By Serret J.-A.

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The question then arises: does α∈A have any members? At first glance, it appears that it must; since each Bα is non-empty, there exists cα in Bα , and we can take (cα )α∈A as an element of α∈A . The problem is that we must do this simultaneously, for all α ∈ A. We require a further axiom to say that this is valid. 9 The axiom of choice 27 Axiom 10: The axiom of choice This states that if (Bα )α∈A is a family of non-empty sets then α∈A (Bα ) is non-empty: there exists a function c, a choice function, from A to ∪α∈A Bα , such that cα = c(α) ∈ Bα for each α ∈ A.

In 1963, Cohen showed that there are models of ZF in which the axiom of choice does not hold. Thus the axiom of choice is independent of the axioms of ZF, and cannot be proved or disproved, starting from ZF. We can add the axiom of choice to obtain a stronger axiom system ZFC. Within this, there are further statements that cannot be proved or disproved, such as the continuum hypothesis (which states that if A is an uncountable subset of the set R of real numbers, then there exists a bijection f : A → R).

M+n=n+m (commutativity) (m + n) + p = m + (n + p) (associativity) if m + n = p + n then m = p (cancellation) if m + n = 0 then m = n = 0. The proof uses induction many times over. (i) We prove this in three steps. First we show that m + 0 = 0 + m for all m. We use induction. Let U = {m ∈ Z+ : 0 + m = m + 0}. Then 0 ∈ U , since 0 + 0 = 0 + 0. Suppose that m ∈ U . Then (m + 1) + 0 = m + 1, and 0 + (m + 1) = (0 + m) + 1, by (∗ ), and (0 + m) + 1 = m + 1. Thus m + 1 ∈ U , and so U = Z+ , by induction.

### Cours de calcul differentiel by Serret J.-A.

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