What is the explanation for the choice intervals

18 Views Asked by At

In my Metric Spaces textbook, I have an example which shows that as set is open by showing the the preimage of an open set is open.

So, the set is $$U=\{(x,y)\in \mathbb{R}:y>0,x/y>7\}$$

They show that it's open in $\mathbb{R}$ the following way:

The function $f(x,y)=y$ and $g(x,y)=x-7y$ are continuous from $\mathbb{R}^2\to \mathbb{R}$, and so $$U=f^{-1}(0,\infty)\cap g^{-1}(0,\infty)$$

is open.

So, I am not completely sure why they choose such intervals for $f$ and $g$ (i.e $(0,\infty)$ and $(0,\infty)$). If somebody could explain that would be great.

1

There are 1 best solutions below

2
On BEST ANSWER

We have that $U=f^{-1}(0,\infty)\cap g^{-1}(0,\infty)$ .

Since $f$ is continuous, $f^{-1}(0,\infty)$ is open.

Since $g$ is continuous, $g^{-1}(0,\infty)$ is open.

Hence $U$ is open.