$Z=\{(x,y)\in\mathbb{R}^2:x>0,y=\frac{1}{x}\}$ closed in $\mathbb{R}^2$? 
The set  $Z=\{(x,y)\in\mathbb{R}^2:x>0,y=\frac{1}{x}\}$, Z is closed in $\mathbb{R}^2$.

How can I show that Z is closed in $\mathbb{R}^2$?
Question:
I know that all the positive $(x,y)$ belong to $Z$, the first quadrant. In the infinity y=0, but x will never equal  point zero so $Z$ is not the first quadrant.  How is that closed under $\mathbb{R}^2$?
 A: HINT:It is the intersection of the closed set 
$$\{ (x,y) \ | \ x y -1 = 0\}$$ with the closed half plane $\{ (x,y) \ | x \ge 0\}$. 
$\bf{Added:}$ It is maybe puzzling that the graph of a continuous function $\phi$ defined on $(0,\infty)$ ( not $[0, \infty)$ !) is closed. The important thing is  $\phi(x) \to \infty $ as $x \searrow 0$. Then the function $\psi(x) = \frac{1}{\phi(x)}$ has limit $0$ as $x \searrow 0$. Now we can define $\psi(0) = 0$ and $\psi$ is continuous on $[0, \infty)$. Moreover, the graph of $\phi$ equals $\{ (x,y)\ | \ x \ge 0 \textrm{ and } \psi(x) \cdot y -1 = 0 \}$. 
A: We are going to show $Z$ is closed under taking limits. Let $((x_n,y_n))$ be a sequence in $Z$ that converges to some $(x_0, y_0) \in \mathbb{R^2}$. Then we have $y_n=\frac{1}{x_n}$ and since
$$|x_n - x_0| \leq ||(x_n,y_n)-(x_0,y_0)||$$
we see that $x_n \rightarrow x_0$. By the same argument $y_n \rightarrow y_0$. Now $x_0$ cannot be zero since else $y_n=\frac{1}{x_n}$ would get arbitrarily large. By the same argument $y_0$ is not zero. Now it follows 
$$\lim y_n=\lim \frac{1}{x_n}=\frac{1}{\lim x_n}=\frac{1}{x_0}.$$
So $(x_0,y_0)=(x_0,\frac{1}{x_0}) \in Z$.
A: In the comments you say:  "An open set is the union of open sets which have as subsets open balls. The closed set is the negation by Morgan Laws. "
So... consider $a \not \in Z$.  If you can show that $a \in$ some open ball that is entirely within $Z^c$, then every point in $Z^c$ is in such an open ball.  The $Z^c$ is open.  And $Z$ therefore is closed.
So let $a \not \in Z \implies a \in Z^c \implies a = (x,y)$ where either $x\le 0$ or $x > 0$ but $y \ne \frac 1x$.
Case 1:  $x < 0$ then let $r = |x| = -x$.  Then open ball of $B_r(a)=\{b|d(a,b) < r= |x|\}$ will consist entirely of points $b = (x_b, y_b)$ where $(x_b - x)^2 + (y_b - y) < r^2 = x^2$.  From that you can show $x_b < 0$ so $b \in Z^c$.
Case 2a: $x=0$.  $y< 0$.  Let $r= |y|$ and do the same argument as above.
Case 2b: $a = (0,0)$.  Let $r < 1$ then $B_1((0,0)) = \{b=(x_b,y_b)|x_b^2 + y_b^2 < 1\}$.  From that you can show that $(v,w) \in B_1((0,0))$ with $v > 0$ then $w = \frac 1v > 1$ is impossible.  So $B_1((0,0)) \subset Z^c$.
Case 2c: $x = 0$. $y > 0$.  Let $r = \min(y, \frac 1y) \le 1$. then $B_r(a) = \{b=(x_b, y_b)|x_b^2 + (y_b - y)^2 < \min(y, \frac 1y)^2\}$.  You can prove $B_r(a) \subset Z^c\$.
Case 3a: $x >0$ and $y \ne \frac 1x$. Let $r =\min (|y-\frac 1x|, |x - \frac 1y|)$.  You can prove $B_r(a) \subset Z^c\$.
Thus $Z^c$ is open.  So $Z$ is closed.
A: Yet another version: Note that $Z= \{ (x,y) | x+y \ge 0 \} \cap \{  (x,y) |  xy = 1 \}$.
