proof of closed subset

**tsang** · September 3rd 2010, 02:32 AM

Hi,can anyone please help me with this question? I cannot even start it.

Let f: $R \rightarrow R$ be a continuous function and let
$C=\{x \in R \mid f(x)=0\}$
Prove that C is a closed subset of $R$ .

**Ackbeet** · September 3rd 2010, 02:58 AM

What ideas have you had so far?

**tsang** · September 3rd 2010, 04:21 AM

Originally Posted by Ackbeet

What ideas have you had so far?

All I can think about is go back to definition, so I should try to show that R\C is open, which means C is closed. Am I right? But I really don't know how to even start. So C is a constant function, and R\C should be R\{0}, am I right so far?

**Plato** · September 3rd 2010, 04:31 AM

That is a good approach.
Recall that for continuous functions if $f(x_0)\ne 0$ then $\left( {\exists \delta > 0} \right)$ such that $x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$

**tsang** · September 3rd 2010, 05:13 AM

Originally Posted by Plato

That is a good approach.
Recall that for continuous functions if $f(x_0)\ne 0$ then $\left( {\exists \delta > 0} \right)$ such that $x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$

Hi, thanks for your reply. However, I actually still don't know how to continue, could you pleas;e give me a bit more details? I just started learning Real Analysis and I'm still trying to improve it at this moment. I will really appreciate for your time. Thanks a lot.

**Ackbeet** · September 3rd 2010, 05:18 AM

How do you normally show that a set is open?

**tsang** · September 3rd 2010, 05:25 AM

Originally Posted by Ackbeet

How do you normally show that a set is open?

To show that for any point in the set, can always find $\epsilon$ neighborhood which makes $V_\epsilon \in$ set?

**Ackbeet** · September 3rd 2010, 05:51 AM

Exactly. So pick a point in $\mathbb{R}\setminus C$ and try to get your neighborhood. How would you proceed?

**Plato** · September 3rd 2010, 06:23 AM

@stang, Do you realize that $\left( {x_0 - \delta ,x_0 + \delta } \right)$ is an open set containing $x_0?$

**Iondor** · September 4th 2010, 11:10 AM

Alternatively you could also prove it with the sequences definition of closed sets in $\mathbb{R}$ :
$A \subseteq \mathbb{R}$ is closed iff the limit of every convergent sequence $\{x_{n}\}_{n\in \mathbb{N}}$ , with $x_{n} \in A$ , stays in A.

Let $\{x_{n}\}_{n\in \mathbb{N}}$ be any sequence in C, which has a limit x.
For all n $x_{n} \in C => f(x_{n})=0$
=> $\lim_{n\to \infty} f(x_{n})=0=f(x)$ , because f is continous => $x \in C$ .
qed

Btw. more generally for any continous function f, it is true that $f^{-1}(A)$ 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 $A=\{0\}$ , which is closed and so $C=f^{-1}(\{0\})$ must be closed too.

**Ares_D1** · September 4th 2010, 09:16 PM

Originally Posted by Plato

That is a good approach.
Recall that for continuous functions if $f(x_0)\ne 0$ then $\left( {\exists \delta > 0} \right)$ such that $x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$

I'm just trying to think of a reasonable condition on $\delta$ such that this would be true. Perhaps $\delta = min \{x_0 - c, c - x_0\}$ , where $c \in \mathbb{R}$ is the closest element of $C$ to $x_0$ ? Just a guess; I'm not too sure.

**tsang** · September 5th 2010, 12:11 AM

Originally Posted by Iondor

Let $\{x_{n}\}_{n\in \mathbb{N}}$ be any sequence in C, which has a limit x.
For all n $x_{n} \in C => f(x_{n})=0$
=> $\lim_{n\to \infty} f(x_{n})=0=f(x)$ , because f is continous => $x \in C$ .
qed

Hi,thank you for your time. But I just wonder that what is the purpose to prove this $\lim_{n\to \infty} f(x_{n})=0=f(x)$ ? Could you please just give me a bit more details that why you are doing this? Thank you.