Closed & boundedness, sequentially compactness and Completeness
$begingroup$
Let $(X,||·||)$ be a normed vector space and S be a non-empty subset
of $X$. Which of the following conditions always implies the other two
conditions? Prove one of the two implications. Also, give one example
illustrating that one condition does not imply some other.
- S is closed and bounded.
- S is sequentially compact
- S is complete.
My Answer:
- (2) implies both (1) and (3):
Proof:
$(2)rightarrow(1)$ Suppose $S$ is sequentially compact. Then we have shown that compactness will imply both closed and bounded properties for any metric space (Converse of the Heine Borel theorem is true for any metric space).
$quadquad(2)rightarrow(3)$
begin{align*}
S text{ is Sequentially Compact} &Rightarrow text{ Any Cauchy Sequence in S has a convergent subsequence}\
&Rightarrow text{Cauchy Seqence converges [*]}\
&Rightarrow text{ S is complete}
end{align*}
[Proof of *: Cauchy sequence is convergent iff it has a convergent subsequence ]
- (1) does not imply (2)
Proof: Consider the metric space $X=(0,1)$ with the Euclidean. That is the open interval. Since this is the metric space, it is closed. (Any metric space is clopen). Also it is bounded because it is a bounded interval.
Now consider the sequence $(frac{1}{n})_{ninmathbb{N}}$ from $X$. And as we have shown that in the set of real numbers with the Euclidean metric, this sequence converges to 0. Thus all sub sequences should also converge to 0. And as we haven't change the metric for $X$, now we have a sequence from $X$ which does not have a subsequence converging to a point in $X$. Thus it is not sequentially compact. (And also similar argument can be used to say that (1) doesnot imply (3))
I would like some suggestions and feedback on my proposed answer. Moreover I'm wondering whether there is any possible proper subset of a metric space which can be used to say that (1) doesnot imply (2)
real-analysis general-topology metric-spaces
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
$endgroup$
add a comment |
$begingroup$
Let $(X,||·||)$ be a normed vector space and S be a non-empty subset
of $X$. Which of the following conditions always implies the other two
conditions? Prove one of the two implications. Also, give one example
illustrating that one condition does not imply some other.
- S is closed and bounded.
- S is sequentially compact
- S is complete.
My Answer:
- (2) implies both (1) and (3):
Proof:
$(2)rightarrow(1)$ Suppose $S$ is sequentially compact. Then we have shown that compactness will imply both closed and bounded properties for any metric space (Converse of the Heine Borel theorem is true for any metric space).
$quadquad(2)rightarrow(3)$
begin{align*}
S text{ is Sequentially Compact} &Rightarrow text{ Any Cauchy Sequence in S has a convergent subsequence}\
&Rightarrow text{Cauchy Seqence converges [*]}\
&Rightarrow text{ S is complete}
end{align*}
[Proof of *: Cauchy sequence is convergent iff it has a convergent subsequence ]
- (1) does not imply (2)
Proof: Consider the metric space $X=(0,1)$ with the Euclidean. That is the open interval. Since this is the metric space, it is closed. (Any metric space is clopen). Also it is bounded because it is a bounded interval.
Now consider the sequence $(frac{1}{n})_{ninmathbb{N}}$ from $X$. And as we have shown that in the set of real numbers with the Euclidean metric, this sequence converges to 0. Thus all sub sequences should also converge to 0. And as we haven't change the metric for $X$, now we have a sequence from $X$ which does not have a subsequence converging to a point in $X$. Thus it is not sequentially compact. (And also similar argument can be used to say that (1) doesnot imply (3))
I would like some suggestions and feedback on my proposed answer. Moreover I'm wondering whether there is any possible proper subset of a metric space which can be used to say that (1) doesnot imply (2)
real-analysis general-topology metric-spaces
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
$endgroup$
2
$begingroup$
You arguments and examples are correct, but you did not answer the question. $X= (0,1)$ is not a normed vector space. But you may consider $X = mathbb{R}$ to show (3) does not imply (1) nor (2).
