General way to show that a set is closed I'm having trouble looking for a guideline on how to prove a set is open/closed:


*

*Show that $A = \{(x,y) \in \mathbb{R}^{2} \mid x^{2} + y + 2x = 3\}$ is closed by showing that every
limit point of A is in A.

*a) Let $S = \{x \in \mathbb{R} \mid x \not\in \mathbb{Q}\}$, is S closed?
b) Show that $S = \{(x,y) \in \mathbb{R}^{2} \mid xy > 0\}$ is open
c) Let $A,B \subset \mathbb{R}$ with $A$ open, and defined $AB = \{xy \mid (x \in A)\wedge(y \in B)\}$, is $AB$ necessarily open?
Please help me. Thanks!
 A: To show that the set given in 1) is closed, consider a point (p,q,r) with the property that for every $\epsilon>0$, there exists a point $(a(\epsilon), b(\epsilon), c(\epsilon))$ in the set so that  $d((p,q,r),(a(\epsilon), b(\epsilon), c(\epsilon)))<\epsilon$.  Then, show that $(p^{2}+q+2r-3)=0$.  That will complete the proof that the set is closed.  You can fill in the details. A similar method works for the others.
A: Let's start with Problem 2:
(a) Here $S$ is the set of irrational numbers. Does $S$ contain all of its limits points? For example, we know $0$ is a rational number and hence not in $S$. Is $0$ a limit point of $S$?
Alternatively, we could take the complement of $S$ and ask if it's open. But the complement of $S$ is just $\mathbb{Q}$. Is it true that the rationals are an open set (given the subspace topology from $\mathbb{R}$)? In other words, given a rational number, can you put an open interval around it completely contained in $\mathbb{Q}$?
(b) This is just the points $(x,y)$ in the first or third quadrant (but not on the $x$ or $y$ axis). Given such a point, can you draw an open ball around it contained entirely in its respective quadrant? (Do you see why I would pose such a question?)
