Show that the set$ \{(x, y, z) : x^2 − y^2 + \sin z ≤ 3\}$ is closed in $ \mathbb R^3$ This is a homework question for an introduction to Real Analysis class. I know the following things, but I'm not sure how to start to put them together to prove that this set is closed in R^3.
I know that convergence in R^N is component wise. I also know that a set A is closed if and only if all convergent sequences in A that converge, converge to a value that is back in A.
Its easy to show that A_x, A_y, and A_z = R. From there, we can show that any sequence of A_x (or A_y or A_z) is closed, since its limit must be back in A_x, which is R. 
However, I feel like I'm missing a step in being able to backtrack that to A as a whole.
 A: Let $A= \{(x, y, z) : x^2 − y^2 + \sin z ≤ 3\}$ and let $((x_n,y_n,z_n))$ be a convergent sequence in $A$ with limit $(x_0,y_0,z_0).$
Then we have $x_n \to x_0, y_n \to y_0$ and $z_n \to z_0.$
From $x_n^2 − y_n^2 + \sin z_n ≤ 3$ for all $n$ we get with $n \to \infty$ that $x_0^2 − y_0^2 + \sin z_0 ≤ 3$, since the functions $x \to x^2, y \to y^2$ and $ z \to \sin z$ are continuous.
$x_0^2 − y_0^2 + \sin z_0 ≤ 3$ now shows that $(x_0,y_0,z_0) \in A.$
A: Let $f(x,y,z)= x^2-y^2 +\sin z$ and $A= \{(x, y, z) : x^2 − y^2 + \sin z ≤ 3\}$. Then
$$ \mathbb R^3 \setminus A=f^{-1}((3, \infty)).$$
$(3, \infty)$ is open and $f$ is continuous, hence $\mathbb R^3 \setminus A$ is open, thus $A$ is closed.
A: Let $f: \Bbb R^3 \to R$ be given by $f(x,y,z)=x^2 - y^2 + \sin z$ and note that when we know addition/substraction are continuous on $\Bbb R^2$ and $x \to x^2$ is continuous on $\Bbb R$ and $\sin z$ likewise (and the projection functions are too), $f$ is continuous as a composition of these maps.
Then your set is just $f^{-1}[(-\infty,3]]$, which is closed as the inverse image of a closed set under a continuous function.
A: Yes, you are missing a crucial step. Consider for example 
$$B=\{(x,y) \in \mathbb R^2\;|\; x+y > 0\}.$$
Then $B_x=B_y=\mathbb R$, but $B$ is not closed. That's because for example the sequence $(\frac1n,\frac1n)_{n\ge 1}$ lies in $B$, but it's limit, $(0,0)$ doesn't lie in $B$.
So while $0 \in B_x$ and $0 \in B_y$, so both components of the limit lie in $B_x$ and $B_y$ respectively, it doesn't mean at all that $(0,0)$ lies in $B$ (which this example shows).
See Fred's answer with the limits to see what you actually need to do to prove closeness using the limit of a sequence of points in the set.
