crdavis1219
New member
- Joined
- Mar 26, 2019
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We recursively define a sequence of subsets of [MATH]\mathbb Z[/MATH] as follows:
Let [MATH]S_0=\{0\}[/MATH], and let [MATH]S_{n+1}=\{2m: m \in S_n\} \cup \{2m+1: m \in S_n\}[/MATH] for all [MATH]n \geq 0[/MATH]. (So [MATH]S_1=\{0,1\}[/MATH], [MATH]S_2=\{0,1,2,3\}[/MATH],...)
1. Find [MATH]S_3[/MATH]. This part I have already solved. My solution: [MATH]S_3=\{0,1,2,3,4,5,6,7\}[/MATH]2. I claim that [MATH]7 \in S_n[/MATH] for all [MATH]n \geq 3[/MATH]. It turns out to be easier to prove the following stronger statement: "[MATH]\{0,1,3,7\} \subseteq S_n[/MATH] for all [MATH]n \geq 3[/MATH]". Prove by induction.
3. Now consider the infinite union, [MATH]\bigcup_{n=0}^{\infty} S_{n}=S_0 \cup S_1 \cup S_2 \cup ...[/MATH] Find this set (List its elements, nicely, possibly using "..."). Briefly explain why your answer is correct.
I'd greatly appreciate it if someone could help me solve parts 2 and 3!
Let [MATH]S_0=\{0\}[/MATH], and let [MATH]S_{n+1}=\{2m: m \in S_n\} \cup \{2m+1: m \in S_n\}[/MATH] for all [MATH]n \geq 0[/MATH]. (So [MATH]S_1=\{0,1\}[/MATH], [MATH]S_2=\{0,1,2,3\}[/MATH],...)
1. Find [MATH]S_3[/MATH]. This part I have already solved. My solution: [MATH]S_3=\{0,1,2,3,4,5,6,7\}[/MATH]2. I claim that [MATH]7 \in S_n[/MATH] for all [MATH]n \geq 3[/MATH]. It turns out to be easier to prove the following stronger statement: "[MATH]\{0,1,3,7\} \subseteq S_n[/MATH] for all [MATH]n \geq 3[/MATH]". Prove by induction.
3. Now consider the infinite union, [MATH]\bigcup_{n=0}^{\infty} S_{n}=S_0 \cup S_1 \cup S_2 \cup ...[/MATH] Find this set (List its elements, nicely, possibly using "..."). Briefly explain why your answer is correct.
I'd greatly appreciate it if someone could help me solve parts 2 and 3!