Hi,can anyone please help me with this question? I cannot even start it.
Let f: $\displaystyle R \rightarrow R$ be a continuous function and let
$\displaystyle C=\{x \in R \mid f(x)=0\}$
Prove that C is a closed subset of $\displaystyle R$.
Hi,can anyone please help me with this question? I cannot even start it.
Let f: $\displaystyle R \rightarrow R$ be a continuous function and let
$\displaystyle C=\{x \in R \mid f(x)=0\}$
Prove that C is a closed subset of $\displaystyle R$.
That is a good approach.
Recall that for continuous functions if $\displaystyle f(x_0)\ne 0$ then $\displaystyle \left( {\exists \delta > 0} \right)$ such that $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$
Alternatively you could also prove it with the sequences definition of closed sets in $\displaystyle \mathbb{R}$ :
$\displaystyle A \subseteq \mathbb{R}$ is closed iff the limit of every convergent sequence $\displaystyle \{x_{n}\}_{n\in \mathbb{N}}$, with $\displaystyle x_{n} \in A$, stays in A.
Let $\displaystyle \{x_{n}\}_{n\in \mathbb{N}}$ be any sequence in C, which has a limit x.
For all n $\displaystyle x_{n} \in C => f(x_{n})=0$
=> $\displaystyle \lim_{n\to \infty} f(x_{n})=0=f(x)$ , because f is continous => $\displaystyle x \in C$.
qed
Btw. more generally for any continous function f, it is true that $\displaystyle f^{-1}(A)$ is always a closed set, when A is a closed set and is always an open set, when A is an open set.
In your exercise, we just have the special case $\displaystyle A=\{0\}$, which is closed and so $\displaystyle C=f^{-1}(\{0\})$ must be closed too.
I'm just trying to think of a reasonable condition on $\displaystyle \delta$ such that this would be true. Perhaps $\displaystyle \delta = min \{x_0 - c, c - x_0\}$, where $\displaystyle c \in \mathbb{R}$ is the closest element of $\displaystyle C$ to $\displaystyle x_0$? Just a guess; I'm not too sure.