Vakil Rmk 5.2.2: Identify the following space with $\operatorname{Spec}(\mathbb C[x])$: $$X:=\operatorname{Spec}(\mathbb C[x,y_1, y_2,\ldots]/(y_1^2, y_2^2,\ldots, (x-1)y_1, (x-2)y_2,\ldots))$$ and then show that the nonreduced points of $X$ are precisely the closed points corresponding to positive integers and that the complement of this set is not Zariski open.
Attempt:
First let's consider $$X':=\operatorname{Spec}(\mathbb C[x,y_1, y_2,\ldots]/(y_1^2, y_2^2, \ldots))$$
Because we are modding out $\operatorname{Spec}(\mathbb C[x,y_1, y_2, \ldots])$ by the ideals $(y_i^2)$, we get $\mathbb A^1$ with a fuzz that is the intersection of the fuzz of each one of the $(y_i^2)$. This intersection is parallelepiped around the $x$-axis. So we see that we get a map $X' \to \operatorname{Spec}(\mathbb C[x])$, which implies that we also have $X \to \operatorname{Spec}(\mathbb C[x])$.
Now we look at what happens when we further mod out by $((x-i)y_i)$. These ideals are a cross of hyperplanes $x = i$ with $y_i = 0$. They surely add some fuzz at the integer points, but I can't really visualize it. Algebraically, they are adding 1-order differential information, since now we can consider a Taylor expansion around $(i, 0_1, 0_2,\ldots y_i, 0_{i+1}, \ldots)$ up to terms $a_0 + a_xx + a_yy$.