• Uncategorized

Mathematical Real Analysis

MathematicalReal Analysis

Question101

  1. To prove that ∑ is compact,

LetH∞be the set of real sequences such that each element in each sequencehas |an|≤1.The metric is defined as

d({an},{bn})=∑n=1∞|an−bn|2n.

Toprove that H∞is a compact metric space, we have

Ifwe have a sequence {{an}(k)}in H∞,then for all i,the real sequence {a(k)i}has a convergent subsequence, since it is bounded by 1. So we can geta convergent subsequence {a(kj)1},and then a convergent subsequence of {a(kj)2},and continue taking subsequences of subsequences until we have aconvergent subsequence of (a1,a2,a3,…,an)(k)with nsome positive integer if we stop taking subsequences at the nthsubsequence this gives a sequence {xn}where xnis the limit of the nthconvergent subsequence of {a(k)n}.Therefore we show that the sequence in H∞converges to {xn}.

  1. To prove that ∑ is homeomorphic to the cantor set, we have:

Foreach positive integer nlet Dn={0,1}with the discrete topology, also when we let

X=∏n=1∞Dnwiththe product topology.

Elementsof Xare infinitesequences of 0’s and 1’s, so (0,0,0,1) and 0,1,1,1,1,1,1) are notelements of X padding these with an infinite string of 0’s to get(0,0,0,1,0,0,0,0,…) and (0,1,1,1,1,1,1,0,0,0,0,…), however, youdo get points of X.A more interesting point of Xis the sequence (pn)n,where pn=1if nis prime, and pn=0if n

isnot prime.

Weare therefore to show that C,with the topology that it inherits from R, is homeomorphic to X.To do that, we must find a bijection h:C→Dsuch that both hand h−1are continuous. The suggestion that you found is to let h(∑n=1∞an3n)=(a12,a22,a32,…).

Weought to bear in mind that an2= {0, 1, if&nbspan=0if&nbspan=2,so this really does define a point in X.This really is a bijection: if b=(bn)n∈X,h−1 (b)=∑n=1∞2bn3n.

Question109

  1. “The triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side. To show that the ultra metric property simply implies the triangle inequality,” we have

“If|x + y| ≤ max {|x|, |y|}, then the equality occurs when |x| ≠|y|.”

“Assumingthat |x| ˃ |y|, it implies that |x + y| ≤ |x|”

“Also,|x| = | (x+ y) – y| ≤ max {| x+ y|, |y|}”

“Thevalue of max {| x+ y|, |y|} ≠ |y|, we therefore have |x| ≤ |y|which is contrary to our initial assumption above.”

Thus,max {| x+ y|, |y|} = |x+ y| and |y| ˃ |x| and from the initialinequality, we have tat | x| ≤ |x + y| ≤|x| and so | x+ y| = |x|

  1. “An isosceles triangle is one that has two sides of equal length. In some cases, it is specified as having at least two sides of equal length, the latter version thus including the equilateral triangle as a special case.”

“Toshow that all triangles in the ultrametric space are isosceles,” wehave

“Thisis d( x , y ) = d ( y , z ) {displaystyle d(x,y)=d(y,z)} d=(x, y) = d{x, z) or d(x, z) = d(y, z) or d(x, y) or d(x, y) = d(x,z)”d( x , y ) = d ( z , x ) {displaystyle d(x,y)=d(z,x)} d ( x , z ) = d( y , z ) {displaystyle d(x,z)=d(y,z)} d ( x , y ) = d ( z , x ){displaystyle d(x,y)=d(z,x)}

  1. “A topological space X is totally disconnected if the only connected subsets Y of X is of the form Y = {y}.”

“Forany two distinct elements xand yof X,there exist disjoint open neighborhoods Uof xand Vof ysuch that Xis the union of Uand V. Since the set is separated, it is totally disconnected.”

  1. To prove that ∑ is homeomorphic to the cantor set, we have:

Foreach positive integer nlet Dn={0,1}with the discrete topology, also when we let

X=∏n=1∞Dnwiththe product topology.

Elementsof Xare infinitesequences of 0’s and 1’s, so (0,0,0,1) and 0,1,1,1,1,1,1) are notelements of X padding these with an infinite string of 0’s to get(0,0,0,1,0,0,0,0,…) and (0,1,1,1,1,1,1,0,0,0,0,…), however, youdo get points of X.A more interesting point of Xis the sequence (pn)n,where pn=1if nis prime, and pn=0if n

isnot prime.

Weare therefore to show that C,with the topology that it inherits from R, is homeomorphic to X.To do that, we must find a bijection h:C→Dsuch that both hand h−1are continuous. The suggestion that you found is to let h(∑n=1∞an3n)=(a12,a22,a32,…).