariski open neighborhood of $p$. Contradiction with the hypothesis $T$ nowhere dense in $Z$.