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sols21 - 21 NOVEMBER 17TH CARDINALITY November 17th...

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Unformatted text preview: 21. NOVEMBER 17TH: CARDINALITY III 49 21. November 17th: Cardinality III 21.1. Complete the alternative proof of Lemma 20.3 sketched in §21.1. That is: with notation as in the end of §21.1, prove that ψ does not equal ϕs for any s ∈ S . Answer. Recall that ψ : S → 0, 1 is defined by ￿ 1 if ϕs (s) = 0 ψ ( s) = 0 if ϕs (s) = 1 Now let s ∈ S . Then either ϕs (s) = 0, or ϕs (s) = 1. • If ϕs (s) = 0, then ψ (s) = 1, so ψ (s) ￿= ϕs (s). This shows that ψ ￿= ϕs in this case. • If ϕs (s) = 1, then ψ (s) = 0, so ψ (s) ￿= ϕs (s). This shows that ψ ￿= ϕs in this case as well. This shows that ϕs ￿= ψ for all s ∈ S , which was the last step to verify in order to conclude the alternative proof of Lemma 20.3. ￿ Of course this argument mimics precisely the analogous part of the proof of Lemma 20.3. Whether you saw how to do this or not, please make sure you understand that the ‘diagonal argument’ is just another way to formulate the idea behind the original proof of Lemma 20.3. 21.2. Let A, B be sets. Prove that |A| ≤ |B | if and only if there exists a surjective function B → A. (Argue as in the beginning of the proof of Proposition 21.3.) Answer: By definition, |A| ≤ |B | if and only if there is an injective function f : A → B. If f : A ￿→ B is injective, then it has a left-inverse g : B → A (Theorem 12.11); as g ◦ f = idA , g has a right inverse, hence it is surjective. Thus, if |A| ≤ |B |, then there is a surjective function B ￿ A. Conversely, if g : B ￿ A is surjective, then it has a right-inverse f : A → B . As g ◦ f = idA , f has a left inverse, hence it is injective. This shows that if there is a surjective function B → A, then there is an injective function A ￿→ B , and therefore |A| ≤ |B |. ￿ 21.3. Explain why R ∼ 2N . (You don’t need to be too rigorous, but you should = provide a convincing sketch of an argument.) Answer: First recall that R ∼ (0, 1) (Example 14.4); proving that R ∼ 2N is then = = N equivalent to proving that (0, 1) and 2 have the same cardinality. Now, every real number between 0 and 1 has a binary expansion r = 0.a0 a1 a2 a3 a4 · · · where each ai is either 0 or 1. So r determines a function N → {0, 1}, defined by n ￿→ an . Conversely, every function N → {0, 1} determines a real number between 0 and 1, by providing its binary expansion. Making this sketch into a real proof requires a bit of care, due to the same problem raised after the proof of Theorem 21.5: for example, the binary expansions 0.1000000 · · · , 0.01111111 · · · 50 determine the same real number (1/2, as it happens). And the empty set would correspond to 0.000 · · · = 0, which is not in (0, 1). On the other hand, these problems only affects a countable infinity of numbers, so they do not change the result. ￿ 21.4. Prove that any finite product A1 × · · · × Ar of ￿ countable sets is countable. Question: is a countable product of countable sets, n∈N An , necessarily countable? Answer: —We can view A1 × · · · × Ar−2 × Ar−1 × Ar as (· · · (((A1 × A2 ) × A3 ) × A4 ) × · · · ) × Ar . By the argument at the end of Example 21.7, A1 × A2 is countable if A1 and A2 are countable. But then (A1 × A2 ) × A3 is also countable, since it is the product of two countable sets, again by the first part. And then ((A1 × A2 ) × A3 ) × A4 ) is the product of two countable sets, so it is countable. Etc. ‘Etc.’ does not sound very formal, and indeed the proper way to write up this solution is by using induction. I’ll write a more formal version of the same argument here, as an illustration. The claim is that if Ai , i = 1, 2, . . . are all countable sets, then for all n ≥ 1, A1 × · · · × An is countable. The case n = 1 needs no proof, and the case n = 2 is covered by the end of Example 21.7. Now argue by induction: n = 2 works as seed; assume that n ≥ 2, and that A1 × · · · × An is countable, and prove that A1 × · · · × An+1 is countable. We have that An+1 is countable by hypothesis, and A1 × · · · × An+1 ∼ (A1 × · · · × An ) × An+1 = (write out a bijection if you don’t recall this!). This shows that A1 × · · · × An+1 is isomorphic to the product of two countable sets, and it follows that it is countable (Example 21.7 again) as needed. This verifies the induction steps, concluding the proof. —An infinite countable product of countable sets need not be countable. Indeed, 2N is such a product (it may be viewed as infinite sequences of 0s and 1s, that is, as ￿ N n∈N An , where each An is the set {0, 1}. We have proved that 2 is not countable, Corollary 21.4. 21.5. We have seen that the set 2N of subsets of N is uncountable (Corollary 21.4). Since Q ∼ N, it follows that the set 2Q of all subsets of Q is uncountable. = Let’s throw away ‘most’ of these subsets: consider the set of subsets A ￿ Q with the property that x < y for all x ∈ A and all y ∈ A. Is this set countable or / uncountable? Answer: This set is uncountable, because it contains an uncountable set. Indeed, consider the set of A ⊆ Q as in the statement, but required to be nonempty and to not have a maximum. For every such set A, letting B = Q ￿ A gives a Dedekind cut (A|B ). Conversely if (A|B ) is a Dedekind cut, then the set A satisfies all the given requirements. The set of Dedekind cuts is uncountable, because it is isomorphic to R. Therefore, the set specified in the problem is uncountable, because it contains a copy of D. ...
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