Show sets not connected

**thaopanda** · September 27th 2009, 08:31 AM

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

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

**redsoxfan325** · September 27th 2009, 09:05 PM

Originally Posted by thaopanda

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

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

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

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

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

--------------

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

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

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

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