To determine if set is open , closed Define a function $f : \mathbb{R}^{2} \rightarrow  \mathbb{R} $ by 
$$f(x, y) =\begin{cases}1  & \text{if  $xy=0$} \\
 2& \text{otherwise}
\end{cases}$$
If $S  = \{(x, y): f \text{ is continous at point $( x, y)$}\}$, then set $S$ is
A. Open
B. Connected
C. Closed
D. Empty
As $f$ is not continuous on the axes and is continuous at all other points. So set $S \neq \phi$. But how do I choose between other options?
 A: Consider all the points $(x,y)$ for which $xy=0$. Clearly, this can only happen if $x=0$ or $y=0$, so $f(x,y) = 1$ along the coordinate axes, and is $2$ everywhere else. It should also be clear that $f$ is continuous everywhere except the axes; the axes are the only place where a discontinuity occurs. What kind of set is $S$ then going to be?
EDIT: In terms of how to think about the other options:


*

*An (open) subset of $\mathbb{R}^n$ is connected iff it is path connected, i.e., between any two points in the set, I can draw a line between them that stays inside the set. $S$ can't be path connected since there is no way to draw a line from $(1,-1)$ to $(1,1)$, since doing so would require me to cross an axis and then go outside the set. Admittedly, this isn't helpful if we don't know that $S$ is open to begin with, so here's the actual definition for you to puzzle through: a subset of $\mathbb{R}^n$ is connected if it is not a subset of the disjoint union of two open sets. I recommend googling for additional resources to make this more clear.

*A subset of $\mathbb{R}^2$ is open if, for every point $(x,y)$, I can find an open disk $D$ (i.e., the interior points without the circumference) containing $(x,y)$  that that disk is in $S$. In symbols, there is a disk $D$ such that $(x,y)\in D\subset S$. It may not be clear visually if this is true for $S$ right now (though it is), so we'll set it aside and move on.

*A subset of $\mathbb{R}^2$ is closed if it contains all of its limit points. A limit point of $S$ has the property that any disk about that point has a nonempty intersection with $S$. This can be a little tricky to think about if you haven't done much topology. I recommend googling for images/additional resources. Looking at our particular set $S$, if we draw a disk about a point of the axis (the origin, say), then this clearly intersects $S$. But the origin is not in $S$! Therefore, $S$ is not closed, so it must be open.


EDIT 2: fixed flub with open definition
EDIT 3: Fixed connectedness explanation.
A: Think about the function as a height function.  The coordinate axes would be one meter above the floor.  The quadrants would all be two meters above the floor.  If you are on the axes, you cannot go in all directions and stay on the graph without making a jump.  But, if you are walking on the quadrants, you can always go a small distance in any direction and stay on the graph.
So, you just need to answer if the union of all the quadrants is open, connected, or closed.
A: It's the whole plane minus the two axes.
A set is open if for any point in it, there is a neighborhood contained in the set.
This is obviously true for the set $S$. So it's open.
A: You can answer the question by considering only the 4 types of open subsets of $\mathbb R$ and then seeing if their respective inverses are open in $\mathbb R^2$. These sets must be open and contain $1,2$ or both. so you can use, e.g., $(3/2,3), (1/2, 5/4)$ and $(0,3)$, but you need to generalize these cases to $(a, \infty); 1<a<2 , (b,c) ; 1<b<2 , (d,e); 1<d<e , e>2$ , etc. 
EDIT : you can also work with sequences and consider the case where a given point $(x,y)$ is either on the axes or not. Then $f$ is continuous at $(x,y)$ iff* $$(x_n,y_n) \rightarrow (x,y) \implies f(x_n,y_n) \rightarrow f(x,y) $$


*

*This is true for all 1st countable spaces.

A: We have, for given $f(x,y)$ :
$c=\{(x,y) : f(x,y) =1\} = (\{0\} \times \mathbb{R})\bigcup(\mathbb{R} \times \{0\}) $
Clearly $f(\vec{x})$ is discontinuous $\forall \vec{x} \in c$, thus $S= \mathbb{R}^2 -c$ 
Consider the compliment of $S$, which is in fact $c$. $\forall \vec{x} \in c$, if $U$ is an open rectangle such that $\vec{x} \in U$, then $U \bigcap S \neq \emptyset$, thus $c$ cannot be open, which implies $S$ cannot be closed. We then have $S$ is neither empty nor closed. Furthermore, we may partition $S$ into any two open subsets of our choosing, hence $S$ cannot be connected. We must then have that $S$ is open.
