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Given the set $C = \left\{(x_1, x_2, x_3) \in \mathbb{R}^3: \cos(x_1x_2x_3) < −1/2\right\}$.

Determine if $C$ is an open or closed set and also determine if the set is bounded or unbounded.

Also determine if the set is sequentially compact or not.

This set is making me struggle to work out the 3 questions asked so any help will be appreciated.

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    For boundedness, note $\cos(x_{1}x_{2}x_{3})<-\frac{1}{2}$ if $\frac{2\pi}{3}+n\pi$n\in \mathbb{Z}$. Taking $x_{1}=x_{2}=1$, then is there an upper bound for the value of $x_{3}$? – 2017-02-20
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    Im pretty sure there isnt an upper bound. Is that correct?. @yung_Pabs – 2017-02-20
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    ???? @yung_Pabs – 2017-02-21

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For openness, hint:

We can rewrite $$\{ (x_{1}, x_{2}, x_{3}) \in \Bbb R^{3} : \cos(x_{1}x_{2}x_{3}) < -\frac{1}{2} \}$$ as $$h^{-1}( (-\infty, - \frac{1}{2}) ),$$ where $h(x_{1}, x_{2}, x_{3}) := f(g(x_{1}, x_{2}, x_{3}))$, and $$f(x) = \cos(x),$$ and $$g(x_{1}, x_{2}, x_{3}) = x_{1}x_{2}x_{3}.$$ Since $g : \Bbb R^{3} \to \Bbb R$ and $f: \Bbb R \to \Bbb R$ are both continuous (why?), their composition $f \circ g$ is continuous, so $h := f \circ g$ is continuous. What does that imply about $h^{-1}( (-\infty, - \frac{1}{2}) )$?

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    Im pretty sure it implies its an open set. Is that correct? – 2017-02-20
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    @B.Smith sorry for the late reply. Yes you are right. Under a continuous function, the preimage of any open set is open, so your set is open. Now you'll also have to think about if it's closed, since a set can be both open and closed. – 2017-02-21