**Question:**

Let \(f: \mathbb{R} \to \mathbb{R} \) be a continuous function. Which of the following can not be the image of \((0,1]\) under \(f\)?

A. {0}

B. \((0,1)\)

C. \([0,1)\)

D. \([0,1]\)

**Discussion:**

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that \(f((0,1])=(0,1)\) . Then \(f((0,1))=(0,1)- \{f(1)\} \). Now since \(f(1)\in (0,1) \) the set \( (0,1)- \{f(1)\} \) is not connected. But \((0,1)\) is connected, and we know that continuus image of a connected set is connected. This gives a contradiction. So \((0,1)\) can not be the image of \((0,1]\) under f.

Define \(f(x)=1-x\). Then \(f((0,1])= [0,1)\).

Define \(f(x)=0\) for \(x\in [0,\frac{1}{2}] \) and \(f(x)= 2(x-\frac{1}{2}) \) for \(x\in [\frac{1}{2} ,1 ] \). \(f\) is continuous on \((0,\frac{1}{2}] \) and \( [\frac{1}{2} ,1 ] \) and \(f\) agrees on the common points, by pasting lemma \(f\) is continuous on \( [0,1] \) . And image of \((0,1] \) is \([0,1]\).