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math104-f21:hw5

HW 5

0. Correct your mistakes in midterm 1, you don't need to submit the correction.

1. Prove that there is a sequence in R\R, whose subsequential limit set is the entire set R\R.

2. Prove that lim sup(an+bn)lim sup(an)+lim sup(bn)\limsup(a_n+b_n) \leq \limsup(a_n) + \limsup(b_n) , where (an)(a_n) and (bn)(b_n) are bounded sequences in R\R.

3. Give an explicit way to enumerate the set Z\Z, and then Z2\Z^2.

4. Prove that the set of Z\Z-coefficient polynomials is countable.

5. Prove that the set of maps {f:N{0,1}}\{f: \N \to \{0,1\}\} is not countable.

Solution

1. Let (an)(a_n) be any sequence that enumerate of Q\Q. For any xRx \in \R, to show that xx is a subsequential limit of ana_n, we only need to show that for any ϵ>0\epsilon>0 and N>0N>0, there is a n>Nn>N, with anx<ϵ|a_n - x| < \epsilon. Suppose there is no such an nn, i.e., for all n>Nn > N, we have anx>ϵ|a_n - x|>\epsilon, then that means there are at most NN rational numbers within [xϵ,x+ϵ][x-\epsilon, x+\epsilon], which is absurd.

2. Let An=sup{ammn}A_n = \sup \{ a_m | m \geq n\}, and Bn=sup{bmmn}B_n = \sup \{ b_m | m \geq n\}, and Cn=sup{am+bmmn}C_n = \sup \{a_m + b_m \mid m \geq n\}. We claim that CnAn+BnC_n \leq A_n + B_n. Indeed, for any mnm \geq n, am+bmAn+Bna_m + b_m \leq A_n + B_n, hence An+BnA_n + B_n is an upper bound of the set {am+bmmn}\{a_m + b_m \mid m \geq n\}, thus CnAn+BnC_n \leq A_n + B_n. Hence, taking limit nn\to \infty, we get limnCnlimAn+limBn\lim_n C_n \leq \lim A_n + \lim B_n.

3. To enumerate Z\Z, we can do 0,1,1,+2,2,0, 1, -1, + 2, -2, \cdots .

To enumerate Z2\Z^2, we define a function f:Z2Nf: \Z^2 \to \N, f(a,b)=a+bf(a,b) = |a| + |b|. For nNn \in \N, let An=f1(n)={(a,b)Z2a+b=n}A_n = f^{-1}(n) = \{(a,b) \in \Z^2 \mid |a|+|b|=n\}. Then Z2=nNAn\Z^2 = \sqcup_{n \in \N} A_n, and each AnA_n is a finite set. We can enumerate A0A_0, then A1A_1, …

4. Let PnP_n denote the set of integer coefficient polynomials of at most degree nn, Pn={a0+a1x++anxaiZ}P_n = \{a_0 + a_1 x + \cdots + a_n x \mid a_i \in \Z \}. Then the set of integer coefficient polynomial P=n=0nPnP = \cup_{n=0}^n P_n, each each Pn=Zn+1P_n = \Z^{n+1}. Since finite product of countable set is countable, we have PnP_n is countable for all nn. Since countable union of countable sets are countable, we have PP being countable.

5. This is done is class. See Rudin Thm 2.14.

math104-f21/hw5.txt · Last modified: 2022/01/11 18:31 by 24.253.46.239