Show sets not connected

• Sep 27th 2009, 08:31 AM
thaopanda
Show sets not connected
Show that the following sets are not connected:
(a) A := $\displaystyle {(x,y) \in R^2 : x^2 + y^2 \neq 1}$;
(b) B := $\displaystyle {(x,y) \in R^2 : xy = 1}$.

For this problem, do I just need to show that A $\displaystyle \cap$ B is the empty set?
If so, I would need to prove that for any x,y $\displaystyle \in$ A cannot equal any x,y $\displaystyle \in$ B, right?
• Sep 27th 2009, 09:05 PM
redsoxfan325
Quote:

Originally Posted by thaopanda
Show that the following sets are not connected:
(a) A := $\displaystyle {(x,y) \in R^2 : x^2 + y^2 \neq 1}$;
(b) B := $\displaystyle {(x,y) \in R^2 : xy = 1}$.

For this problem, do I just need to show that A $\displaystyle \cap$ B is the empty set?
If so, I would need to prove that for any x,y $\displaystyle \in$ A cannot equal any x,y $\displaystyle \in$ B, right?

You are not asked to show that $\displaystyle A$ and $\displaystyle B$ are disconnected from each other. (It seems like that's the way you interpreted the problem.) You are asked to show that $\displaystyle A$ is disconnected. You are also asked to show that $\displaystyle B$ is disconnected.

A set X is disconnected if there exist two open sets, U and V such that:

$\displaystyle X\subset U\cup V$ and $\displaystyle cl(U)\cap V=U\cap cl(V)=\emptyset$

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For (a), take $\displaystyle U=\{(x,y):x^2+y^2<1\}$ and $\displaystyle V=\{(x,y):x^2+y^2>1\}$. It's obvious that $\displaystyle A\subset U\cup V$. It is also clear that $\displaystyle cl(U)\cap V = U\cap cl(V)=\emptyset$ (simply because $\displaystyle cl(U)=V^c$ and vice-versa).

For (b), take $\displaystyle U=\{(x,y):x>0,y>0\}$ and $\displaystyle V=\{(x,y):x<0,y<0\}$. Then proceed to show that:

1. $\displaystyle B\subset U\cup V$
2. $\displaystyle cl(U)\cap V=U\cap cl(V)=\emptyset$