Elipsoide in the three-dimensional space, with the graph of the ellipsoid the set is open, but how do I prove it?
Asked
Active
Viewed 59 times
0
-
1Let $f:X \to Y$ be continuous. If $U$ is an open subset of $Y$, then $f^{-1}(U)$ is an open subset of $X$. Does this help? – ThePortakal Oct 16 '17 at 19:25
-
But what is $f^{-1}((-\infty, 0))$ and how does one show that is open? – fleablood Oct 16 '17 at 19:36
-
@fleablood. The topological definition of continuous is inverse images of open sets are open. – William Elliot Oct 16 '17 at 19:54
-
Brainfart. Idiotically I was have trouble getting that $f^{-1}((-\infty,0)) = S$. Don't know why I was haven't the glitch.... – fleablood Oct 16 '17 at 20:21
1 Answers
1
If $f(x,y,z)$ is a polynomial of three variables, then $f$ is continuous. Since $f$ is continuous, what can we say about $f^{-1}(U)$ when $U\subset\Bbb R$ is open?
Dave
- 13,568