# 2 disjoint clsed subsets of (R^2)

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• Aug 27th 2011, 04:23 PM
rqeeb
2 disjoint clsed subsets of (R^2)
Q:Find two disjoint clsed subsets of (R^2) which are zero distance a part?

my answer is A & B where

A={(1/n) : n belong to z+}X{0}
B={[-1,0]XR}

Is it right

• Aug 27th 2011, 06:31 PM
hatsoff
Re: 2 disjoint clsed subsets of (R^2)
No, it is not correct. Notice that $\displaystyle \{1/n:n\in\mathbb{Z}^+\}=\mathbb{R}^+$. So $\displaystyle A=\{(r,0):r\in\mathbb{R}^+\}$, and this set is not closed since $\displaystyle (0,0)$ is a boundary point not in $\displaystyle A$.

However, consider the sets $\displaystyle C=\{(x,1/x):x\in\mathbb{R}^+\}$ and $\displaystyle D=\{(x,0):x\in\mathbb{R}\}$.
• Aug 27th 2011, 09:35 PM
rqeeb
Re: 2 disjoint clsed subsets of (R^2)
Quote:

Originally Posted by hatsoff
However, consider the sets $\displaystyle C=\{(x,1/x):x\in\mathbb{R}^+\}$ and $\displaystyle D=\{(x,0):x\in\mathbb{R}\}$.

Thank u for your respond
but $\displaystyle C=\{(x,1/x):x\in\mathbb{R}^+\}$ looks like is not closed since $\displaystyle D=\{(x,0):x\in\mathbb{R}\}$ is boundry point for the set C which not conained in D. Am I right or wrong
• Aug 28th 2011, 04:09 AM
hatsoff
Re: 2 disjoint clsed subsets of (R^2)
D is a set, not a point. So D could not possibly *be* a boundary point.

Think of it graphically: C is just the curve y=1/x in the first quadrant of the plane. D is the line y=0.