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--- math104-f21:hw5 [2021/09/24 16:19]
pzhou created
+++ math104-f21:hw5 [2022/01/11 18:31] (current)
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 . Prove that there is a sequence in  $\R$ , whose subsequential limit set is the entire set  $\R$ .
-. Prove that  $\limsup(a_n+b_n) \leq \limsup(a_n) + \limsup(b_n)$ .
+. Prove that $\limsup(a_n+b_n) \leq \limsup(a_n) + \limsup(b_n)  $, where$ (a_n) $and$ (b_n) $are bounded sequences in$ \R$.
 . Give an explicit way to enumerate the set  $\Z$ , and then  $\Z^2$ .
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 . Prove that the set of maps  $\{f: \N \to \{0,1\}\}$  is not countable.
+===== Solution =====
+. Let  $(a_n)$  be any sequence that enumerate of  $\Q$ . For any  $x \in \R$ , to show that  $x$  is a subsequential limit of  $a_n$ , we only need to show that for any $\epsilon>0 $and$ N>0 $, there is a$ n>N $, with$ |a_n - x| < \epsilon $. Suppose there is no such an$ n $, i.e., for all$ n > N $, we have$ |a_n - x|>\epsilon$, then that means there are at most  $N$  rational numbers within $[x-\epsilon, x+\epsilon]$, which is absurd.
+. Let  $A_n = \sup \{ a_m | m \geq n\}$ , and  $B_n = \sup \{ b_m | m \geq n\}$ , and  $C_n = \sup \{a_m + b_m \mid m \geq n\}$ . We claim that  $C_n \leq A_n + B_n$ . Indeed, for any  $m \geq n$ ,  $a_m + b_m \leq A_n + B_n$ , hence  $A_n + B_n$  is an upper bound of the set  $\{a_m + b_m \mid m \geq n\}$ , thus  $C_n \leq A_n + B_n$ . Hence, taking limit  $n\to \infty$ , we get  $\lim_n C_n \leq \lim A_n + \lim B_n$ .
+. To enumerate  $\Z$ , we can do  $0,  1, -1,  + 2, -2,  \cdots$ .
+To enumerate  $\Z^2$ , we define a function  $f: \Z^2 \to \N$ ,  $f(a,b) = |a| + |b|$ . For  $n \in \N$ , let   $A_n = f^{-1}(n) = \{(a,b) \in \Z^2 \mid |a|+|b|=n\}$ . Then  $\Z^2 = \sqcup_{n \in \N} A_n$ , and each  $A_n$  is a finite set. We can enumerate  $A_0$ , then  $A_1$ , ...
+. Let  $P_n$  denote the set of integer coefficient polynomials of at most degree  $n$ ,   $P_n = \{a_0 + a_1 x + \cdots + a_n x \mid a_i \in \Z \}$ . Then the set of integer coefficient polynomial  $P = \cup_{n=0}^n P_n$ , each each  $P_n = \Z^{n+1}$ . Since finite product of countable set is countable, we have  $P_n$  is countable for all  $n$ . Since countable union of countable sets are countable, we have  $P$  being countable.
+. This is done is class. See Rudin Thm 2.14.
+{{math104-f21:pasted:20211001-220110.png}}

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