Interior of $T=\{(x,y) \in \mathbb R^2 :1>y>x^2\}\cup \{(1,0),(-1,0)\}$ How can I show the set is open without the two addition.

Question

Problem is I cannot show that $T=\{(x,y) \in \mathbb R^2 :1>y>x^2\}$ is an open set.

Attempt:

Let take $(x_0,y_0)$ so at this center, how should I choose the radius, I can just say that $r(x)=\inf\{||x-z||\in\mathbb R:z \in T \}$ and there exists $B(x,r(x))$ st $B\subset T$ but I guess I should give concrete $r,B$ because I know T and I should talk about the example, not too general.

Moreover, how can I visualize that kind of sets? Given in the form of inequalities of two functions.

Answer 1

Hint: note that $T$ can be written as

$$ T = \{(x,y) \in \mathbb{R}^2 : 1 > y \} \cap \{(x,y) \in \mathbb{R}^2 : y - x^2 > 0 \}. $$

Now, recall that the sets $(-\infty , 1)$ and $(0, +\infty)$ are open, and the functions $$ f(x,y) = y, \\ g(x,y) = y -x^2 $$ are continuous.