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Solutions to homework 2, SF2736, fall 10.

The solutions shall be delivered at latest on November 15. Nothing else than hand written solutions will be accepted.

You get one bonus point to the written final exam to this course if you solve three out of the four problems below. These bonus points will be added to the points of the first part of the exam. The first part of the exam will consist of five problems, each rated three points, and so in total you can get at most 15 points on part 1. You will pass the exam if you get 15 points or more. (Part 2 of the exam will consist of three problems rated 3 or 4 points each and part 3 will consist of two problems rated 4 or 5 points.)

Give short explanations to the answers you give to the problems below!

1. Let A be an infinite set that is not a countable infinite set, and let B be a subset of A which is countable infinite. Can then A \ B be a countable infinite set.

Solution: Let A1 = B = {b1, b2,...} and A2 = A \ B. If A2 were countable infinite A2 = {a1, a2,...} then, as A = A1 ∪ A2 we would get an enumeration of A by the bijection ϕ from N to A by ½ a if n = 2k , ϕ(n) = k bk if n = 2k + 1 . However, as A is not countable infinite, we can conclude that it cannot be the case that A2 is countable infinite. 2. Is the set of all equivalence relations on the set of natural numbers N a countable infinite set?

Solution: We will use the fact that the set of subsets of N that contain 1 is an infinite set that is not countable. We first consider the set of equivalence relations R that gives just two equivalence classes C1 and C2 = N\C. One of these sets will contain the element 1, and without loss of generality we assume that 1 ∈ C1. Hence, to every such equivalence relation we may associate a unique subset of N that contains the element 1. So these type of relations will not be countable infinite. Trivially, a subset of a countable infinite set must be either finite or countable infinite. We can conclude that the set of all equivalence relations cannot be countable infinite.

3. Let F(A, B) denote the set of functions from A to B and let N denote the set of natural numbers.

(a) Is the set F({1, 2, 3}, N) a countable infinite set?

Solution: To simplify, and without loss of any generality, we may assume that 0 6∈ N. Any member f of F({1, 2, 3}, N) can be described as a 3-tuple

(f(1), f(2), f(3)) ∈ N3 , 2

and any element in N3 will in this way, uniquely, describe an element in F({1, 2, 3}, N). In the same way as we derived an enumeration of the rational numbers, we can get an enumeration of the set N2 and then similarly use this enumeration to get an enumeration of the set

N3 = N2 × N .

So the conclusion is that as N3 is countable infinite, the same must hold for F({1, 2, 3}, N). (b) Is the set F(N, {1, 2, 3}) a countable infinite set?

Solution: The set A = F(N, {1, 2}) is a subset of F(N, {1, 2, 3}). However, A is infinite but not countable, compare Problem 2, and hence F(N, {1, 2, 3}) cannot be countable infinite.

4. Is the set of all injective functions from N to N a countable infinite set?

Solution: Every injective function from N to N can be described as an infinite sequence a1a2a3 ...,

where {ai | i ∈ N} = N and ai 6= aj for i 6= j. Now, suppose that the set of injective functions were countable and consider an enumeration of them: b1 = a11a12a13a14 ..., b2 = a21a22a23a24 ..., b3 = a31a32a33a34 ..., b4 = a41a42a43a44 ..., . . However, we now can define a sequence, that does not belong to this enumeration, by b = a1a2a3 ...,

where a1 is the first natural number that is distinct from a11, and recursively, ak is the least integer in the set

N \{a11, a22, . . . , ak−1,k−1} .

So, the above enumeration of injective functions can impossibly be an enumeration of all injective functions, and it will thus be impossible to enumerate all injective functions.