Is f(x) a closed interval if f is continuous and onto on a bounded interval?

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SUMMARY

If a function f is continuous and onto a closed interval, then the image f(x) is also a closed interval, as established by the Intermediate Value Theorem. Conversely, if f is continuous on an open bounded interval, the range can be closed, open, or neither. Specifically, there is no continuous function mapping from the closed interval [0,1] to the open interval (0,1). However, functions can be constructed to map from (0,1) onto [0,1], such as using polynomial interpolation or sine functions like sin(4x).

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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

Homework Equations


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.
 
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True, that is a theorem. f continuous on a closed interval then the image of f is a closed interval. If f is continuous on an open interval then the range can be closed, open or neither. Can you find an example of each?
 
Dick said:
True, that is a theorem. f continuous on a closed interval then the image of f is a closed interval. If f is continuous 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]
 
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.
 
Dick said:
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.
 
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.
 
Can't you linearly interpolate between the values I gave? You could also fit it to a polynomial. You could scale a sine function. Any number of things you could do to get an explicit example. You did sketch one, right? That's all I care about.
 
Dick said:
Can't you linearly interpolate between the values I gave? You could also fit it to a polynomial. You could scale a sine function. Any number of things you could do to get an explicit example. You did sketch one, right? That's all I care about.

sin(4x) works

Thx
 

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