For a set K in a metric space X, the first two conditionsare equivalent and imply the third.
1) Every sequence in K has a subsequence converging to apoint of K.
2) K is compact: every open cover has a finite subcover
3) K is closed and bounded
Prove that 1) ==> 3)
Prove by not bounded or not closed.
For a set K in a metric space X, the first two conditionsare equivalent and imply the third.
1) Every sequence in K has a subsequence converging to apoint of K.
2) K is compact: every open cover has a finite subcover
3) K is closed and bounded
Prove that 1) ==> 3)
Prove by not bounded or not closed.
Solution
Proof: Suppose that K is not closed. A set is closed if andonly if it contains all of its limit points.
Therefore, there is apoint p in X that is a limit point of K, but is not in K. Because pis a limit
point of K, there exists a sequence in K that convergesto K, called {a_n}. A sequence that
converges to a point q cannothave a subsequence that converges to a point that is not
q.Therefore, all convergent subsequences of {a_n} converge to p,which is not in K. Therefore K
is not compact, and this is acontradiction.
Theorem: K is bounded.
Proof: Suppose that K is not bounded. Consider a point p in K.No ball centered at p can contain
K.
Furthermore, every ball centered at p contains points in K,and if a>b, a ball at p of radius a
contains a point in K thatis not contained in a ball of radius b at p.
Consider a sequence of balls b_n centered at p, with radiia_n, such that a_n = n and a_1 = 1. And
consider a sequence ofpoints in K, p_n, such that p_n is in b_n but not b_n+1. p_n is atleast
distance 1 from p_n+2, for all n. Therefore no subsequence ofp_n can converge.
Alternate proof (uses statement 2): Consider the openballs centered upon a common point, with
any radius. This can coverany set, because all points in the set are some distance away fromthat
point. Any finite subcover of this cover must be bounded,because all balls in the subcover are
contained in the largest openball within that subcover. Therefore, any set covered by
thissubcover must also be bounded..
For a set K in a metric space X, the first two conditionsare equiv.pdf
1. For a set K in a metric space X, the first two conditionsare equivalent and imply the third.
1) Every sequence in K has a subsequence converging to apoint of K.
2) K is compact: every open cover has a finite subcover
3) K is closed and bounded
Prove that 1) ==> 3)
Prove by not bounded or not closed.
For a set K in a metric space X, the first two conditionsare equivalent and imply the third.
1) Every sequence in K has a subsequence converging to apoint of K.
2) K is compact: every open cover has a finite subcover
3) K is closed and bounded
Prove that 1) ==> 3)
Prove by not bounded or not closed.
Solution
Proof: Suppose that K is not closed. A set is closed if andonly if it contains all of its limit points.
Therefore, there is apoint p in X that is a limit point of K, but is not in K. Because pis a limit
point of K, there exists a sequence in K that convergesto K, called {a_n}. A sequence that
converges to a point q cannothave a subsequence that converges to a point that is not
q.Therefore, all convergent subsequences of {a_n} converge to p,which is not in K. Therefore K
is not compact, and this is acontradiction.
Theorem: K is bounded.
Proof: Suppose that K is not bounded. Consider a point p in K.No ball centered at p can contain
K.
Furthermore, every ball centered at p contains points in K,and if a>b, a ball at p of radius a
contains a point in K thatis not contained in a ball of radius b at p.
Consider a sequence of balls b_n centered at p, with radiia_n, such that a_n = n and a_1 = 1. And
consider a sequence ofpoints in K, p_n, such that p_n is in b_n but not b_n+1. p_n is atleast
distance 1 from p_n+2, for all n. Therefore no subsequence ofp_n can converge.
Alternate proof (uses statement 2): Consider the openballs centered upon a common point, with
2. any radius. This can coverany set, because all points in the set are some distance away fromthat
point. Any finite subcover of this cover must be bounded,because all balls in the subcover are
contained in the largest openball within that subcover. Therefore, any set covered by
thissubcover must also be bounded.
