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Show that a subset of a countable set is countable. 4-3. Let X be a countable set. If Y is a finite set, show that the cartesian product X × Y = {(x, y) : x ∈ X, y ∈ Y } is countable. 10(d) or a modification of its proof to show that this is still true if Y is countably infinite.

B) This time we proceed by Induction on m. Consider the statement Q(m) : For n ∈ N0 , if there is a surjection m −→ n then m n. When m = 0, there is exactly one function ∅ −→ ∅ (the identity function) and this is a bijection; if n > 0 there are no surjections ∅ −→ n. So Q(0) is true. Suppose that Q(k) is true for some k ∈ N0 and let f : k + 1 −→ n be a surjection. , f (j) = f (j) for j ∈ k. There are two cases to deal with: (i) f is a surjection, (ii) f is a not a surjection. (i) By the assumption that Q(k) is true, k n which implies that k + 1 n.

Then |H| Proof. The idea is to divide up G into disjoint subsets of size H. We do this by defining for each x ∈ G the left coset of x with respect to H, xH = {g ∈ G : x−1 g ∈ H} = {g ∈ G : g = xh for some h ∈ H}. We need the following facts. i) For x, y ∈ G, xH ∩ yH = ∅ ⇐⇒ xH = yH. This is seen as follows. If xH = yH then xH ∩ yH = ∅. Conversely, suppose that xH ∩ yH = ∅. If yh ∈ xH for some h ∈ H, then x−1 yh ∈ H. For k ∈ H, x−1 yk = (x−1 yh)(h−1 k), which is in H since x−1 yh, h−1 k ∈ H and H is a subgroup of G.

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