Let $f:[0,1]\to \mathbb{R}$ be a bounded function. Which of the following $may$ be true statement(s)?
(a) $f$ attains infimum but not supremum.
(b) $f$ attains supremum but not infimum.
(c) $f$ attains neither a infimum nor a supremum.
(d) $f$ attains its supremum and infimum on any non-empty open interval $A \subset [0,1].$
Solution:
Since the function $f$ is given to be bounded, it should have both- an upper bound and a lower bound, that means infimum and supremum must exist. Going by my claims, (a),(b),(c) would have been incorrect. But the answer key does not agree.
Also, i don't know how to verify option (d), why is it that $f$ attains its supremum and infimum on any non-empty open interval $A \subset [0,1].$
The answer key says:
(a),(b),(c),(d)
Each of $(a),(b),(c),(d)$ $\it{may}$ be true. For $(d)$ just consider the constant function $f(x)=1$.
For $(a)$ consider $f(x)=\sin(\pi x)\;\; (x\neq \frac{1}{2})$ with $f(\frac{1}{2})=0$. Here $f$ does not attain its supremum value $1$ but $f$ attains the infimum $0$ at $x=0,1,\frac{1}{2}$. For $(b)$ note that $-f$ attains its supremum of $0$ at $x=0,1,\frac{1}{2}$ but never attains its infimum of $-1$.
For $(c)$ define $f$ so that $f(x)=\sin(\pi x)\;\;(x\neq 0,\frac{1}{2},1)$ and $f(0)=f(\frac{1}{2})=f(1)=\frac{1}{2}$. Here, $f$ does not attain its supremum of $1$ or its infimum of $0$.