Definition*: Let $X$ be a set of points in $\mathbb{A}^n(k)$ ($n-$dimensional affine space over the field $k$).
The ideal of $X$ is given by $$I(X) = \{ P \in k[x_1, \dots, x_n]: P(a_1,\dots,a_n)=0\text{ for all }(a_1, \dots, a_n) \in X \} $$
Conjecture: Let $A(X)$ denote the smallest algebraic set containing $X \subset \mathbb{A}^n(k)$. Then $$I(X) = I(A(X)) $$
The result is trivial when $X$ is algebraic, because then obviously $X= A(X)$.
Clearly, because $X \subset A(X)$, we have to have that $I(A(X)) \subset I(X)$. So proving the conjecture reduces to showing that $$I(X) \subset I(A(X)). $$
The simplest examples suggesting that the conjecture is correct are in $\mathbb{A}^1(\mathbb{C})$, because then any infinite set which is not equal to $\mathbb{C}$ is not algebraic, and its ideal then is clearly $\{ 0 \}$ by the fundamental theorem of algebra, and of course $I(\mathbb{C})=\{0\}$.
Then I thought of an arc on the unit circle in $\mathbb{A}^2(\mathbb{R})$ and it seemed clear to me that the ideal of any such arc should be equal to the ideal of the entire unit circle itself. But I don't know how to prove it.
I don't know of a clear analog of the fundamental theorem of algebra for arbitrary affine spaces over arbitrary fields. I don't even know how I could apply Bezout's theorem in special cases since that is stated for algebraic curves which are special cases of algebraic sets, whereas I need to show a result involving non-algebraic sets.
I tried to prove the result by contradiction as follows:
Let $X$ be non-algebraic. Let $p$ be such that $p(x)=0$ for all $x \in X$. Assume by means of contradiction that $p(x)\not=0$ for some $x \in A(X) \setminus X$.
It might be an issue of me not knowing how to describe $A(X)$ using set-builder notation which is preventing me from progressing further. However, I feel like this is an "intuitively clear" result, so it is bothering me that I can neither prove nor disprove it right now.
*(4.2.2. of p. 202 of Garrity et al's, Algebraic Geometry: A Problem-Solving Approach)