# Union of sets in R^n

• Jan 11th 2012, 04:15 PM
I-Think
Union of sets in R^n
The question states:
For $\displaystyle A,B\subset{{R^n}}$, prove that $\displaystyle \overline{A\cup{B}}=\overline{A}\cup{\overline{B}}$

Shouldn't it be $\displaystyle \overline{A}\cap{\overline{B}}$ by De Morgan's Laws?
Also, I thought of this counter example in $\displaystyle {R{^2}}$

Let $\displaystyle A,B\subset{{R{^2}}}$
$\displaystyle x\in{A}$ if $\displaystyle x_1$ even, $\displaystyle x_2=3k, k\in{\mathbb{Z}}$
$\displaystyle x\in{B}$ if $\displaystyle x_1$ even, $\displaystyle x_2=4k, k\in{\mathbb{Z}}$

Now consider $\displaystyle y=(4,8)$. $\displaystyle y\in{\overline{A}}$, thus $\displaystyle y\in{\overline{A}\cup{\overline{B}}}$.
But $\displaystyle y$ not in $\displaystyle \overline{A\cup{B}}$
Thus $\displaystyle \overline{A\cup{B}}\neq{\overline{A}\cup{\overline {B}}}$

Where is my mistake or is this the question's mistake?
• Jan 11th 2012, 04:48 PM
Plato
Re: Union of sets in R^n
Quote:

Originally Posted by I-Think
The question states:
For $\displaystyle A,B\subset{{R^n}}$, prove that $\displaystyle \overline{A\cup{B}}=\overline{A}\cup{\overline{B}}$

Shouldn't it be $\displaystyle \overline{A}\cap{\overline{B}}$ by De Morgan's Laws?

I think you are miss-reading the question.
$\displaystyle \overline{A}$ is not the complement of $\displaystyle A$. It is the closure of $\displaystyle A.$

The theorem is: The closure of a union is the union of closures.