# Continuous onto function

1.Is it true that if f is continuous onto function on a closed interval then f(x) must also be a closed interval. How about the other way around. f is continuous and onto on a open bounded interval and f(x) is a closed interval

f:[0,1]-->(0,1)
f:(0,1)-->[0,1]

## The Attempt at a Solution

There is a theorem that says that if f is continuous on a closed and bounded interval then set of f(x) is a closed and bounded interval.

Dick
Homework Helper
True, that is a theorem. f continuous on a closed interval then the image of f is a closed interval. If f is continous on an open interval then the range can be closed, open or neither. Can you find an example of each?

True, that is a theorem. f continuous on a closed interval then the image of f is a closed interval. If f is continous on an open interval then the range can be closed, open or neither. Can you find an example of each?

Does this mean that f:[0,1]-->(0,1) does not exist. If both are closed intervals there are several examples, including the trivial f(x)=x

I don't have an example for continuous onto f:(0,1)-->[0,1]

Dick
Homework Helper
Yes, there is no continuous function f mapping [0,1]->(0,1). You cited the theorem. For mapping (0,1) into [0,1], can't you think of a function where f(0)=1/2, f(1/3)=1, f(2/3)=0 and f(1)=1/2? Surely you can draw one.

Yes, there is no continuous function f mapping [0,1]->(0,1). You cited the theorem. For mapping (0,1) into [0,1], can't you think of a function where f(0)=1/2, f(1/3)=1, f(2/3)=0 and f(1)=1/2? Surely you can draw one.

It has to be an onto function. I am not aware of one as you have described.

Dick
Homework Helper
Onto what? It's certainly onto [0,1].

Yes. But I don't have an example as you have described. Sorry. I am always bad with examples.

Dick