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.

As you can see from the different kind of answers the argument is depending on your definition of a continuous function. It would be helpful to include it. — Maik Pickl, Oct 01 '18 at 15:02

score 4 · Accepted Answer · answered Oct 01 '18 at 14:22

4

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?

answered Oct 01 '18 at 14:22

Marc

6,861

How do you know that a convergent sequence $(x_n)_{n\in\mathbb{N}}$ exist in $A$? – Aqqqq Oct 01 '18 at 15:05
1

Let $x\in A$. Then the sequence $(x,x,x,\ldots)$ is a convergent sequence for example. – Marc Oct 01 '18 at 16:46

score 4 · Answer 2 · answered Oct 01 '18 at 14:29

4

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

answered Oct 01 '18 at 14:29

score 3 · Answer 3 · answered Oct 01 '18 at 14:28

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.

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

3 Answers3