Proving $A$={$(x,y,z)\in \mathbb{R}^3 \mid |x^3+y+z|>3$} is open Having little trouble showing this, even though intuitively it is pretty clear. Definition of open set: 

$U$ is open if every point in $U$ has a neighborhood contained in $U$.

We have done few of these proofs, but they were bounded and this is clearly unbounded set, so I have no idea how to choose distance or anything. 
 A: Let's prove the function is continuous at each point in $A$.
Let $f(x,y,z) = x^{3} + y + z$.  Let $(x_0,y_0,z_0) \in \Bbb R^{3}$.  We want to show $f$ is continuous at $(x_0,y_0,z_0)$.
Okay, so for each $\epsilon > 0$, we must find $\delta > 0$ such that $|| (x,y,z) - (x_0,y_0,z_0)|| \leq \delta$ implies $|f(x,y,z) - f(x_0,y_0,z_0)| < \epsilon$.
Note that the norm $||\cdot||$ we are using on the domain here can be any norm on $\Bbb R^{3}$ since for any fixed natural number $n$, all norms on $\Bbb R^{n}$ are equivalent.  The norm on the range $| \cdot |$ is just the normal Euclidean norm on $\Bbb R$ (basically, it's absolute value).
Okay, so let's choose the norm $||(a,b,c) - (d,e,f)|| = |d - a| + |e- b| + |f - c|$, which is the sum of the (positive) differences in components.
So, let $\epsilon > 0$.  Does there exist a $\delta > 0$ such that $|x-x_0| + |y-y_0| + |z-z_0| < \delta$ implies $|x^{3} + y + z - (x_0^{3} + y_0 + z_0)| < \epsilon$?
Well, we know by the triangle inequality that $|x^{3} + y + z - (x_0^{3} + y_0 + z_0)| \leq |x^{3} - x_0^{3}| + |y - y_0| + |z - z_0|$.  
Also, $h(x) = x^{3}$ is continuous.
So, let $y$ satisfy $|y - y_0| < \epsilon/3$ and $z$ satisfy $|z - z_0| < \epsilon/3$. We know by the continuity of $x^{3}$ that there is some $\delta_0>0$ such that $|x - x_0| < \delta$ implies $|x^3 - x_0^3| < \epsilon/3$.
So, finally, after all that work, let's choose $\delta = \min\{\delta_0, \epsilon/3 \}$.
Then $||(x,y,z) - (x_0, y_0, z_0)|| \leq \delta$ implies $|f(x,y,z) - f(x_0,y_0,z_0)| < \epsilon$.
So, $f(x,y,z)$ is continuous on $\Bbb R^{3}$.  Also, $r(x,y,z) := |f(x,y,z)|$ is continuous since the absolute value function is continuous, and so $r$ is the composition of continuous functions. Now that we know $r$ is continuous, we can rewrite the set $A$ in the question as $r^{-1}(3,\infty)$, i.e., the preimage of $(3,\infty)$.  Since $(3,\infty)$ is open and $r$ is continuous, this means $A$ is open.
A: A definition of continuity is: For two metric spaces $X$ and $Y$, a function $f: X\to Y$ is continuous if and only if for every open set $V\subseteq Y$, $f^{-1}(V)\subseteq X$ is an open set of $X$. 
If we set $f: \mathbb{R}^3\to \mathbb{R}$ to $f(x,y,z)=|x^3+y+z|$ this is obviously an continuous function since $x^3+y+z$ is a polynomial and thus continuous and since $|\cdot|$ is continous, $f$ is a composition of two continous functions and thus continuous. Since $(3,\infty)$ is an open set,
$$\{(x,y,z)\,|\, x^3+y+z >3\}=f^{-1}((3,\infty))$$
is open.
A: That the set is unbounded doesn't really matter.  Where things are interesting is what is going on around the inner boundary.
For any $(x,y,z) \in A, |x^3 + y+ z|>3$
let $\delta = \min (\frac 18 ||x^3 + y+ z|-3|, 1)$
Let $U$ be the open ball around $(x,y,z)$ of radius $\delta$
Every element of $U$ is in $A$
