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0 Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup _{n=1}^infty I_nright}$ . Let $E=bigcup_{n=1}^infty E_n$ . Suppose $m^*(E)=0$ . Prove that $m^*(E_n)=0$ for every $n$ . My attempt: Since $m^*(E)=0$ , $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$ . QED Is this correct? real-analysis proof-verification share | cite | improve this question asked Dec 3 '18 at 0:04 Thomas 713 4 16 ...

31 12 Let ${cal P}$ be the set of prime numbers. Define a subset ${cal P}'={p_1,p_2,p_3,cdots}$ of ${cal P}$ by setting $p_1=2$ and defining $p_{n+1}$ to be the smallest element of ${cal P}$ dividing $1+p_1cdots p_n$ . Is there any obstruction to ${cal P}'={cal P}$ ? nt.number-theory prime-numbers algorithms share | cite | improve this question edited Dec 4 '18 at 0:58 Konstantinos Kanakoglou 3,150 2 11 33 asked Dec 2 '18 at 22:19 Zidane 271...