数学分析原理习题第二章

1. prove that the empty set is a subset of every set.

for a element x in a empty set, x is null , so x is a element of every other set since x is null. so the empty set is a sub set of every set .

 

2. prove that the set of all algebaric numbers is countable.

( a complex number z is said to be algebraic if there are integers, not all zero ,such that

  )

hint: for

,

for every positive N, this equation is finite.

we collect these equations:

 

as set , since these equations are finite, so  is countable , so  are countable.  for every algebraic number ,we can form a equation in , so all algebraic numbers are countable.

 

3. prove that there exist real numbers which are not algebraic.

since R is not countable , if all real numbers are algebraic , then R is countable, a contradiction.

 

4. is the set of all irrational real numbers countable?

if irrational real numbers is countable , then R-Q is countable , then R = (R-Q)Q  is countable ,a contradiction . thus all irrational  real numbers is not countable.

 

5. construct a bouded set of real numbers with exactly three limit points.

 

6. let  be the set of all limit points of a set E. prove that  is closed.

for any point p that is in , p is not a limit point of E,so there exists a neighborhood of p such that q is not in E , so p is a interior point of ,so  is open, so  is closed.

prove that  and   have the same limit points.(hint :=)

if p is the limit point of , then p is the limit point of ,since =.

if p is the limit point of , then the neighborhood of p contains a point q that q , if q , then we have p is a limit point of .

if q is not in ,

Do  and always have the same limit points?

no

 

7.let ...be subsets of a metric space'

(a) if , prove that , for n=1,2,3,...

 

so,  

in particular(in reserve )

 

8.is every point  of every open set  a limit point of ? answer the same question for cclosed sets in .

 

9.let  denote the set of all interior points of a set .

(a) pove that  is always open.

for , there exists ,

then for , there exists  where s=min{d(p,q), r-d(p,q)},

so , so q is a interior point of ,

so ,so  is open.

(b) prove that  is open if and only if

this statement is pretty straightforward

(c) if  and  is open, prove that

p is an interior point of G since G is open.

, so

so

(d) prove that the complement of  is the closure of the complement of

(e) Do  and  always have the same interiors

 

10.define

              

prove that this is a metric.

 

11. define

                  

determine ,for each of these , whether it is a metric or not.

this is pretty straightforward . use the definition of metric.

 

12. let  consist of 0 and the numbers 1/n, for n=1,2,3,...prove that k is compact directly from the definition.

contstruct a finite subcover of k

13. construct a compact set of real numbers whose limit points form a countable set.

 

14. Give an example of an open cover of the segment (0,1),which has no finite subcover.

 

15.show the theorem 2.36 and ts corollary are wrong if the word "compact " is replace by "closed " or by "bounded "

for closed:

for bounded :

 

16. regard Q, the set of all rational numbers, as a metric space, with d(p,q）=|p-q|. let E be the set of all pQ, such that  

show that E is closed an bounded in Q, but that E is not compact. is E open in Q.

E ={} where , so E is bounded in Q

since the set of rational numbers is dense, so Q is dense in R,every limit point of Q is in Q,so every limit point of E is in E ,so E is closed.

if E is compact, we construct a open couvering set {} of E :

{}={}

E is compact ,then there are finitely indices  such that

[p+r,] can not be covered since Q is dense ,a contradiction.

 

17. let E be the set of all  whose decimal expansion contains only the digits 4 and 7. is E countable ? is E dense in [0,1]?  is E compact ? is E perfect ?