# Proving a set is open using pre image and continuity of a function

I was tasked with determining whether the set {(x,y,z)$\in R^{3}: \sqrt{x^2+y^2}\leq z\leq 1$} is open or closed. I am wondering whether the following approach is valid/ formally okay. I am going to use the theorem: A function R^n -> R^m is continuous iff all open sets in R^m have a pre image that is open in R^n.

Define a function $f(x,y,z)= z - \sqrt{x^2+y^2}$. Consider $U_1$=[0, inf), a closed set in R. Thus, the pre image $f^{-1}(U_1)$ is closed since f(x,y,z) is continuous. Similarly, define f(x,y,z) = z, a continuous function. Consider $U_2$ = (-inf,1], a closed set. Thus, the pre image $f^{-1} (U_2)$ is closed. Thus, since both pre images are closed, it follows that the pre image of their intersection must be closed too. Thus, the set in question is a closed set.

Was the above approach valid?

• The approach is valid, but you need to take $U_2=(-\infty,1]$. – drhab Mar 21 '18 at 11:23
• @drhab Oh yes, sorry that a typo. Do you have any other advice on how I could make this more formal? – user542938 Mar 21 '18 at 11:28
• Not really. I think it is formal enough. – drhab Mar 21 '18 at 11:33
• Can it be proven that if a set in Euclidean space is defined with a series of non-strict inequalities, it is closed? – bipll Mar 21 '18 at 11:36