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10, 000! > 2 · 10 ... , , ... In Chapter VI we derive a formula which can be used to estimate this rapid growth. 14(a). 2 Verify the following equalities using induction: (a) n k=0 k = n(n + 1)/2, n ∈ N. (b) n k=0 k2 = n(n + 1)(2n + 1)/6, n ∈ N. 3 Verify the following inequalities using induction: (a) For all n ≥ 2, we have n + 1 < 2n . (b) If a ∈ N with a ≥ 3, then an > n2 for all n ∈ N. 4 Let A be a set with n elements. Show that P(A) has 2n elements. 44 I Foundations 5 (a) Show that m! (n − m)!
The fiber f −1 (y) is simply the solution set x ∈ X ; f (x) = y of the equation f (x) = y. This could, of course, be empty. 8 Proposition The following hold for the set valued functions induced from f : (i) A ⊆ B ⊆ X = ⇒ f (A) ⊆ f (B). (ii) Aα ⊆ X ∀ α ∈ A = ⇒ f α Aα = α f (Aα ). (iii) (iv) (i ) (ii ) ⇒ f α Aα ⊆ α f (Aα ). Aα ⊆ X ∀ α ∈ A = c A⊆X= ⇒ f (A ) ⊇ f (X)\f (A). A ⊆B ⊆Y = ⇒ f −1 (A ) ⊆ f −1 (B ). Aα ⊆ Y ∀ α ∈ A = ⇒ f −1 α Aα = α f −1 (Aα ). ⇒ f −1 (iii ) Aα ⊆ Y ∀ α ∈ A = α −1 Aα = c α f −1 (Aα ).
Then there is a bijective function from {1, . . , m} to {1, . . , n} if and only if m = n.