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 ?