Is the set $\{ (x,y) \in \mathbb{R}^2 : xy=1 \}$ open or closed in $\mathbb{R}^2$ 
Determine whether the following sets are open or closed in
  $\mathbb{R}^2$ endowed with the eucledian metric
$1. \{ (x,y) \in \mathbb{R}^2 : xy=1 \}$
$2. \{ (x,y) \in \mathbb{R}^2 : xy\le1 \}$
$3. \{ (x,y) \in \mathbb{R}^2 : xy<1 \}$

The first two are not open because there is no ball that is contained in those sets and that doesn't contain an element from their complements. I'm having trouble expressing their complements as some type of union of open sets so I can prove that they are closed. The third one is a bit more interesting since the answer in my book says it's an open set. It's odd to visualize how there exists an open ball that is contained in $xy<1$ but doesn't contain any other points.
I would like to prove each of these a bit more rigorously and I have limited knowledge of topology/metric spaces. Hoping to prove it using only definition of open sets, (no sequences, continuity etc). Any hints or full answers are appreciated.
 A: Consider the complement of the second set $\{(x,y) : xy > 1\}$. We will show that $\{(x,y) : xy > 1\}$ is open.
Let $(x_0,y_0) \in \{(x,y) : xy > 1\}$. Then $x_0y_0 > 1$ so we can pick $\varepsilon > 0$ such that $x_0y_0 - \varepsilon > 1$.
Now consider the open ball $B((x_0, y_0), \delta)$ of radius $\delta > 0$ and let $(x,y) \in B((x_0, y_0), \delta)$.
We have $$|x| \le \|(x,y)\| \le \|(x,y) - (x_0,y_0)\| + \|(x_0,y_0)\| < \delta + \|(x_0, y_0)\|$$
so
\begin{align}
|xy - x_0y_0| &= |x(y-y_0)+y_0(x - x_0)| \\
&\le |x||y-y_0|+|y_0||x - x_0| \\
&\le \sqrt{|x|^2 + |y_0|^2}\sqrt{|x-x_0|^2 + |y-y_0|^2}\\
&< \sqrt{(\delta + \|(x_0, y_0)\|)^2 + |y_0|^2} \cdot \delta
\end{align}
Pick $\delta$ small enough such that the last expression is $< \varepsilon$ and we will get $$xy = (xy - x_0y_0) + x_0y_0 \ge x_0y_0 - |xy - x_0y_0| \ge x_0y_0 - \varepsilon > 1$$
We conclude that $B((x_0, y_0), \delta) \subseteq \{(x,y) : xy > 1\}$ so $\{(x,y) : xy > 1\}$ is open.
An analogous argument shows that the third set $\{(x,y) : xy < 1\}$ is also open.
Now we have 
$$\{(x,y) : xy = 1\}^c = \{(x,y) : xy > 1\} \cup \{(x,y) : xy < 1\}$$
so the complement of the first set is also open.
A: Let, $S=\{ (x,y) \in R^2 : xy<1 \}$
Suppose, $(x_0,y_0)\in S$.
If $x_0y_0<0$ then it's easy to see that $S$ contains a nbd of $(x_0,y_0)$.
Suppose, $x_0y_0\ge 0$
Let, $\epsilon<\min\{1,\frac{1-|x_0y_0|}{1+|x_0|+|y_0|}\}$.
Then, for $|x-x_0|,|y-y_0|<\epsilon$ we have,
$xy-1\le |x||y|-1<(|x_0|+\epsilon)(|y_0|+\epsilon)-1=|x_0y_0|-1+\epsilon(|x_0|+|y_0)+\epsilon^2<|x_0y_0|-1+\epsilon(|x_0|+|y_0|+1)<0$.
Hence, $S$ contains a nbd of $(x_0,y_0)$.
Hence $S$ is open.
A: 
" It's odd to visualize how there exists an open ball that is contained in xy<1 but doesn't contain any other points."

It's actually very easy.
$x_0y_0 < 1$ so $(x_0,y_0)$ is not on the curve $xy = 1$.  So there is some distance between the point $(x_0, y_0)$ and the curve $xy=1$.  Take a ball whose diameter is smaller than that distance.  All the points in that  ball aren't far enough to be that distance. so they are all not on the curve.
I guess you are thinking, if $x_0y_0 < 1$ it can get arbitrarily close to the curve and it can get close enough to the curve that nothing can get closer.  Well, in that case get out a magnifying glass and zoom in so that that distance can appear large.  Take a radius half that distance.  Space (at least in the Euclidean $\mathbb R^n$ universe) is continous and there is no such thing as "so close nothing can be closer".  We just have to zoom in with a magnifying glass and a distance of $\frac {1}{10^{100}}$ might as well be a million miles.
====
Okay to do these formally.
1) $A = \{(x,y)| xy =1\}$
$A^c = \{(x,y)|xy < 1\} \cup \{(x,y)|xy > 1\}$.
that's the union of two sets call them $M = \{(x,y)|xy < 1\}$ and $N = \{(x,y)|xy > 1\}$.  If we can prove they are both open we are done.
Any my instinct as described above says that they are.
2) $B =  \{(x,y)| xy \le 1\}$.
$B^c = \{(x,y)|xy > 1\} = N$.
So if we prove $N$ is open we are done.
3) $M = \{(x,y)| xy < 1\}$.
That's just $M$.  If we prove it is open we are done.
So...
So we just have to prove $M$ and $N$ are open.
A: Lets start with $xy<1$ is open.
for any $x_0y_0$ in the set. $x_0y_0 < 1$
or there is a $\epsilon >0$ such that $\epsilon = 1 - x_0y_0$
Let $\delta  = \min (\frac {\epsilon}{4x_0},\frac {\epsilon}{4y_0})$
Then all $x,y$  in the ball of radius $\delta$
Next $xy \le 1$
The compliment of this set is $\{x,y|xy > 1\}$
By an identical argument, show that this set is open.
and finally $xy = 1$
The compliment is $\{x,y|xy<1\} \cup \{x,y|xy>1\}$ which is the union of two open sets.
