Mathematical Real Analysis

  • Uncategorized

MathematicalReal Analysis

MathematicalReal Analysis



infS =−1. Let 1/n−1/m be an arbitrary element in S Then,1/n−1/m≥1/n−1 &gt−1. So −1 is a lower bound for S. Let ²&gt 0. There exists n0 ∈N such that 1/n0 &lt ². Now, 1/n0−1 &lt ²−1 =−1 + ² and1/n0−1∈S.Thus, −1 = inf S. We claim supS = 1. Note that S = −S and inf S =−sup{−s : s ∈S}. We get −1 = inf S = −sup{−s : s∈S}=−supS,which implies supS = 1.


LetS be a nonempty bounded set in R. Thus, S has an infimum and asupremum. Let v = supS. We need to show that bv = inf S. Let bs be anarbitrary element of bS. Then, s ∈S and so s ≤ v. But this implies that bs ≥ bv. Thus, bv is alower bound for bS. Let r be an arbitrary element of R such that bv &ltr. Then v &gt r/b. Since v = supS, there exists an element s0 ∈Ssuch that s0 &gt r/b. But this implies that for the element bs0 ∈bS,bs0 &lt r. Thus, we have shown that bv = inf bS = bsupS.

Letw = inf S. We need to show that bw = supS. Let bs be an arbitraryelement of bS. Then, s ∈S and so s ≥ v. But this implies that bs ≤ bw. Thus, bw is alower bound for bS. Let r be an arbitrary element of R such that bw &gtr. Then w &lt r/b. Since w = inf S, there exists an element s0 ∈S such that s0 &lt r/b. But this implies that, for the element bs0 ∈bS, bs0 &gt r. Thus, we have shown that bw = supbS = binf S.Hencethe proof.


Letα be any Dedekind cut. Let x∈α + 0∗. Then there exist r ∈α and s∈0∗with x = r + s. Since s &lt 0, we have x &lt r. since α is closedunder &lt, we have x∈α.This argument shows α + 0∗⊆α. Let x ∈α. Since α has no least rational number, there must be somerational number c ∈α with x &lt c. Then x−c is rational and less than 0, so x−c ∈0∗. We have x = c + (x−c), so x∈α+ 0∗. This argument shows α⊆α+ 0∗. Hence α + 0∗ = α


Whenn=1,the inequality becomes:u-y=(u-y)(1+uy+………….+1)

Whenn=2,the inequality becomes:u2-y2=(u-y)(u+y+………..y)

Sincethe first equation is the square root of the second one, ourassertion holds.


First,we prove by induction that for any positive integer n and for any b &gt0, there exists 2 n √b. This is true for n = 1.The set E = {t &gt0 |t2 &lt a} has a least upper bound x = supE ≥ 0. We claim thatx2 = a. Indeed, let 0 &lt ε &lt 1 and define δ = ε/ (2x + 1).Clearly, 0 &lt δ ≤ ε &lt 1. There exists an element t ∈E such that x−δ &lt t ≤ x. We have x2 &lt (t + δ) 2 = t2 +2tδ + δ2 &lt a + (2t + 1) δ ≤ a + ε. Since the inequality x2 &lta + ε holds for any ε ∈(0, 1), we conclude that x2 ≤ a. On the other hand, x + δ / ∈E, so a ≤ (x + δ) 2&lt x2 + (2x + 1) δ = x2 + ε. Since theinequality x2 + ε &gt a holds for any ε ∈(0, 1), we conclude that x2 ≥ a. We have proved the equality x2 =a. In particular, x =√a &gt 0. Suppose that for some positiveinteger k the number 2k √c is defined for every c &gt 0. Let b bea positive number by inductive hypothesis, for c = √b there existsy &gt 0 such that y2k = √b. From this it follows that b = √b2 =y2·2k = y2k+1

andthen y = 2 k+1 √b that is also true for n = k + 1. Finally, weshall prove that for any positive integer n≥ 2 and for any a &gt 0there exists n √a &gt 0. We proceed by induction on n. Our theoremis true for n = 2 since any positive number a, has a square root.Suppose that n−1 √c exists for any c &gt 0 and for some n ≥ 3.Let a &gt 0 and define the set E = {t ≥ 0 |tn &lt a}. By this,set has a supremum, say x = supE ≥ 0. Define m = 2n −n. Then m +n = 2n and m is a positive integer, since 2n &gt n. There exists apositive y for which y2n = ym+n = axm. Observe that yn = y−mym+n =y−maxm =x ym a (2.1) We claim that xn = a. Indeed, suppose that xn&lt a. Then xm+n &lt axm = ym+n. This implies that x &lt y. On theother hand, from (2.1), yn &lt a and then y ∈E, so y ≤ x. But y &gt x and we get a contradiction. Now, supposethat xn &gt a. Then xm+n &gt axm = ym+n and this implies that x &gty. On the other hand, (2.1) implies that yn &gt a. Now, if t ∈E, then tn &lt a &lt yn. This inequality implies that t &lt y forany t ∈E. This says us that y is an upper bound of E that is less than x =supE and we again get a contradiction. We have proved the equality xn= a. In particular x = n √a &gt 0.


Supposethat (xn) is a sequence such that xn → x and xn → x0 as n →∞.Let &gt 0. Then there exist N,N0 ∈Nsuch that |xn −x| &lt 2 for all n &gt N, |xn −x0| &lt2 for alln &gt N0.

Chooseany n &gt max{N,N0}. Then, by the triangle inequality,|x−x0|≤|x−xn|+|xn −x0| &lt 2 + 2 &lt. Since thisinequality holds for all &gt 0, we must have |x−x0| = 0 (otherwisethe inequality would be false for = |x−x0|/2 &gt 0), so x = x0.



Thelimit exists since it is a sequence


Let` l= limsupan. This number is well defined since it is the limit ofa decreasing sequence of bounded numbers. We first claim that `satisfies the defining properties of a. Let Mn = sup{ak : kgen.Choose N so that ` l≤ Mn &ltl`+. Then, there are no numbers akwith k &gt N with ak &gtl`+. Hence, there are finitely many n suchthat an &gtl` + . Next, if we choose N large enough, Mn &gt `−lwhen n &gt N. Hence, for each n &gt N, there must be a number kn ≥n with akn &gtl`−. This gives an infinite sequence of numbers aknwith akn &gt `−. So there is such a number (`). Suppose there wasanother such number, call it b. Then for large enough n, an ≤ b + and hence ` ≤ b + . But also there are infinitely many n so thatan &gt b−. Hence ` &gt b−. But is arbitrary so b = l` and thenumber a = b = l` in the theorem is unique.


Inthis case, the limit does not exist

Close Menu