Is the domain of $f(x,y) = \dfrac{xy}{x^2 - y^2}$ an open set? And does the limit at $(0,0)$ exist? I am not all too sure if I get the whole concept of open and closed sets. 
So as an example I have this function:  $f(x,y) = \dfrac{xy}{x^2 - y^2}$.
I do understand that this is an open set because $G= \{(x,y) \in \Bbb R^2 : x \neq y \text{ and } x \neq -y\}$, meaning that every point of $G$ is an interior point. The boundaries are not included. Is this a correct way of thinking? 
Also the next question would be about the $\lim_{(x,y) \to (0,0)} f(x,y)$ not not existing. How can you show this? 
On a mini side note, the next set is closed because the boundaries are included in the set right? 
$$S = \{x \in \Bbb R^2 : x_1 ≥ 0, x_2 ≥ 0, x_1 + x_2 = 2\}. $$
I might be overthinking this but any help would be nice! 
 A: One of the classical method that you can show that this kind of limits dont exist is this, you use $x=my^k$ or $y=mx^k$ where $m$ is variable and $k$ is integer. After substitution  $x=my^k$ or $y=mx^k$ in the limit, if the limit be dependent of the variable $m$, it means, it dosent limit. There are three note one) this method is acceptable in just in the case $(0,0)$. second) after substitution $x=my^k$ or $y=mx^k$, when you can say It dosent limit that in the limit dosent exist $x$ and $y$ and three) this method is necessary condition not sufficient, I mean with this method just you can show there is no limit not exist of the limit.For your question you should choose $y=mx$ and substitute it in the limit:
$$\frac{x(mx)}{x^2-(mx)^2}=\frac{mx^2}{x^2-m^2x^2}=\frac{m}{1-m^2}$$ 
we can see, the limit is dependent to $m$ and so there is no limit in $(0,0)$.
A: Regarding "open set," your reasoning is correct. I personally would prefer this line of reasoning: A set is open if and only if its complement is closed. If $f:X \rightarrow Y$ is continuous and $C \subseteq Y$ is closed then $f^{-1}(C)$ is closed. (You can work out a short proof directly from the definition of continuous). And a set of one point is of course closed in $\mathbb R$.
I think this proof stays much closer to the definitions and is honestly shorter, provided you take the fact that your denominator function is continuous. You can, of course, use the fact that a finite union of closed sets is closed, to work with the lines $x = \pm y$ instead of the quadratic function.
A: $G$ is the union of $4$ open sets. I list one and you can list the other $3$ for yourself. Define $G_1 = \{(x,y): x+y > 0, x-y > 0 \}$. You prove $G_1$ is open since $G_1 = f^{-1}((0,\infty))\cap g^{-1}((0,\infty))$ whereas $f(x,y) = x+y, g(x,y) = x-y$ are both continuous functions on $\mathbb{R^2}$. Thus $G$ being a finite union ($4$ only) of open sets hence is open. The limit issue is not that hard to find $2$ different paths that give different limits.
A: It's undefined on the lines $y=x$ and $y=-x$, which are closed sets. So their union is also a closed set. Hence the function is defined on the complement of a closed set; i.e., it's domain is open.
If you approach the origin along the line $y=x$, you get
$\displaystyle \lim_{x\to 0} \dfrac{x^2}{0}$ which is undefined.
If you approach the origin along the line $(2x,x)$, you get
$\displaystyle \lim_{x\to 0} \dfrac{2x^2}{3x^2} = \dfrac 23$.
Since the value of the limit dependes upon which direction you approach the origin, the limit does not exist.
A: A set is open precisely if every point in the set is an interior point.  A point $x$ is an interior point if some open ball about $x$ is entirely included within the set.
That is the same as saying the set does not contain any of its boundary points.  A point $x$ is a boundary point if every open ball centered at $x$, no matter how small, intersects both the set and its complement.
A set is closed precisely if it contains all of its boundary points.
A useful thing to know is that inverse-images of open sets under continuous functions are open sets and inverse-images of closed sets under continuous functions are closed.  The set $\{ x : f(x) \in A \}$ is the inverse-image of $A$. If $A$ is open and $f$ is continuous, then the inverse-image of $A$ under $f$ is open. If $A$ is closed and $f$ is continuous, then the inverse-image is closed.  So look at the set
$$
S = \{x \in \Bbb R^2 : x_1 ≥ 0, x_2 ≥ 0, x_1 + x_2 = 2\}.
$$
The function $(x_1,x_2)\mapsto (x_1,x_2,x_1+x_2)$ is continuous, and the set $[0,\infty)\times[0,\infty)\times\{2\}$ is closed.  Therefore $S$ is closed.
