1
$\begingroup$

I was hoping someone could check if my solution to this question is correct and hopefully offer some feedback if the answer is not correct.

Question: Is $U = \{(x,y)\in \mathbb{R}^2 : sin(\frac{x^3}{x^4 + y^2}) > 0 \}$ an open subset of $\mathbb{R}^2$?

I think that it is open, because the functions $f(x,y) = x^4 + y^2$ and $g(x,y) = sin(\frac{x^3}{x^4+y^2})$ are continuous and so

$U = \{(x,y)\in \mathbb{R}^2 : x^4+y^2 > 0, sin(\frac{x^3}{x^4 + y^2}) > 0 \} = (f^{-1}(0,\infty) \cap g^{-1}(0,\infty))$

Since, $f,g$ are continuous $f^{-1}(X), g^{-1}(X)$ are open for any open set X. So $U$ is open since it is a finite intersection of open sets.

  • 0
    This is perfectly correct !! – 2017-02-07
  • 1
    By the way $x^4 + y^2 > 0$ iff $(x,y) \neq (0,0)$ so you can simply write $f^{-1}(0, \infty)$ as $\mathbb R^2 \backslash \{(0,0)\}$. Your argument works **perfectly** but here there is no need to use such function and argument. – 2017-02-07
  • 1
    You might want to mutter something on why $\sin(\frac{x^3}{x^4 + y^2})$ is continuous ? Invoke theorems that you covered? – 2017-02-07
  • 1
    As already pointed out, your answer is correct. As an aside, when using trig functions in mathmode you can place a backslash before them to remove the italics, e.g., \sin. This results in $\sin$ rather than $sin$. – 2017-02-07

0 Answers 0