Chapter 4.1, Problem 2E

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.

1. Claim. Let $A\subseteq ℝ.$ If $i=\inf \left(A\right)$ and $\epsilon =0,$ then there is $y\in A.$ such that $y<i+\epsilon .$
   "proof." Let $y=i+\frac{\epsilon }{2}.$ Then $i<y,$ so $y\in A.$ By construction of $y,y<i+\epsilon .$
2. Claim. Let $A\subseteq ℝ.$ If A is bounded above, then $A^c$ is bounded below.
   "proof." If A is bounded above, then sup(A) exists (because $ℝ$ is complete). Because $\sup \left(A\right)=\inf \left(A^c\right)$ (see the figure), $\inf \left(A^c\right)$ exists. Thus, $A^c$ is bounded below.

Claim. If $A\subseteq B\subseteq ℝ,$ and if sup(A) and sup(B) both exist, then $\sup \left(A\right)\le \sup \left(B\right).$
"proof." Assume that $\subseteq B$ and $\sup \left(A\right)>\sup \left(B\right).$ We choose $\epsilon =\frac{1}{2}\left(\sup \left(A\right)-\sup \left(B\right)\right).$ Then $\epsilon >0$ and $\sup \left(B\right)<\sup \left(A\right)-\epsilon <\sup \left(A\right).$ By part (ii) of Theorem 7.1.1, there is $y\in A$ such that $y>\sup \left(A\right)-\epsilon .$ Then $y\in B$ and $y>\sup \left(B\right).$ This is impossible. Therefore, $\sup \left(A\right)\le \sup \left(B\right).$

Claim. If f: $ℝ\to ℝ$ and A is a bounded subset of $ℝ,$ then the image setf(A) is bounded.
"proof." Let m be an upper bound for A. Then $a\le m$ for all $a\in A.$ Therefore, $f\left(a\right)\le f\left(m\right)$ for all $a\in A.$ Thus, $f\left(m\right)$ is an upper bound for $f\left(A\right).$,/li>

Claim. The set $\mathbb{N}$ is not bounded above.
"Proof." Suppose that $\mathbb{N}$ is bounded above by the natural number m. Then $m+1.$ is a natural number larger than m. This is a contradiction.