How to prove a set of solution of a continuous function is closed?

Question

E.g. if the set is $S=\{(x,y), g(x,y) = 0\}$, $g$ is a continuous function. How to prove that S is closed?

It seems that the relationship between continuous function and convergent sequence and the Lemma that "A set A in a metric space is closed iff the limit of every convergent sequence in A belongs to A" can be used here. I am not sure how the relationship between continuous function and convergent sequence is applicable here.

Answer 1

Suppose $X$ is a general metric space, let $g:X\rightarrow\mathbb{R}$ be a real continuous function and let $A=g^{-1}(0)$. Choose a convergent sequence $(x_n)_{n\in\mathbb{N}}$ in $A$. Then by continuity $$ g\left(\lim_{n\rightarrow\infty}x_n\right) = \lim_{n\rightarrow\infty}g\left(x_n\right). $$ Can you finish?

Answer 2

Take $(x,y)\in S^c$, then $g(x,y)\neq 0$, say $ g(x,y) = a>0$ (if $ g(x,y) = a<0$ we can do similar things). Then take b,c real number such that $0<b<a<c$. Since $g$ is continuous, $g^{-1}[(b,c)]$ must be open. And we can easly see that $(x,y)\in g^{-1}[(b,c)] \subset S^c$. So $S^c$ is open then $S$ is closed.

Answer 3

Or you could just note that $\{0\}$ is a closed set. Hence $S=g^{-1}(0)$ is closed.