How to quickly recognize closed sets and open sets in $\mathbb{R}^2$ Let $S$ be any set in $\mathbb{R}^2$.
Given a definition $S$, are there tell-tale signs that can be used to conclude that $S$ is closed? For example, in $\mathbb{R}$, I know that any set defined in terms of weak inequalities will be a closed set. Is there a similar rule in $\mathbb{R}^2$.
I am wondering about this because a lot of the solutions to a lot of the exercises that asked to determine whether $S$ is closed seem to imply that it is obvious to see. 
For example:

Determine if $$S=\left\{ \left( x,y \right): 1\le x^2y\le 2 \right\}$$
  is closed.

The solution simply says:

The set $S$ is closed and this will be clear from sketching it.

So, here's the sketch:

But it is not obvious to me based on the definition and sketch of $S$, why $S$ must be closed. What am I missing?
For reference, I know the following about closed sets:


*

*A set,$S$, is closed if its complement is open.

*A set, $S$,  is closed if for any convergent sequence in $S$, the limit of that sequence is also in $S$.


I also have a similar query in the case of open sets in $\mathbb{R}^2$ i.e. is there a way to quickly recognize that a subset of $\mathbb{R}^2$ is open given its definition and or sketch?
 A: The $\le$-sign here is a hint that the set might be closed. In this case, we can write the set as $S = f^{-1}[[1,2]]$, where $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x,y) = x^2y$. One notes that $f$ is continuous (because multiplication is, essentially). So we have $S$ as the inverse image of a closed set $[1,2] \subset \mathbb{R}$ under a continuous function, and so closed (an alternative formulation of the usual "inverse image of open is open" definition).
Of course we have to be careful here, but usually, if the conditions are continuous in the coordinates and involve only $\le$-signs, chances are that the set is closed.
To see another example: $\{(x,y) : x \ge 0 \} = (\pi_1)^{-1}[[0, \infty)]$, where $\pi_1(x,y) = x$ is just the projection onto the first coordinate, which is continuous.
A: It is the $\le$ sign which is the critical clue.
Since your set includes the boundary cases, it's closed.
A: We have:
$$S^c = U_1 \cup U_2\\
U_1 = \{(x,y) : x^2 y < 1\}\\
U_2 = \{(x,y) : x^2 y > 2\}$$
These two sets are open and then $S^c$ is open.
Consider the point $(1,1) \in S$. Then for any $\epsilon>0$, $(1-\frac{\epsilon}{2},1- \frac{\epsilon}{2}) \in B_\epsilon(1,1)$ but $(1-\frac{\epsilon}{2})^2(1 - \frac{\epsilon}{2}) < 1$ and so it is not an interior point of $S$.
