Determine whether the set $\{(,,):++=2 \}$ is open or closed

Question

I'm trying to figure out whether this set is open or closed $\{(x,y,z):x+y+z=2\}$, but I'm having a hard time to figure it out in three dimensions.

Can I say that its complement is open and it is therefore closed?

Answer 1

Hint:

Show that any point $(x,y,z)$ in the set $\;A=\bigl\{(x,y,z)\mid x+y+z\ne 2\bigr\}\;$ has a small neighbourhood contained in the set. Two cases: either $x+y+z>2$, or $x+y+z<2$.

What can you conclude from this property?

Answer 2

Also, letting $f:\mathbb R^3\to \mathbb R$ be the continuous function $f(x,y,z)=x+y+z$, then $A=f^{-1}(\{2\})$. The set $\{2\}$ is closed, and inverse images of closed sets are closed.