Let $f:\Bbb R \to \Bbb R$ be a function with \begin{equation*} f(x) = \begin{cases} 0, &\text{if} \ x \lt -1 \\ x^2+1, &\text{if} \ x \in [-1,1] \\ \frac{1}{2x}, &\text{if} \ x \gt 1 \end{cases} \end{equation*}
- Is it true that the set $A=\{f(x) \mid x \in \Bbb R\}$ bounded? Explain it.
- Find the infimum and the supremum of the set $B = \{f(x)+2 \mid x \in \Bbb R\}$.
Attempt: For $1$, the first think I do is to partition $f$ into three cases as follows.
Case 1. $x \in (-\infty,-1)$. Let $C = \{f(x) \mid x \in (-\infty,-1)\}$. By definition, $f$ is a constant (zero) function. Then, $C = \{0\}$ and so, $C$ is bounded.
Case 2. $x \in [-1,1]$. Let $D = \{f(x) \mid x \in [-1,1]\}$. Then, $D = [1,2]$ and so, $D$ is bounded.
Case 3. $x \in (1, \infty)$. Let $E = \{f(x) \mid x \in (1,\infty)\}$. Then, $E = \left(0,\frac{1}{2}\right)$ and so, $E$ is bounded.
Hence, from three cases above, we conclude that $A$ is bounded.
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For $2$, notice that $B = \{2\} \cup [3,4] \cup \left(0,\frac{5}{2}\right)$. Hence, $\inf(B) = 0$ and $\sup(B) = 4$.
Am I true?
Your answer to the first question is correct. Bu the other on is not. In fact, $B=\{2\}\cup[3,4]\cup\left(2,\frac52\right)$. Therefore, $\inf(B)=2$ and $\sup(B)=4$.