Prove that function $g(x×y)=x+y^2$ is a open map

Question

I have this definition: A map $f$ is open if the image of any open set $\mathcal{O}$ is open. I.e., if $f(\mathcal{O})$ is open. But I don't know how to use it, could you please guide me

I thank you very much

Answer 1

For $g:\Bbb R^2\ni (x\times y)\longmapsto x+y^2\in \Bbb R$ we have $$g\big((a,b)\times (c,d)\big)=\begin{cases}(a,b)+\big(\min\{|c|,|d|\}^2,\max\{|c|,|d|\}^2\big) & \text{ if } 0\not \in(c,d)\\ (a,b)+\big[0,\max\{|c|,|d|\}^2\big)& \text{ if } 0 \in(c,d).\end{cases}$$

Now, the sum of an open set with any other subsets of $\Bbb R$ is again an open set: for $U\subseteq_\text{open}\Bbb R$ and $A\subseteq \Bbb R$ we have $U+A=\{u+a:u\in U,a\in A\}=\bigcup_{a\in A}\big(a+U\big)=$ union of open seubsets of $\Bbb R$. Note that any translation is a homeomorphism of $\Bbb R$ so sends open set to an open set.

Also, each open subsets of $\Bbb R^2$ can be written as a countable union of open sets of the form $(a,b)\times (c,d)$.

Thanks to Brian Moehring for finding out a mistake.

Answer 2

You just need to show that if $O$ is open (in $\mathbb{R}^2$) then $g(O)$ is open in $\mathbb{R}$.

So, for each $(x,y) \in O$ we need to find some $\epsilon >0$ such that $(g(x,y)-\epsilon,g(x,y)+\epsilon) \subset g(O)$.

Since $O$ is open, there is some $\epsilon>0$ such that $W=(x-\epsilon,x+\epsilon) \times \{y\} \subset O$. We see that $g(W) = (g(x,y)-\epsilon,g(x,y)+\epsilon)$ which is open, and so $g(O)$ is open.