Prove that A= { (x,1/x) ; x>0 } is closed: trying to show closure(A) is subset of A

[pisrationalhahaha]

Prove that A= { (x,1/x) ; x>0 } is closed: trying to show closure(A) is subset of A

Prove that A= { (x,1/x) ; x>0 } is closed

I was trying to prove that $\displaystyle Cl(A)\subseteq A$

So if I take $\displaystyle (a,b)\in Cl(A)\Rightarrow B((a,b),r)\cap A\neq \phi$ for any r>0
Then I should prove that this couple belongs to A
Well here is the problem, I don't know how to continue

 

[pka]

I was trying to prove that $\displaystyle Cl(A)\subseteq A$
So if I take $\displaystyle (a,b)\in Cl(A)\Rightarrow B((a,b),r)\cap A\neq \phi$ for any r>0
Then I should prove that this couple belongs to A
Well here is the problem, I don't know how to continue

Lets use $\displaystyle \overline{A\;}$ as the closure of $\displaystyle A$. It actually easier to show the the completement $\displaystyle {A}^c$ is open.
If you are allowed to do that, note that if $\displaystyle P: (p,q)\notin A$ then show the distance $\displaystyle \mathscr{D}(P,A)>0$
If not allowed, please tell us.

 

[pisrationalhahaha]

Lets use $\displaystyle \overline{A\;}$ as the closure of $\displaystyle A$. It actually easier to show the the completement $\displaystyle \overline{A\;}^c$ is open.
If you are allowed to do that, note that if $\displaystyle P: (p,q)\notin A$ then show the distance $\displaystyle \mathscr{D}(P,A)>0$
If not allowed, please tell us.

No not allowed
I don't even know such information yet

 

[pka]

No not allowed. I don't even know such information yet

What a strange response? What kind of course are you taking?
In any course in which the concept of a closed set comes up then so does the idea of an open set.
Moreover, it is well known that the complement of an open set is a closed set & visa-versa.

 

[pisrationalhahaha]

What a strange response? What kind of course are you taking?
In any course in which the concept of a closed set comes up then so does the idea of an open set.
Moreover, it is well known that the complement of an open set is a closed set & visa-versa.

No I know this
But what you've said about the distance was unfamiliar

 

[pisrationalhahaha]

Okay I'll try to prove that the complement of A is open
So there would be two cases to consider
Case 1: y>1/x
Case 2: y<1/x (for any x>0)
If y>1/x:
Let $\displaystyle \forall (x,y)\in A^{c}$
B((x,y),r) for some r>0 be given
If $\displaystyle (a,b)\in B((x,y),r)$
then
$\displaystyle (x-a)^2+(y-b)^2< r^2$
Knowing that $\displaystyle (y-b)^2<(x-a)^2+(y-b)^2$.
then $\displaystyle (y-b)^2<r^2$ $\displaystyle \Leftrightarrow -r<y-b<r$
So now we have y-r<b<r+y
So if we choose r=y-1/x then we get that b>1/x, therefore (a,b) belongs to the complement of A
Is what I am doing right ?