I have problem with proving using double inclusion that it's an image of function where we have open interval for instance:
Find image $f[A]$ where $A=(0,2) \times (1,3)$ of $f(x,y)=|x-y|$.
My try:
we have $0<x<2$ and $1<y<3$ so $-3<x-y<1$ hence $0\le|x-y|<3$ so I proved that $f[A] \subseteq [0,3)$ and I have problem with the other inclusion since I can't take the edge value of $x$ and $y$.
It helps to draw a picture.
From the picture, we expect the larger values of $f$ to be at the upper left and lower right. Since $f(x,y) \ge 0$ and $({3 \over 2},{3 \over 2}) \in A $, we expect that $f(A) = [0,3)$.
It is easy to show that $y \le 3$ if $y \in f(A)$, however $3 \notin f(A)$. From the picture, we can see that $(x, 3-x) \in f(A)$ for $x \in [{3 \over 2},0) $ and since $f(x,3-x) = |3-2x|$, we see that $[0,3) \subset f(A)$.
Here is another approach:
First consider the range of values of $x-y$: We have $0<x<2$ and $1<y<3$ so since $-3 < -y < -1$, we have $-3 < x-y < 1$. In particular, the range of values that $x-y$ takes in $A$ is exactly $(-3,1)$. Now look at the range of values that the function $t \mapsto |t|$ can take on $(-3,1)$. This gives $[0,3)$.