why the set $\{(x,y,0) : x^2 +y^2 < 1\}$ is neither open nor closed?

Question

why the set $\{(x,y,0) : x^2 +y^2 < 1\}$ is neither open nor closed ?

My attempt : it will not closed because it is locally compact

i don't know how to proved that it will not open ?

Any hints/solution will be appreciated

thanks u

Answer 1

Denote $S:=\{(x,y,z)\in\mathbb R^3:z=0,~x^2+y^2<1\}$.

The set is clearly not closed because one can easily produce a sequence in $S$ which converges to a point outside of $S$ (indeed, see bof's comment).

To see why it is not open, consider the definition of an open set. Take a point $(x,y,0)\in S$. Then any open ball of radius $r>0$ around $(x,y,0)$ contains points of the form $(x,y,z)$ where $z>0$, and these points are not in $S$.

Answer 2

I will call the set mentioned in the question $A$.

To show a set is not closed, it suffices to show that a set does not contain all of its limit points. So try to find a point not in $A$ such that all of its neighborhoods intersect $A$.

To show a set is not open, one can show that no basis element is contained within it. Note that basis elements of $\mathbb{R}^3$ can be of the form $(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)$ where $a_i < b_i$ for $i$ in $\{1, 2, 3\}$.