$endgroup$
– Paul Frost
Jan 5 at 17:48
$begingroup$
But then what should I need to choose as the complete subset of X? can I take $[0,infty)$
$endgroup$
– Charith Jeewantha
Jan 5 at 18:48
1
$begingroup$
Just $S=X=mathbb{R}$ itself is complete. And not bounded nor sequentially compact.
$endgroup$
– Henno Brandsma
Jan 5 at 22:29
1
$begingroup$
As for (1) not implying (2) : take $S$ as the closed unit ball in say $X=ell_2$ (or any inifnite-dimensional Banach space), closed and bounded but not sequentially compact. It also shows (3) does not imply (2).
$endgroup$
– Henno Brandsma
Jan 5 at 22:30
add a comment |
$begingroup$
Let $(X,||·||)$ be a normed vector space and S be a non-empty subset
of $X$. Which of the following conditions always implies the other two
conditions? Prove one of the two implications. Also, give one example
illustrating that one condition does not imply some other.
- S is closed and bounded.
- S is sequentially compact
- S is complete.
My Answer:
- (2) implies both (1) and (3):
Proof:
$(2)rightarrow(1)$ Suppose $S$ is sequentially compact. Then we have shown that compactness will imply both closed and bounded properties for any metric space (Converse of the Heine Borel theorem is true for any metric space).
$quadquad(2)rightarrow(3)$
begin{align*}
S text{ is Sequentially Compact} &Rightarrow text{ Any Cauchy Sequence in S has a convergent subsequence}\
&Rightarrow text{Cauchy Seqence converges [*]}\
&Rightarrow text{ S is complete}
end{align*}
[Proof of *: Cauchy sequence is convergent iff it has a convergent subsequence ]
- (1) does not imply (2)
Proof: Consider the metric space $X=(0,1)$ with the Euclidean. That is the open interval. Since this is the metric space, it is closed. (Any metric space is clopen). Also it is bounded because it is a bounded interval.
Now consider the sequence $(frac{1}{n})_{ninmathbb{N}}$ from $X$. And as we have shown that in the set of real numbers with the Euclidean metric, this sequence converges to 0. Thus all sub sequences should also converge to 0. And as we haven't change the metric for $X$, now we have a sequence from $X$ which does not have a subsequence converging to a point in $X$. Thus it is not sequentially compact. (And also similar argument can be used to say that (1) doesnot imply (3))
I would like some suggestions and feedback on my proposed answer. Moreover I'm wondering whether there is any possible proper subset of a metric space which can be used to say that (1) doesnot imply (2)
real-analysis general-topology metric-spaces
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
$endgroup$
Let $(X,||·||)$ be a normed vector space and S be a non-empty subset
of $X$. Which of the following conditions always implies the other two
conditions? Prove one of the two implications. Also, give one example
illustrating that one condition does not imply some other.
- S is closed and bounded.
- S is sequentially compact
- S is complete.
My Answer:
- (2) implies both (1) and (3):
Proof:
$(2)rightarrow(1)$ Suppose $S$ is sequentially compact. Then we have shown that compactness will imply both closed and bounded properties for any metric space (Converse of the Heine Borel theorem is true for any metric space).
$quadquad(2)rightarrow(3)$
begin{align*}
S text{ is Sequentially Compact} &Rightarrow text{ Any Cauchy Sequence in S has a convergent subsequence}\
&Rightarrow text{Cauchy Seqence converges [*]}\
&Rightarrow text{ S is complete}
end{align*}
[Proof of *: Cauchy sequence is convergent iff it has a convergent subsequence ]
- (1) does not imply (2)
Proof: Consider the metric space $X=(0,1)$ with the Euclidean. That is the open interval. Since this is the metric space, it is closed. (Any metric space is clopen). Also it is bounded because it is a bounded interval.
