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Chapter 2 Set Theory

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** Discrete Mathematics **

CHAPTER 2

Let the universal set be Z. Let A={1,2,3,4},B={2k+1:kZ} (the set of odd integers) and C={nZ:5n0}. Write out the following:
(i) AB,AC,BC
(ii) AB,AC.
(iii) AB,BC.
(iv) Bc,Cc.

Show that (AB)(AB)=(AB)(BA).

Calculate AB where A is the unit disk ( AC ) and B={f(z):zC,|z|<1,f(z)=az1a¯z}

Let f and g be the functions from the set of integers to the set of integers defined by
f(x)=ex and g(x)=sinx. What is the composition of f and g ? What is the composition of g and f?

Let E={x:f(g(x))logsinx3log2,x(0,π/2)} , find E.

Let E denote the set of even integers and O denote the
set of odd integers. Determine each of these sets.

  1. EO
    b) EO
    c) ZE
    d) ZO

Show that R(0,1)

Show that (0,1)[0,1] .

A function f:AB is invertible if and only if f is a bijection.

Show that X and Y are countable sets, then XY is also a countable set.

Let nN, and let X1,X2,,Xn be nonempty countable sets. Then i=1nXi=X1×X2××Xn is countable

Show that the set of irrational numbers is an uncountable set.

Show that every open set in Rn is measurable.

Show that m(E)=0 where E is a countable set

A countable union of measurable sets is measurable.

Suppose E is a measurable subset of Rd. Then, for every ϵ>0,show that
(i) There exists an open set O with EO and m(OE)ϵ
(ii) There exists a closed set F with FE and m(EF)ϵ.
(iii) If m(E) is finite, there exists a compact set K with KE and m(EK)ϵ
(iv) If m(E) is finite, there exists a finite union F=j=1NQj of closed cubes such that
m(EF)ϵ

In the unit interval [0,1] consider a subsetE={x in the decimal expansion of x there is no 4}Show that E is measurable and calculate its measure.

Solution

** Solution of 3:**
The answer is the unit disk.

** Solution of 5:**
We will prove that exlogx>3log2. Let p(x)=exlogx=ex2x+2xlogx
Then h(x)=ex2x,j(x)=2xlogx,we have h(x)22log2,g(x)1+log2. h(x)=2log2 iffx=log2,g(x)=1+log2 iff x=1/2. Then we have h(x)+j(x)>3log2. Hence esinxlogsinx>3log2,the answer is .

** Solution of 7:**

** Solution of 9:**

Assume that f is invertible with inverse f1. Let f(a1)=f(a2) for some a1,a2A. Then f1(f(a1))=f1((f(a2)) and so a1=a2. Hence f is injective.

Let bB with a=f1(b)∈∣A. Then f(a)=b and so f is surjective. Therefore f is a bijection.
assume that f is a bijection. This means that for every bB there is a unique aA with f(a)=b and we can define f1(b)=a. This defines the inverse function f1:BA, as required.

** Solution of 10:**

Suppose that X and Y are countable. In fact , there exists surjective functions
f:NX and g:NY . 
Define a function h:N(XY) by
h(k)={f(k2)k is even g(k+12)k is odd 
Since f and g are surjective, we have h(N)=f(N)g(N)=XY, and thus h is a surjective function. Hence, we have XY is countable.

** Solution of 14:**

      
By assumption, we can enumerate Δ={cjjN}. Now for any ϵ>0 choose the sequence of radii rj=2j1ϵ, then clearly jNBrj(cj) contains Δ (since it contains every cjΔ as centre of a ball) and we have j=1rj=k=02kϵ4=ϵ2<ϵ.

** Solution of 15:**

Suppose E=i=1Ej, where each Ej is measurable. Given ϵ>0, we choose for each j an open set Oj with EjOj and m(OjEj)ϵ/2j. Then O=j=1Oj is open, EO, and (OE)j=1(OjEj), so monotonicity and sub-additivity of the
outer measure imply
m(OE)j=1m(OjEj)ϵ

** Solution of 17:**
Consider the complement of E. It has a digit 4 in the n -th place of the decimal expansion of x. Then such numbers consists of the whole half interval of length (1/10)n. There are countably many such intervals. The union of them is measurable. Hence E is measurable.
The complement of E has measure 1/10+9(1/10)2+ =1, then we have m(E)=0

投稿日:2021411
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  1. ** Discrete Mathematics **
  2. CHAPTER 2
  3. Solution