1) Every sequenc in K has a subsequence converging to a pointof K. 2) K is compact ; every open cover has a finitesubcover. 3) K is closed and bounded. Please prove 1) ==> 3) in metric space. 1) Every sequenc in K has a subsequence converging to a pointof K. 2) K is compact ; every open cover has a finitesubcover. 3) K is closed and bounded. Please prove 1) ==> 3) in metric space. Solution Assume 1) is true. Let x be a limit point of K. Since x is a limit point of K,for every n > 0, we can choose some x_n in K so that d(x, x_n)< 1/n, i.e., we can create a sequence in K converging tox. By 1), there is a convergent subsequence that converges tosome element of K. But since the whole sequence converges tox, so must the subsequence. Thus x is in K and K isclosed. If K were not bounded, construct a sequence as follows. Choose x_1 arbitrarily, and proceed inductively. Havingchosen points x_1, ..., x_n, choose x_(n + 1) such that d(x_(n +1), d_(i)) > 1 for every i from 1 to n. If this were notpossible, then every point of K would be a distance <= 1 fromone of a finite set of points, from which it would follow that Kwas bounded. By 1) such a sequence has a convergentsubsequence, which must then be Cauchy. But by construction,the distance between any 2 elements in the sequence is at least 1,so it can\'t be Cauchy. This contradiction proves theresult..

1. 1) Every sequenc in K has a subsequence converging to a pointof K.
2) K is compact ; every open cover has a finitesubcover.
3) K is closed and bounded.
Please prove 1) ==> 3) in metric space.
1) Every sequenc in K has a subsequence converging to a pointof K.
2) K is compact ; every open cover has a finitesubcover.
3) K is closed and bounded.
Please prove 1) ==> 3) in metric space.
Solution
Assume 1) is true. Let x be a limit point of K. Since x is a limit point of K,for every
n > 0, we can choose some x_n in K so that d(x, x_n)< 1/n, i.e., we can create a sequence in K
converging tox. By 1), there is a convergent subsequence that converges tosome element of K.
But since the whole sequence converges tox, so must the subsequence. Thus x is in K and K
isclosed. If K were not bounded, construct a sequence as follows. Choose x_1 arbitrarily, and
proceed inductively. Havingchosen points x_1, ..., x_n, choose x_(n + 1) such that d(x_(n +1),
d_(i)) > 1 for every i from 1 to n. If this were notpossible, then every point of K would be a
distance <= 1 fromone of a finite set of points, from which it would follow that Kwas bounded.
By 1) such a sequence has a convergentsubsequence, which must then be Cauchy. But by
construction,the distance between any 2 elements in the sequence is at least 1,so it can't be
Cauchy. This contradiction proves theresult.