Topology Question Prove the following statements SolutionPro

Topology Question:

Prove the following statements

Solution

Proof for set of Integers complete

Complete means if we start with a cauchy sequence, it actually converges. So, let xn be a cauchy sequence in Z. Then, for for any >0, NN,n,mN,|xnxm|<. So, lets pick an . Say. 110. Then after a certain point, |xnxm|<110. So we have 110<xnxm<110

, hence

xm110<xn<xm+110

. But integers are at least 1 apart, so the only way for this to be true is for xn=xm. Thus, nN,xn=xN

So your sequence is eventually constant, and thus converges to that constant.

Proof For Rtaional Numbers Set Incomplete:

Construct a sequence of rationals that would converge to, say, 2–. For example, x1=1, xn+1=xn+2/xn2=x2n+22xn

.

Then, if this converges to a rational, let the limit be r=a/b

, and let xn=an/bn. Then an+1/bn+1=an/bn+2bn/an2=a2n+2b2n2anbn so an+1=a2n+2b2n and bn+1=2anbn

.

a2n+12b2n+1=(a2n+2b2n)22(2anbn)2=a4n+4a2nb2n+4b4n8a2nb2n=a4n4a2nb2n+4b4n=(a2n2b2n)2

If we start with x1=1

, then a1=b1=1, so (a212b21)2=1. Therefore a2n2b2n=1 for all n>1

.

Since r

is rational, r=a/b and 2=r2=a2/b2

. Therefore

1=a2n2b2n=a2n(a2/b2)b2n=a2nb2a2b2nb2=(anbabn)(anb+abn)b2(anb+abn)b2since (anbabn)1>anbb2=anb

Therefore, anb

. But, since an+1=a2n+2b2n, an gets arbitrarily large. This is a contradiction, so the assumption that the sequence converges to a rational is false