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Lecture 6: real numbers One extremely useful property of R that follows from the least upper bound property is Theorem 0.1 (Archimedean property of R). Given x, y ∈ R with x 6= 0, there exists n ∈ Z such that nx > y . Proof. First let x, y ∈ R such that x, y > 0 and assume that there is no such n. Then the set {nx : n ∈ N} is bounded above by y. As it is clearly nonempty, it has a supremum s. Then s − x < s, so s − x cannot be an upper bound, giving the existence of some m ∈ N such that s − x < mx . However this implies that s < (m+1)x, so s was actually not an upper bound, contradiction. This proves the statement for the case x < y. The other cases can be obtained from this one by instead considering −x and/or −y. The Archimedean property implies Corollary 0.2 (Density of Q in R). Let x, y ∈ R with x < y. There exists q ∈ Q such that x < q < y. Proof. Apply the Archimedean property to y − x and 1 to find n ∈ Z such that n(y − x) > 1. We can also find m1 > nx and m2 > −nx, so −m2 < nx < m1 . It follows then that there is an m ∈ Z such that m − 1 ≤ nx < m. Finally, nx < m ≤ 1 + nx < ny . Dividing by n we get x < m/n < y. Now we return to countability. Theorem 0.3. The set Q is countable, whereas R is uncountable. Proof. We already know that N × N is countable: this is from setting up the array (1, 1) (2, 1) (3, 1) · · · (1, 2) (2, 2) (3, 2) · · · (1, 3) (2, 3) (3, 3) · · · ··· 1 and listing the elements along diagonals. On the other hand, there is an injection f : Q+ → N × N , where Q+ is the set of positive rationals. One such f is given by f (m/n) = (m, n), where m/n is the “reduced fraction” for the rational, expressed with m, n ∈ N. Therefore Q+ is countable. Similarly, Q− , the set of negative rationals, is countable. Last, Q = Q+ ∪Q− ∪{0} is a union of 3 countable sets and is thus countable. To prove R is uncountable, we will use decimal expansions for real numbers. In other words, we write x = .a1 a2 a3 . . . where ai ∈ {0, . . . , 9} for all i. Since we have not proved anything about decimal expansions, we are certainly assuming a lot here, but this is how things go. Note that each real number has at most 2 decimal expansions (for instance, 1/4 = .2500 . . . = .2499 . . .). Assume that R is countable. Then as there are at most two decimal expansions for each real number, the set of decimal expansions is countable (check this!) Now write the set of all expansions in a list: 1 .a0 a1 a2 . . . 2 .b0 b1 b2 . . . 3 .c0 c1 c2 . . . ··· ··· We will show that no matter what list we are given (as above), there must be a sequence that is not in the list. This implies that there can be no such list, and thus R is uncountable. Consider the diagonal element of the list. That is, we take a0 for the first digit, b1 for the second, c2 for the third and so on: .a0 b1 c2 d3 . . . We now have a rule to transform this diagonal element into a new one. We can use many, but here is one: change each digit to a 0 if it is not 0, and replace it with 9 if it is 0. For example, .0119020 . . . −→ .9000909 . . . Note that this procedure changes the diagonal number into a new one that differs from the diagonal element in every decimal place. Call this new expansion A = .â0 â1 . . . Now our original list contains all expansions, so it must contain A at some point; let us say that the n-th element of the list is A. Then consider the n-th digit ân of A. On the one hand, by construction, ân is not equal to the n-th digit of the diagonal element. On the other hand, by the position in the list, ân equals the n-th digit of the diagonal element. This is a contradiction. 2 Rn for n ≥ 2 A very important extension of R is given by n-dimensional Euclidean space. Definition 0.4. For n ≥ 2, the set Rn is defined as Rn = {~a = (a1 , . . . , an ) : ai ∈ R for all i} . Addition of elements is defined as ~a + ~b = (a1 , . . . , an ) + (b1 , . . . , bn ) = (a1 + b1 , . . . , an + bn ) and multiplication of elements by numbers is c~a = c(a1 , . . . , an ) = (ca1 , . . . , can ), c ∈ R . Note that this definition gives us R for n = 1. On Rn we place a distance, but to do that, we need the existence of square roots. We will take this for granted now, since we will prove it later using continuity. Lemma 0.5. For each √ x ∈ R with x ≥ 0 there exists a unique y ∈ R such that y 2 = x. This element is written y = x. Definition 0.6. On the set Rn we define the norm q |~a| = |(a1 , . . . , an )| = a21 + · · · + a2n and inner product ~a · ~b = (a1 , . . . , an ) · (b1 , . . . , bn ) = a1 b1 + · · · + an bn . Theorem 0.7. Suppose ~a, ~b, ~c ∈ Rn and c ∈ R. Then 1. |~a| ≥ 0 with |~a| = 0 if and only if ~a = 0. 2. |c~a| = |c||~a|. 3. (Cauchy-Schwarz inequality) |~a · ~b| ≤ |~a||~b|. 4. (Triangle inequality) |~a + ~b| ≤ |~a| + |~b|. 5. |~a − ~b| ≤ |~a − ~c| + |~c − ~b|. Proof. Next time. 3