How do we decide if a set is open. How to use continuity property? $S=\{(x,y,x^3y^3)\in\mathbb R^3 | x^2+y^2<1\}$ open?

Question

Is $S=\{(x,y,x^3y^3)\in\mathbb R^3 | x^2+y^2<1\}$ open?

I want to use If $C$ is closed(open) then $f^{-1}(C)$ is closed(open) too.

Lets convert the given set with this idea:

$S=\{(x,y,z)\in \mathbb R^3: x^2+y^2<1 \;\wedge\; z=x^3y^3\}=\{(x,y,z)\in \mathbb R^3: x^2+y^2<1\}\cap\{(x,y,z)\in \mathbb R^3: z=y^3x^3\}$

since we can find $y^3,x^3$ for every $z$ with $z=y^3x^3$ so actually $\{(x,y,z)\in \mathbb R^3: z=y^3x^3\}$ is the whole $\mathbb R^3$ (why?)

So $S$ is actually $S=\{(x,y,z)\in \mathbb R^3: x^2+y^2<1\}$

and $S=f^{-1}((-\infty,1))$ with continuous $f(x,y)=x^2+y^2$ we got $S$ is open too?

Even If my answer is correct how can I do this everytime? How can I check if a set is open or not written in this kind of functions form?

Answer 1

Taking $x=1,y=0$ we see that $(1,0,0) \in S$. Since $x<1$ for all $(x,y,z) \in S$ it follows that $(1+\epsilon,0,0) \notin S$ for any $\epsilon >0$. Hence $S$ is not open.

Answer 2

$S$ is not open in $\mathbb R^3$. In fact, let $D = \{(x,y)\in \mathbb R^2: x^2+y^2 < 1 \}$ which is open in $\mathbb R^2$. Then $S$ is the graph of the continuous function $\phi : D \to \mathbb R, \phi(x,y) = x^3y^3$. If $S$ were open in $\mathbb R^3$, then $S \cap (\{( 0,0) \} \times \mathbb R) = \{( 0,0,0) \}$ would be open in $\{( 0,0) \} \times \mathbb R$, and this is not true.

By the way, the map $i(\phi) : D \to S, i(\phi)(x,y) = (x,y,\phi(x,y)), $ is homeomorphism; its inverse is $p: S \to D, p(x,y,z) = (x,y)$.