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