By Terence Tao
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Extra info for Analysis I (Volume 1)
4, this means that at least one of x E AU B or x E Cis true. We now divide into two cases. 4 again we have x E AU (B U C). 4 again x E A or x E B. 4 we have x E B U C and hence x E A U (B U C). Thus in all cases we see that every element of (AU B) U C lies in AU (B U C). A similar argument shows that every element of A U (B U C) lies in (A U B) U C, and so (AU B) U C =AU (B U C) as desired. 0 Because of the above lemma, we do not need to use parentheses to denote multiple unions, thus for instance we can write AUBUC instead of (AU B) U C or AU (B U C).
Esired. This closes the induction. 0 We now discuss how addition interacts with positivity. 7 (Positive natural numbers). A natural number n is said to be positive iff it is not equal to 0. l). 8. 4). Proof. We use induction ·on b. If b = 0, then a + b = a + 0 = a, which is positive, so this proves the base case. Now suppose inductively that a+ b is positive. 3, and is hence positive. This 0 closes the induction. 9. If a and bare natural numbers such that a+b = 0, then a = 0 and b = 0. 31 2. 2.
Now let us suppose inductively that a(b +c) = ab + ac, and let us prove that a(b + (c++)) = ab + a(c++ ). The left-hand side is a((b +c)++)= a(b+c) +a, while the right-hand side is ab+ac+a = a(b+c) +a by the induction hypothesis, and so we can close the induction. 5 (Multiplication is associative). For any natuml numbers a, b, c, we have (ax b) x c =ax (b x c). Proof. 3. 6 (Multiplication preserves order). If a, b are natuml numbers such that a < b, and c is positive, then ac < be. 3. Multiplication 35 Proof.
Analysis I (Volume 1) by Terence Tao