The question's like this：

Show that a nonempty set E of real numbers is an interval if and only if every continuous real-valued function on E has an interval as its image.

Thank you so much for helping me out!

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- Oct 11th 2011, 05:37 PMviolet8804Interval and its image
The question's like this：

**Show that a nonempty set E of real numbers is an interval if and only if every continuous real-valued function on E has an interval as its image.**

Thank you so much for helping me out! - Oct 12th 2011, 01:50 AMSlipEternalRe: Interval and its image
The statement is an if and only if. So, first prove one direction, then the other. To prove the first direction, assume that E is an interval. Then show that it follows that every continuous real-valued function on E has an interval as its image.

Do you need help with that proof? To get you started: If the function is a constant function, then it is trivially true, as every constant $\displaystyle c$ can be expressed as the interval $\displaystyle \[c,c\]$.

Next, assume that every continuous real-valued function on E has an interval as its image. Try to prove that E is an interval. This should be trivial to prove. Hint: Let $\displaystyle f: E \to E$ be the identity function.