# Thread: proof of closed subset

1. ## proof of closed subset

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 $\displaystyle R$.

2. What ideas have you had so far?

3. 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?

4. 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 $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$

5. Originally Posted by Plato
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 $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$

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.

6. How do you normally show that a set is open?

7. 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 $\displaystyle \epsilon$ neighborhood which makes $\displaystyle V_\epsilon \in$ set?

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

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

10. 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.

11. Originally Posted by Plato
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 $\displaystyle x \in \left( {x_0 - \delta ,x_0 + \delta } \right)\, \Rightarrow \,f(x) \ne 0.$
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.

12. Originally Posted by Iondor
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

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

13. He proved that the set C is closed using the definition of sequences for a closed set, which he said at the start of his post.

14. Originally Posted by Ares_D1
I'm just trying to think of a reasonable condition on $\displaystyle \delta$ such that this would be true.
If $\displaystyle f$ is continuous then so is $\displaystyle |f|$ and $\displaystyle |f|(x_0)>0$.
So in the definition take $\displaystyle \varepsilon = |f|(x_0 )$ then find the corresponding $\displaystyle \delta.$