Why is the infinite union of algebraic sets not necessarily algebraic? Let $\left\{Y_i\right\}_{i \in \mathcal{I}}$, where $\mathcal{I}$ is infinite, be a family of algebraic sets of $k^n$, where $k$ is an algebraically closed field. Then $Y_i = \mathcal{Z}(T_i)$, i.e. each $Y_i$ is the zeros of some subset $T_i$ of $A=k[x_1,\cdots,x_n]$. Then what is the problem with saying that 
$\cup_{i \in \mathcal{I}} Y_i = \mathcal{Z}\left(\cap_{i \in \mathcal{I}} T_i \right)$?
 A: First, even for finite unions, you can't just intersect subsets of equations.  For example the point $0$ is the zero set of $X$ and the point $1$ is the zero set of $X-1$.  Then $\{X\} \cap \{X - 1\} = \emptyset$, so the intersection of these two sets of equations defines all of $\mathbb{C}$, not just the union of the two points.
To remedy this, use the ideal $I(Y)$ of all polynomials that vanish at the given algebraic set $Y$.  Then $\{ 0, 1 \} = \mathcal{Z}((X) \cap (X-1)) = \mathcal{Z}((X (X-1)))$.
Even so, the two sets $\cup_{i \in I} Y_i$ and $\mathcal{Z}(\cap_{i \in I} I(Y_i))$ are not necessarily the same when $I$ is infinite.  Suppose $Y_n$ is the single point $n \in \mathbb{N} \subseteq \mathbb{C}$, so that $I(Y_n) = (X - n)$.  Then $\cap_{n = 1}^{\infty} I(Y_n) = (0)$ (why?), so $\mathcal{Z}(\cap_{n=1}^{\infty} I(Y_n)) = \mathbb{C}$, not $\cup_{n=1}^{\infty} Y_n = \mathbb{N}$.
In fact, $\mathbb{N}$ is not an algebraic subset of $\mathbb{C}$, so there is no ideal $I$ such that $\mathcal{Z}(I) = \mathbb{N}$.  You can prove this by using your knowledge of what all the ideals of $\mathbb{C}[X]$ are.
