Set A open, but f(A) not open

**thaopanda** · September 27th 2009, 08:34 AM

Let $f : R^2 \rightarrow R$ given by $f(x,y) := x^2$ . Find a set A open in $R^2$ such that f(A) in not open in R.

I don't get how to do this...

help..?

**Moo** · September 27th 2009, 08:49 AM

Hello,

Is it really $x^2$ ?

If so, you can just take $A=(-a,a) \times \mathcal{O}$ , where $\mathcal{O}$ is any open set in $\mathbb{R}$ and $a>0$

Then $f(A)=[0,a^2)$ which is not an open set.

**thaopanda** · September 27th 2009, 08:57 AM

is that really all there is to it? I just assume all my answers are supposed to be super long, 'cause so far, most of them have been taken up half a page... hehe now I feel silly

Thank you very much!

Nicole

**arbolis** · September 27th 2009, 09:03 AM

Originally Posted by Moo

Hello,

Is it really $x^2$ ?

If so, you can just take $A=(-a,a) \times \mathcal{O}$ , where $\mathcal{O}$ is any open set in $\mathbb{R}$ and $a>0$

Then $f(A)=[0,a^2)$ which is not an open set.

Hey Moo I'm curious. Why did you chose $A=(-a,a) \times \mathcal{O}$ and not $A=(-a,a)$ ? I don't really know what your " $\times$ " product means. I understand it as "multiplying" two sets to get another one... but I don't know if it makes sense.
Wouldn't $A=(-a,a)$ be sufficient?

**Moo** · September 27th 2009, 09:07 AM

Originally Posted by thaopanda

is that really all there is to it? I just assume all my answers are supposed to be super long, 'cause so far, most of them have been taken up half a page... hehe now I feel silly

Thank you very much!

Nicole

I guess that's all. Unless I made a mistake in the definition of an open set in $\mathbb{R}^2$ ^^'

Originally Posted by arbolis

Hey Moo I'm curious. Why did you chose $A=(-a,a) \times \mathcal{O}$ and not $A=(-a,a)$ ? I don't really know what your " $\times$ " product means. I understand it as "multiplying" two sets to get another one... but I don't know if it makes sense.
Wouldn't $A=(-a,a)$ be sufficient?

Because A has to be an open set in $\mathbb{R}^{\color{red}2}$ .
$\times$ stands for the cartesian product. Because we're having two values, one in $\mathbb{R}$ , the other one in $\mathbb{R}$

**arbolis** · September 27th 2009, 09:12 AM

Originally Posted by Moo

Because A has to be an open set in $\mathbb{R}^{\color{red}2}$ .
$\times$ stands for the cartesian product. Because we're having two values, one in $\mathbb{R}$ , the other one in $\mathbb{R}$

Thanks, I missed this part!

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