Let $k=\mathbb{C}$ and decide for each of the following subsets $S \subset \mathbb{A}_{\mathbb{C}}^2$ whether they are closed, open or dense in the Zariski topology:
(a) $S=\{(t, s t) \mid s, t \in \mathbb{C}\}$
(b) $S=\{(s, t) \mid s, t \in \mathbb{Q}\}$
(c) $S=\left\{\left(t, 2^t\right) \mid t \in \mathbb{Z}\right\}$
So the only way I know how to determine whether something is open or closed is whether the set is a zero set of some ideal or set of polynomials. But looking at these three sets, I don't think it is easy to show whether they are the zero sets or complement of some zero sets. How can I proceed? I have just studied mapping between algebraic sets, so maybe it is possible to find some isomorphisms between different sets, but I have read somewhere that they don't preserve topological properties.
Closed: can you find a polynomial (or finite set of polynomials) with complex coefficients that has the set as its zero set?
Dense: when a polynomial vanishes on your set, must the polynomial be zero everywhere? For example, if a polynomial in $\mathbf C[x,y]$ vanishes on a small Euclidean open disc in $\mathbf C^2$ then it vanishes on all of $\mathbf C^2$, which implies the Zariski closure of an open disc in $\mathbf C^2$ is all of $\mathbf C^2$. Thus open discs in $\mathbf C^2$ are Zariski dense in $\mathbf C^2$.
Note a Zariski-closed set is also Euclidean-closed, so the Zariski closure of a set contains the Euclidean closure, or put differently the Zariski closure is at least as big as the Euclidean closure. That's consistent with the example indicated above: the Euclidean closure of an open disc in $\mathbf C^2$ is the closed disc with the same center and radius, while its Zariski closure is all of $\mathbf C^2$.