1. ## Proof Closed set??

So I have f: R -> R be continuous. Let S={x in R: f(x)=0} be nonempty. Prove S is a closed set.

So far I've started with:
Let {x_n} be a sequence in S such that x_n -> x ?? yes? where to next??

Any help would be greatly appreciated!

2. That's how I would start. The next step would be to observe what $f(x_n)$ equals for all $n \in \mathbb{N}$, and then conclude what the value of $f(x)$ will be, using the continuity of $f$ over $\mathbb{R}$.

3. Originally Posted by engineer5373
So I have f: R -> R be continuous. Let S={x in R: f(x)=0} be nonempty. Prove S is a closed set.

So far I've started with:
Let {x_n} be a sequence in S such that x_n -> x ?? yes? where to next??

Any help would be greatly appreciated!
Problem: Let $f:\mathbb{R}\mapsto\mathbb{R}$ be continuous. Define $\mathcal{Z}\left(f\right)=\left\{x\in\mathbb{R}:f( x)=0\right\}$. Prove that $\mathcal{Z}\left(f\right)$ is closed.

Proof: Let $\xi$ be a limit point of $\mathcal{Z}(f)$. Then there exists some sequence $\left\{\xi_n\right\}$ such that $\xi_n\in\mathcal{Z}(f)\quad\forall n$ and $\xi_n\to\xi$. But then $\lim_{n\to\infty}f\left(\xi_n\right)=\lim_{n\to\in fty}0=0$. And since $f$ was continuous as $\xi$ this means that $\lim_{n\to\infty}f\left(\xi_n\right)=\lim_{x\to\xi }f(x)=f(\xi)=0$. Therefore $\xi\in\mathcal{Z}(f)$. Thus $\mathcal{Z}(f)$ contains all its limit points, which consequently means it's closed.

4. See now it makes sense. Thank you both so much!!!

5. Originally Posted by Drexel28
Problem: Let $f:\mathbb{R}\mapsto\mathbb{R}$ be continuous. Define $\mathcal{Z}\left(f\right)=\left\{x\in\mathbb{R}:f( x)=0\right\}$. Prove that $\mathcal{Z}\left(f\right)$ is closed.

Proof: Let $\xi$ be a limit point of $\mathcal{Z}(f)$. Then there exists some sequence $\left\{\xi_n\right\}$ such that $\xi_n\in\mathcal{Z}(f)\quad\forall n$ and $\xi_n\to\xi$. But then $\lim_{n\to\infty}f\left(\xi_n\right)=\lim_{n\to\in fty}0=0$. And since $f$ was continuous as $\xi$ this means that $\lim_{n\to\infty}f\left(\xi_n\right)=\lim_{x\to\xi }f(x)=f(\xi)=0$. Therefore $\xi\in\mathcal{Z}(f)$. Thus $\mathcal{Z}(f)$ contains all its limit points, which consequently means it's closed.
Mathstud, is that you ?

6. Originally Posted by Moo
Mathstud, is that you ?
You're like the third person that's said that?? Why do you think I'm 'Mathstud'?

7. Here is another way to do this problem

Problem: Let $\phi:\mathbb{R}\mapsto \mathbb{R}$ be continuous. Define $\mathcal{Z}(\phi)=\left\{x\in \mathbb{R}:\phi(x)=0\right\}$. Prove that $\mathcal{Z}(\phi)$ is closed.

Proof: Let $\xi$ be a limit point of $\mathcal{Z}(\phi)$. Then $\forall \ell>0$ there exists some $y\in N_{\ell}(\xi)$ such that $\phi(y)=0$. Because $\phi$ is continous at we know that $\forall\varepsilon>0$ there exists some $\delta>0$ such that if

$y\in N_{\delta}(\xi)\implies \phi(y)\in N_{\varepsilon}(\phi(\xi))$. Now assume that $\xi\notin\mathcal{Z}(\phi)$. Then $\phi(\xi)\ne0\implies|\phi(\xi)|\ne0$. But this is a contradiction, for since $|\phi(\xi)|>0$ one may let $\varepsilon=|\phi(\xi)|$ in the above. But no matter

what $\delta>0$ one picks there will be a point $y\in N_{\delta}(\xi)$ such that $\phi(y)=0$, but clearly $0\notin N_{|\phi(\xi)|}(\phi(\xi))$. Therefore, $\xi\in\mathcal{Z}(\phi)$ and the conclusion follows.

Remark: This can be generalized greatly to much more interesting forms, but I digress. Also, these are the times where one wishes they were a topologist. For topologists define a function continuous if whenever a set is open(closed) in the codomain it's inverse image is open(closed) in the domain. And since $\left\{0\right\}$ is closed by definition $\phi^{-1}\left(\left\{0\right\}\right)=\left\{x\in\mathbb {R}:\phi(x)=0\right\}=\mathcal{Z}(\phi)$ is closed.