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!
That's how I would start. The next step would be to observe what $\displaystyle f(x_n)$ equals for all $\displaystyle n \in \mathbb{N}$, and then conclude what the value of $\displaystyle f(x)$ will be, using the continuity of $\displaystyle f$ over $\displaystyle \mathbb{R}$.
Problem: Let $\displaystyle f:\mathbb{R}\mapsto\mathbb{R}$ be continuous. Define $\displaystyle \mathcal{Z}\left(f\right)=\left\{x\in\mathbb{R}:f( x)=0\right\}$. Prove that $\displaystyle \mathcal{Z}\left(f\right)$ is closed.
Proof: Let $\displaystyle \xi$ be a limit point of $\displaystyle \mathcal{Z}(f)$. Then there exists some sequence $\displaystyle \left\{\xi_n\right\}$ such that $\displaystyle \xi_n\in\mathcal{Z}(f)\quad\forall n$ and $\displaystyle \xi_n\to\xi$. But then $\displaystyle \lim_{n\to\infty}f\left(\xi_n\right)=\lim_{n\to\in fty}0=0$. And since $\displaystyle f$ was continuous as $\displaystyle \xi$ this means that $\displaystyle \lim_{n\to\infty}f\left(\xi_n\right)=\lim_{x\to\xi }f(x)=f(\xi)=0$. Therefore $\displaystyle \xi\in\mathcal{Z}(f)$. Thus $\displaystyle \mathcal{Z}(f)$ contains all its limit points, which consequently means it's closed.
Here is another way to do this problem
Problem: Let $\displaystyle \phi:\mathbb{R}\mapsto \mathbb{R}$ be continuous. Define $\displaystyle \mathcal{Z}(\phi)=\left\{x\in \mathbb{R}:\phi(x)=0\right\}$. Prove that $\displaystyle \mathcal{Z}(\phi)$ is closed.
Proof: Let $\displaystyle \xi$ be a limit point of $\displaystyle \mathcal{Z}(\phi)$. Then $\displaystyle \forall \ell>0$ there exists some $\displaystyle y\in N_{\ell}(\xi)$ such that $\displaystyle \phi(y)=0$. Because $\displaystyle \phi$ is continous at we know that $\displaystyle \forall\varepsilon>0$ there exists some $\displaystyle \delta>0$ such that if
$\displaystyle y\in N_{\delta}(\xi)\implies \phi(y)\in N_{\varepsilon}(\phi(\xi))$. Now assume that $\displaystyle \xi\notin\mathcal{Z}(\phi)$. Then $\displaystyle \phi(\xi)\ne0\implies|\phi(\xi)|\ne0$. But this is a contradiction, for since $\displaystyle |\phi(\xi)|>0$ one may let $\displaystyle \varepsilon=|\phi(\xi)|$ in the above. But no matter
what $\displaystyle \delta>0$ one picks there will be a point $\displaystyle y\in N_{\delta}(\xi)$ such that $\displaystyle \phi(y)=0$, but clearly $\displaystyle 0\notin N_{|\phi(\xi)|}(\phi(\xi))$. Therefore, $\displaystyle \xi\in\mathcal{Z}(\phi)$ 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 $\displaystyle \left\{0\right\}$ is closed by definition $\displaystyle \phi^{-1}\left(\left\{0\right\}\right)=\left\{x\in\mathbb {R}:\phi(x)=0\right\}=\mathcal{Z}(\phi)$ is closed.