Are Ideals and Varieties Inclusion Reversing? Let $S_1$, $S_2$ be sets or varieties (I don't think it matters, does it?). Then if $S_1 \subset S_2$, is it always the case that $I(S_2) \subset I(S_1)$ (where I is an ideal)? Also, is it always the case that if $I_1$ $I_2$ are ideals such that $I_1 \subset I_2$, then $V(I_2) \subset V(I_1)$? This seems to be the case, but I have been getting confused.

Yes. This is true and easy to show. The key is to look at the definitions: $$S \subseteq \mathbb A^n : I(S) = \{f \in k[x_1, \ldots, x_n] : f(P) = 0 \text{ for all } P \in S\}$$ and $$I \subseteq k[x_1, \ldots, x_n] : V(I) = \{P \in \mathbb A^n : f(P) = 0 \text{ for all } f \in I\}.$$

Claim: If $S_1 \subseteq S_2$ are subsets of $\mathbb A^n$, then $I(S_2) \subseteq I(S_1)$.
Proof: Let $f \in I(S_2)$, then $f(P) = 0$ for all $P \in S_2$. If $P \in S_1 \subseteq S_2$, then $f(P) = 0$ which means $f \in I(S_1)$. Conclude that $I(S_2) \subseteq I(S_1)$.
Claim: If $I_1 \subseteq I_2$ are subsets of $k[x_1, \ldots, x_n]$, then $V(I_2) \subseteq V(I_1)$.
Proof: Let $P \in V(I_2)$, then $f(P) = 0$ for all $f \in I_2$. If $f \in I_1 \subseteq I_2$, then $f(P) = 0$  which means $P \in V(I_1)$. Conclude that $V(I_2) \subseteq V(I_1)$.
