Hey guys, how do i solve this problem?
Suppose f: R->R is continuous and let R. Prove that {x: f(x) not equal to 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 )
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 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 R and f: E->R. If E, then f is continuous at iff each >0, there is a delta>0 such that if
|x- |<delta, x E, then |f(x)-f( )< . If f is continuous at x for every x 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
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 and having to show that f(x) . And also in like the second to last thing, that's the compliment right? I just wanna make sure sorry.