Proof that subset is open in $\Bbb R^2$ with coefficients in $\Bbb R^3$

Question

Let $A$ be an open subset of $\mathbb{R}^3$. Let $B$ be a subset of $\mathbb{R}^2$ defined by the points $(x,y)$ for which $(x,y,x+y)\in A$. I am struggling a bit how to show that $B$ is open in $\mathbb{R}^2$. I know that if $B$ is open in $A$ iff $B$ is open in $\mathbb{R}^3$ if they were the same dimension but now I am struggling how to combine the 2D situation and the 3D. My main problem visualising is that I see that $x$ and $y$ come from an open set but how do I know $x+y$ does not pass the boundary of that set? And still I don't see how to combine the dimensions.

Answer 1

I think you meant to say $B=\{(x,y): (x,y,x+y) \in A\}$.

$f(x,y)=(x,y,x+y)$ defines a continuous function from $\mathbb R^{2}$ to $\mathbb R^{3}$ and so the inverse image of $A$ under this function is open.

Answer 2

So $A \subseteq \Bbb R^3$ is given.

You define $B = \{(x,y) \in \Bbb R^2: (x,y,x+y) \in A\}= f^{-1}[A]$ where

$$f: \Bbb R^2 \to \Bbb R^3, f(x,y):= (x,y, x+y)$$ It's quite easy to see that $f$ is continuous (it's even a linear map) and so $A$ open implies that $B$ is open by continuity. But taking $A = f[\Bbb R^2]$ we have that $A$ need not be open even when $B$ is.