Thread: How do I prove that {x: f(x) not eqaul to r_0} is an open set?

1. How do I prove that {x: f(x) not eqaul to r_0} is an open set?

Hey guys, how do i solve this problem?

Suppose f: R->R is continuous and let $r_0$ $\in$R. Prove that {x: f(x) not equal to $r_0$ is an open set.

My second midterm is coming up and i'm trying to study for it but i'm having trouble with some of the problems. i feel that when I see it i can learn it. Thanks guys (ps there will probably a few, considering how many questions are on this review sheet )

2. Originally Posted by tn11631
Hey guys, how do i solve this problem?

Suppose f: R->R is continuous and let $r_0$ $\in$R. Prove that {x: f(x) not equal to $r_0$ is an open set.

My second midterm is coming up and i'm trying to study for it but i'm having trouble with some of the problems. i feel that when I see it i can learn it. Thanks guys (ps there will probably a few, considering how many questions are on this review sheet )
What is the inverse image of an open set under a continuous map?

3. Originally Posted by Drexel28
What is the inverse image of an open set under a continuous map?
as in the inverse image of every open set under f is again open?

4. Originally Posted by tn11631
as in the inverse image of every open set under f is again open?
Bingo.

5. hm, I don't recall going in this direction in class. Sorry as you can tell this isn't my favorite math. Do I have to show that for r_0 in f, f is a neighborhood of r_0? Sorry once I learn one proof I'm able to push myself through similar ones, but i just completely draw blanks.

6. Originally Posted by tn11631
hm, I don't recall going in this direction in class. Sorry as you can tell this isn't my favorite math. Do I have to show that for r_0 in f, f is a neighborhood of r_0? Sorry once I learn one proof I'm able to push myself through similar ones, but i just completely draw blanks.
I mean, I'm sorry to report that I am not a mind-reader. I have no idea what definition of continuity you are given. That would be a nice start.

7. Originally Posted by Drexel28
I mean, I'm sorry to report that I am not a mind-reader. I have no idea what definition of continuity you are given. That would be a nice start.
I apologize if i have upset you or offended you in anyway those weren't my intentions i'm just lost. However the definition the books gives is: "Suppose E $\subset$R and f: E->R. If $x_0$ $\in$E, then f is continuous at $x_0$ iff each $\epsilon$>0, there is a delta>0 such that if
|x- $x_0$|<delta, x $\in$E, then |f(x)-f( $x_0$)< $\epsilon$. If f is continuous at x for every x $\in$E, then we say f is continuous"
I don't know if that gives you any help as to where i'm at but i hope it helps. thanks again

8. Originally Posted by tn11631
I apologize if i have upset you or offended you in anyway those weren't my intentions i'm just lost. However the definition the books gives is: "Suppose E $\subset$R and f: E->R. If $x_0$ $\in$E, then f is continuous at $x_0$ iff each $\epsilon$>0, there is a delta>0 such that if
|x- $x_0$|<delta, x $\in$E, then |f(x)-f( $x_0$)< $\epsilon$. If f is continuous at x for every x $\in$E, then we say f is continuous"
I don't know if that gives you any help as to where i'm at but i hope it helps. thanks again
So the basic concept is this. Let $x\in f^{-1}(\mathbb{R}-\{0\})$. Then, $f(x)\in\mathbb{R}-\{0\}$. So, since $\mathbb{R}-\{0\}$ is open there exists some $B_{\varepsilon}(f(x))\subseteq\mathbb{R}-\{0\}$. But, by continuity there exists a $\delta>0$ such that $\left(d(x,y)<\delta\implies d(f(x),f(y))<\varepsilon\right)\Longleftrightarrow \left(f(B_{\delta}(x))\subseteq B_{\varepsilon}(f(x))\right)$. So, $f(B_{\delta}(x))\subseteq\mathbb{R}-\{0\}\implies B_{\delta}(x)\subseteq f^{-1}(\mathbb{R}-\{0\})$. It follows that $x\in f^{-1}\left(\mathbb{R}-\{0\}\right)\implies x\in\left[f^{-1}\left(\mathbb{R}-\{0\}\right)\right]^{\circ}$. Thus, $f^{-1}(\mathbb{R}-\{0\})$ is open.

9. Originally Posted by Drexel28
So the basic concept is this. Let $x\in f^{-1}(\mathbb{R}-\{0\})$. Then, $f(x)\in\mathbb{R}-\{0\}$. So, since $\mathbb{R}-\{0\}$ is open there exists some $B_{\varepsilon}(f(x))\subseteq\mathbb{R}-\{0\}$. But, by continuity there exists a $\delta>0$ such that $\left(d(x,y)<\delta\implies d(f(x),f(y))<\varepsilon\right)\Longleftrightarrow \left(f(B_{\delta}(x))\subseteq B_{\varepsilon}(f(x))\right)$. So, $f(B_{\delta}(x))\subseteq\mathbb{R}-\{0\}\implies B_{\delta}(x)\subseteq f^{-1}(\mathbb{R}-\{0\})$. It follows that $x\in f^{-1}\left(\mathbb{R}-\{0\}\right)\implies x\in\left[f^{-1}\left(\mathbb{R}-\{0\}\right)\right]^{\circ}$. Thus, $f^{-1}(\mathbb{R}-\{0\})$ is open.
hmm. Ok I deff see where and how this proof is working. However I have what i think is a stupid question. did you take the inverse image b/c of the condition $r_0$ $\in$R and having to show that f(x) $\neq$ $r_0$. And also in like the second to last thing, that's the compliment right? I just wanna make sure sorry.

10. Originally Posted by tn11631
hmm. Ok I deff see where and how this proof is working. However I have what i think is a stupid question. did you take the inverse image b/c of the condition $r_0$ $\in$R and having to show that f(x) $\neq$ $r_0$.
I don't know what you mean.

And also in like the second to last thing, that's the compliment right? I just wanna make sure sorry.
When is something every in a set and it's compliment? It stand for interior.

11. Sorry, I was refering back to the original question. I think i'm ok now. Thanks