show that a set is closed

I have a set $\displaystyle A:=\{(x,x)|x\in[0,1]\}\subset [0,1]\times [0,1]$ i.e the diagonal of the unit square. How to show that A is closed??

By definition a set is closed if it's complement is open, so I need to show that $\displaystyle A^{c}$ is an open set. What $\displaystyle A^{c}$ will be in this case?

$\displaystyle A^{c}=\{(x,x)|x\not\in[0,1]\}$?

is $\displaystyle A^{c}=\{(x,x)|0>x>1|y=x\}=\{(x,x)|x\in(-\infty,0)\cup(0,\infty)|y=x\}$?

Re: show that a set is closed

Quote:

Originally Posted by

**rayman** I have a set $\displaystyle A:=\{(x,x)|x\in[0,1]\}\subset [0,1]\times [0,1]$ i.e the diagonal of the unit square. How to show that A is closed??

By definition a set is closed if it's complement is open, so I need to show that $\displaystyle A^{c}$ is an open set. What $\displaystyle A^{c}$ will be in this case?

$\displaystyle A^{c}=\{(x,x)|x\not\in[0,1]\}$?

Consider that $\displaystyle U=[0,1]\times [0,1]$ is the unit square.

If $\displaystyle P: (p,q)\in(U\setminus A)$ then the distance that $\displaystyle P$

is from the line $\displaystyle x-y=0$ is $\displaystyle \frac{|p-q|}{\sqrt2}$.

So let $\displaystyle \delta=\frac{|p-q|}{2\sqrt2}$.

The ball $\displaystyle \mathcal{B}(P;\delta)$ contains $\displaystyle P$ and no point of the line $\displaystyle x-y=0$.

That shows that $\displaystyle A^c$ is open.

BTW. Your notation $\displaystyle A^{c}=\{(x,x)|x\not\in[0,1]\}$ is incorrect. You want $\displaystyle A^{c}=\{(x,y)\in[0,1]\times [0,1]:x\ne y\}$

Re: show that a set is closed

thank you for the help. I have a question.

Just to clarify

is the radius of the open ball $\displaystyle \delta=\frac{|p-q|}{\sqrt{2}}$? (from the formula I get $\displaystyle \sqrt{2}$ in the denominator)

What about the end points of A? Don't we have to find open balls in the neighborhood of these two with the proper radius?

Re: show that a set is closed

Quote:

Originally Posted by

**rayman** thank you for the help. I have a question.

Just to clarify

is the radius of the open ball $\displaystyle \delta=\frac{|p-q|}{\sqrt{2}}$? (from the formula I get $\displaystyle \sqrt{2}$ in the denominator)

I used $\displaystyle \delta=\frac{|p-q|}{2\sqrt{2}}$ because that makes the closure of the ball disjoint from $\displaystyle A$.

That is merely a matter of choice.

Quote:

Originally Posted by

**rayman** What about the end points of A? Don't we have to find open balls in the neighborhood of these two with the proper radius?

I don't know what you mean by "What about the end points of A?"

The proof I gave is in fact the way we show that any *line* in $\displaystyle \mathbb{R}^2$ is a closed set.

Re: show that a set is closed

Thanks again for helping.