Problem 409 (difficulty: 4/10)

Does there exist a monotone function \(\displaystyle f\) such that

 (1)  \(\displaystyle D(f)=[0,1]\), \(\displaystyle R(f)=(0,1);\phantom{\,\cup [2,3]}\)      (2)  \(\displaystyle D(f)=[0,1]\), \(\displaystyle R(f)=[0,1]\cup [2,3]\);
 (3)  \(\displaystyle D(f)=[0,1]\), \(\displaystyle R(f)=[0,1)\cup [2,3]\);      (4)  \(\displaystyle D(f)=[0,1]\), \(\displaystyle R(f)=[0,1)\cup (2,3]\)?

Solution:

(1) No, \(\displaystyle 0<y<f(0)\) cannot be obtained as a value of \(\displaystyle f\).

(2) \(\displaystyle f(x):= \begin{cases} 3x & \text{if $x\leq1/3$ or $x>2/3$}\\ 2 & \text{if $1/3<x\leq 2/3$.} \end{cases}\)

(3) \(\displaystyle f(x):= \begin{cases} 3x & \text{if $x<1/3$ or $x>2/3$} \\ 2 & \text{if $1/3\leq x\leq 2/3$.} \end{cases}\)

(4) No, \(\displaystyle \sup f^{-1}([0,1))\) cannot be obtained as a value of \(\displaystyle f\).

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