Chapter 4.3, Problem 3E

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

1. Claim. For A, $B\subseteq ℝ,\left(A\cup B\right)\text{'}=A\text{'}\cup B\text{'}.$
   "Proof."
1. Because $A\subseteq A\cup B,A\text{'}\subseteq \left(A\cup B\right)\text{'}$ by Exercise 6(a). Likewise, $B\text{'}\subseteq \left(A\cup B\right)\text{'}.$ Therefore, $A\text{'}\cup B\text{'}\subseteq \left(A\cup B\right)\text{'}.$
2. To show that $\left(A\cup B\right)\text{'},A\text{'}\subseteq A\text{'}\cup B\text{'},$ let $x\in \left(A\cup B\right)\text{'}.$ Then for all $\delta >0,N\left(x,\delta \right)$ contains a point of $A\cup B$ distinct from x. Restating this, we say, for all $\delta >0,$ that $N\left(x,\delta \right)$ contains a point of A distinct from x or a point of B distinct from x. Thus, for all $\delta >0,N\left(x,\delta \right)$ contains a point of A distinct from x, or for all $\delta >0,N\left(x,\delta \right)$ contains a point of B distinct from x. But this means $x\in A\text{'}$ or $x\in B\text{'}.$ Therefore, $x\in A\text{'}\cup B\text{'}.$
2. Claim. For $A\subseteq ℝ,\left(A^c\right)\text{'}={\left(A\text{'}\right)}^c.$
   “Proof.” $x\in {\left(A\text{'}\right)}^c$ iff $x\notin A\text{'}$
   iff x is not an accumulation point for A
   iff x is an accumulation point for $A^c$
   iff $x\in \left(A^c\right)\text{'}.$
3. Claim. For A,  $B\subseteq ℝ,\left(A-B\right)\text{'}\subseteq A\text{'}-B\text{'}.$
   “Proof.” $\begin{array}{cc}\left(A-B\right)\text{'}=\left(A\cap B^c\right)\text{'}& \langle definitionofA-B\rangle \end{array}$
   $\begin{array}{cc}\subseteq A\text{'}\cap \left(B^c\right)\text{'}& \langle Exercise7\left(b\right)\rangle \end{array}$
   $\begin{array}{cc}\subseteq A\text{'}\cap \left(B\text{'}\right)C& \langle because\left(B^c\right)\text{'}\subseteq {\left(B\text{'}\right)}^c\rangle \end{array}$
   $=A\text{'}-B\text{'}.$
4. Claim. If A is closed, then $A\text{'}\subseteq A.$
   "Proof." Suppose that A is closed. Then $A^c$ is open. Let $x\in A^c.$ Then x is an interior point of $A^c.$ Therefore, there exists $\delta >0,$ so $N\left(x,\delta \right)\subseteq A^c.$ Hence, $N\left(x,\delta \right)\cap A=\varnothing .$ Thus, x is not an accumulation point for A. Because $x\in A^c$ implies $x\notin A\text{'},$ we conclude $A\text{'}\subseteq A.$
5. Claim. If A is a set with an accumulation point, $B\subseteq A,$ and B is infinite, then B has an accumulation point.
   "Proof." First, A is infinite because $B\subseteq A$ and B is infinite. Because A has an accumulation point, by the Bolzano-Weierstrass Theorem A must be bounded. Because $B\subseteq A,$ this means B is bounded. Hence, by the Bolzano-Weierstrass Theorem again, B has an accumulation point.