Both sides previous revision
Previous revision
Next revision
|
Previous revision
|
math104-f21:hw5 [2021/09/25 16:34] pzhou |
math104-f21:hw5 [2022/01/11 18:31] (current) 24.253.46.239 ↷ Links adapted because of a move operation |
| |
5. Prove that the set of maps {f:N→{0,1}} is not countable. | 5. Prove that the set of maps {f:N→{0,1}} is not countable. |
| |
| ===== Solution ===== |
| |
| 1. Let (an) be any sequence that enumerate of Q. For any x∈R, to show that x is a subsequential limit of an, we only need to show that for any $\epsilon>0andN>0,thereisan>N,with|a_n - x| < \epsilon.Supposethereisnosuchann,i.e.,foralln > N,wehave|a_n - x|>\epsilon$, then that means there are at most N rational numbers within $[x-\epsilon, x+\epsilon]$, which is absurd. |
| |
| 2. Let An=sup{am∣m≥n}, and Bn=sup{bm∣m≥n}, and Cn=sup{am+bm∣m≥n}. We claim that Cn≤An+Bn. Indeed, for any m≥n, am+bm≤An+Bn, hence An+Bn is an upper bound of the set {am+bm∣m≥n}, thus Cn≤An+Bn. Hence, taking limit n→∞, we get limnCn≤limAn+limBn. |
| |
| 3. To enumerate Z, we can do 0, 1,−1, +2,−2, ⋯. |
| |
| To enumerate Z2, we define a function f:Z2→N, f(a,b)=∣a∣+∣b∣. For n∈N, let An=f−1(n)={(a,b)∈Z2∣∣a∣+∣b∣=n}. Then Z2=⊔n∈NAn, and each An is a finite set. We can enumerate A0, then A1, ... |
| |
| 4. Let Pn denote the set of integer coefficient polynomials of at most degree n, Pn={a0+a1x+⋯+anx∣ai∈Z}. Then the set of integer coefficient polynomial P=∪n=0nPn, each each Pn=Zn+1. Since finite product of countable set is countable, we have Pn is countable for all n. Since countable union of countable sets are countable, we have P being countable. |
| |
| 5. This is done is class. See Rudin Thm 2.14. |
| {{math104-f21:pasted:20211001-220110.png}} |
| |