# show that a set is closed

• Sep 23rd 2013, 03:09 AM
rayman
show that a set is closed
I have a set $A:=\{(x,x)|x\in[0,1]\}\subset [0,1]\times [0,1]$ 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 $A^{c}$ is an open set. What $A^{c}$ will be in this case?

$A^{c}=\{(x,x)|x\not\in[0,1]\}$?
is $A^{c}=\{(x,x)|0>x>1|y=x\}=\{(x,x)|x\in(-\infty,0)\cup(0,\infty)|y=x\}$?
• Sep 23rd 2013, 03:42 AM
Plato
Re: show that a set is closed
Quote:

Originally Posted by rayman
I have a set $A:=\{(x,x)|x\in[0,1]\}\subset [0,1]\times [0,1]$ 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 $A^{c}$ is an open set. What $A^{c}$ will be in this case?

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

Consider that $U=[0,1]\times [0,1]$ is the unit square.
If $P: (p,q)\in(U\setminus A)$ then the distance that $P$
is from the line $x-y=0$ is $\frac{|p-q|}{\sqrt2}$.

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

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

That shows that $A^c$ is open.

BTW. Your notation $A^{c}=\{(x,x)|x\not\in[0,1]\}$ is incorrect. You want $A^{c}=\{(x,y)\in[0,1]\times [0,1]:x\ne y\}$
• Sep 23rd 2013, 04:39 AM
rayman
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 $\delta=\frac{|p-q|}{\sqrt{2}}$? (from the formula I get $\sqrt{2}$ 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?
• Sep 23rd 2013, 05:54 AM
Plato
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 $\delta=\frac{|p-q|}{\sqrt{2}}$? (from the formula I get $\sqrt{2}$ in the denominator)

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

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 $\mathbb{R}^2$ is a closed set.
• Sep 23rd 2013, 06:04 AM
rayman
Re: show that a set is closed
Thanks again for helping.