Let $S^1 \subseteq \mathbb{R}^2$ be the unit circle circle $\mathbb{V}(X^2+Y^2-1)$. Let $S$ be an infinite subset of $S^1$. I want to show that $$\overline{S}^{\text{Zariski}}= \mathbb{V}(\mathbb{I}(S))= S^1$$
Is there an easy way to see this?
Clearly $S^1$ itself is Zariski closed, so it contains the Zariski-closure of $S$. However, the other inclusion seems more delicate. Any ideas how I can show this?
On
The ring $\mathbb{R}[x,y]$ has dimension 2 and consequently a prime ideal containing $x^2+y^2-1$ must be a maximal one.
Moreover, the ring $\mathbb{R}[x,y]$ is Noetherian and every ideal has a finite number of associated primes.
Geometrically this two facts mean exactly what you want: a Zarisky closed proper subset of the Circle can only be a finite union of (closed) points.
Sabino Di Trani's answer is good. Here is a solution which is perhaps more hands-on.
Suppose $f\in \Bbb R[x,y]/(x^2+y^2-1)$ vanishes on an infinite subset of $S^1$. We'll show it's actually zero, which shows that the Zariski closure of any infinite subset of $S^1$ is all of $S^1$: $I(S)=(0)$ which implies $V(I(S))=V(0)=S^1$.
We can write $f$ as $p+yq$ for polynomials $p,q\in\Bbb R[x]$ by rewriting every instance of $y^2$ as $1-x^2$. Then $f\cdot(p-yq)=p^2-y^2q^2=p^2+(x^2-1)q^2$ is a polynomial in $x$ vanishing on an infinite set, so it's zero. As $\Bbb R[x,y]/(x^2+y^2-1)$ is a domain, we must have that either $f=p+yq$ or $p-yq$ is zero, both of which give $f=0$.