Open and Closed Sets in $\mathbb E^2$ I'm having problems trying to prove open and closed sets so here goes: 


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*Prove that $\{(x, y) \in \mathbb E^2 : x > y\}$ is an open set

*Prove that $\{(x,y) \in \mathbb E^2 : xy = 1 \}$ is a closed set
Proof 1: 
Let's call the set $D = \{(x,y) \colon x>y\}$. Now to prove that this is an open set we will choose any point in $D$ that there is an open ball with radius $|y-x| > 0$. The radius is $|y-x|$ because the set does not include points in which $(x, x)$ or where $x=y$. Thus, a point $(x_0, y_0)$ in $D$ will have an open ball for any point in $D$. Thus the set is open. 
Proof 2: 
Let's call the set $xy=1$ as $D$. To prove that this is a closed set, we will prove that the complement of $D$ is open. The complement of $D$ will be denoted as $B$. Now to denote distances in relative terms, x is the inverse of $y$, or $x = 1/y$. Thus any point in $B$ denoted $(x_b, y_b)$ can have an open ball around it with radius $|x - 1/x| > 0$. Thus any point $(x_b, x_b)$ in $B$ will have an open ball as such B is open which in turn proves that $D$ is closed. 


*

*I'm guessing that any function is a closed set, given that they do not have discontinuity. 

 A: Let $B(p,r)$ denote the open ball with centre $p$ and radius $r$.
The condition
$$\forall p\in D \text{ there exists } B(p,r), r>0 \text{ such that } B(p,r) \subset D$$
is by definition the meaning of "$D$ is open", so by proving the above condition they prove that $D$ is open. 
Intuition
If the above condition is just mathematical blah-blah for you, try to picture it this way:
Draw your set, then choose a random point and draw a ball that stays inside your set. First you choose the centre of the ball and then the radius.
Being open
If your set is open, you will know because you can always find a radius. If your centre is close to the end of the set, you can still choose a smaller radius. The only condition is that the radius is not zero
Being "not open"
Your set fails to be open when for all radii the ball does not stay in your set. For example, the graph of a function ($G=\{(x,y) \text{ s.t. } (x,y)=(x,f(x))\}$, in your case since $y=\frac{1}{x}$, $f$ would be $f(x)=\frac{1}{x}$) fails to be open because when you centre a ball in $(x,f(x))$, this ball "covers more" than your graph. 
Mathematically, you can see this by noting that the ball contains for instance $(x,f(x)+\frac{r}{2})$. The only way to have such a point in $G$ is if $(x,f(x)+\frac{r}{2})=(x,f(x))$, but that only happens for $r=0$, the only forbidden value for $r$
Remark
Being "not open" is not the same as being closed. There are sets which are both open and closed (e.g. the empty set), and there are sets which are neither open nor closed (something of the form $(a,b]$ in $\mathbb{R}$). The trick is to answer always the two questions: 


*

*Is it open? 

*Is it closed?

