# Thread: Set A open, but f(A) not open

1. ## Set A open, but f(A) not open

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

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

2. Hello,

Is it really $\displaystyle x^2$ ?

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

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

3. 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

4. Originally Posted by Moo
Hello,

Is it really $\displaystyle x^2$ ?

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

Then $\displaystyle f(A)=[0,a^2)$ which is not an open set.
Hey Moo I'm curious. Why did you chose $\displaystyle A=(-a,a) \times \mathcal{O}$ and not $\displaystyle A=(-a,a)$? I don't really know what your "$\displaystyle \times$" product means. I understand it as "multiplying" two sets to get another one... but I don't know if it makes sense.
Wouldn't $\displaystyle A=(-a,a)$ be sufficient?

5. 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 $\displaystyle \mathbb{R}^2$ ^^'

Originally Posted by arbolis
Hey Moo I'm curious. Why did you chose $\displaystyle A=(-a,a) \times \mathcal{O}$ and not $\displaystyle A=(-a,a)$? I don't really know what your "$\displaystyle \times$" product means. I understand it as "multiplying" two sets to get another one... but I don't know if it makes sense.
Wouldn't $\displaystyle A=(-a,a)$ be sufficient?
Because A has to be an open set in $\displaystyle \mathbb{R}^{\color{red}2}$.
$\displaystyle \times$ stands for the cartesian product. Because we're having two values, one in $\displaystyle \mathbb{R}$, the other one in $\displaystyle \mathbb{R}$

6. Originally Posted by Moo

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