Why are these sets closed? Consider the sets in $\mathbb R^2$ defined by 
$$A=\left\{(x,\frac1x)|x>0\right\}\\ B=\left\{\left(x,−\frac1x\right)|x<0\right\}$$
Lets assume $(x,y)$ is a limit point of $A$ but not in $A$. Then it has to be in $B$, right? Then $x$ has to be negative. Is that a contradiction to $(x, y)$ is a limit point of $A$? Why or why not?
 A: Consider $f: \mathbb R^2 \to \mathbb R$ given by $f(x,y)=|x|y-1$.
Then $f$ is continuous and so its zero set $Z$ is closed.
Now, $A= Z \cap Q_1$ and $B=Z \cap Q_2$, where $Q_i$ is the closed $i$-th quadrant.
Therefore, both $A$ and $B$ are closed, being the intersection of two closed sets.
A: 
Lets assume $(x,y)$ is a limit point of $A$ but not in $A$. Then it has to be in $B$, right?

$A$ is a closed set, so there is no point $(x,y)$ which is a limit point of $A$ but is not in $A$.

Technically, of course, that means that the statement 
"If $(x,y)$ is a limit point of $A$, but not in $A$, then $(x,y)$ is in $B$" 
is a true statement, but it also means that the statement 
"If $(x,y)$ is a limit point of $A$, but not in $A$, then $(x,y)$ is not $B$"
Along with that, other true statements include 
"If $(x,y)$ is a limit point of $A$, but not in $A$, then $1+1=3$"
A: Hint if $$\lim_{n\to \infty}x_n =a ~~~and ~~~\lim_{n\to \infty}\frac{1}{x_n} =b$$ both exist then $a\neq 0$ and $b\neq 0.$ and $a=\frac{1}{b}$