Now consider the sequence $(frac{1}{n})_{ninmathbb{N}}$ from $X$. And as we have shown that in the set of real numbers with the Euclidean metric, this sequence converges to 0. Thus all sub sequences should also converge to 0. And as we haven't change the metric for $X$, now we have a sequence from $X$ which does not have a subsequence converging to a point in $X$. Thus it is not sequentially compact. (And also similar argument can be used to say that (1) doesnot imply (3))
I would like some suggestions and feedback on my proposed answer. Moreover I'm wondering whether there is any possible proper subset of a metric space which can be used to say that (1) doesnot imply (2)
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
asked Jan 5 at 17:34
Charith JeewanthaCharith Jeewantha
11110
11110
2
$begingroup$
You arguments and examples are correct, but you did not answer the question. $X= (0,1)$ is not a normed vector space. But you may consider $X = mathbb{R}$ to show (3) does not imply (1) nor (2).
$endgroup$
– Paul Frost
Jan 5 at 17:48
$begingroup$
But then what should I need to choose as the complete subset of X? can I take $[0,infty)$
$endgroup$
– Charith Jeewantha
Jan 5 at 18:48
1
$begingroup$
Just $S=X=mathbb{R}$ itself is complete. And not bounded nor sequentially compact.
$endgroup$
– Henno Brandsma
Jan 5 at 22:29
1
$begingroup$
As for (1) not implying (2) : take $S$ as the closed unit ball in say $X=ell_2$ (or any inifnite-dimensional Banach space), closed and bounded but not sequentially compact. It also shows (3) does not imply (2).
$endgroup$
– Henno Brandsma
Jan 5 at 22:30
add a comment |
2
$begingroup$
You arguments and examples are correct, but you did not answer the question. $X= (0,1)$ is not a normed vector space. But you may consider $X = mathbb{R}$ to show (3) does not imply (1) nor (2).
$endgroup$
– Paul Frost
Jan 5 at 17:48
$begingroup$
But then what should I need to choose as the complete subset of X? can I take $[0,infty)$
$endgroup$
– Charith Jeewantha
Jan 5 at 18:48
1
$begingroup$
Just $S=X=mathbb{R}$ itself is complete. And not bounded nor sequentially compact.
$endgroup$
– Henno Brandsma
Jan 5 at 22:29
1
$begingroup$
As for (1) not implying (2) : take $S$ as the closed unit ball in say $X=ell_2$ (or any inifnite-dimensional Banach space), closed and bounded but not sequentially compact. It also shows (3) does not imply (2).
$endgroup$
– Henno Brandsma
Jan 5 at 22:30
2
2
$begingroup$
You arguments and examples are correct, but you did not answer the question. $X= (0,1)$ is not a normed vector space. But you may consider $X = mathbb{R}$ to show (3) does not imply (1) nor (2).
$endgroup$
– Paul Frost
Jan 5 at 17:48
$begingroup$
You arguments and examples are correct, but you did not answer the question. $X= (0,1)$ is not a normed vector space. But you may consider $X = mathbb{R}$ to show (3) does not imply (1) nor (2).
$endgroup$
– Paul Frost
Jan 5 at 17:48
$begingroup$
But then what should I need to choose as the complete subset of X? can I take $[0,infty)$
$endgroup$
– Charith Jeewantha
Jan 5 at 18:48
$begingroup$
But then what should I need to choose as the complete subset of X? can I take $[0,infty)$
$endgroup$
– Charith Jeewantha
Jan 5 at 18:48
1
1
$begingroup$
Just $S=X=mathbb{R}$ itself is complete. And not bounded nor sequentially compact.
$endgroup$
– Henno Brandsma
Jan 5 at 22:29
$begingroup$
Just $S=X=mathbb{R}$ itself is complete. And not bounded nor sequentially compact.
$endgroup$
– Henno Brandsma
Jan 5 at 22:29
1
1
$begingroup$
As for (1) not implying (2) : take $S$ as the closed unit ball in say $X=ell_2$ (or any inifnite-dimensional Banach space), closed and bounded but not sequentially compact. It also shows (3) does not imply (2).
$endgroup$
– Henno Brandsma
Jan 5 at 22:30
$begingroup$
As for (1) not implying (2) : take $S$ as the closed unit ball in say $X=ell_2$ (or any inifnite-dimensional Banach space), closed and bounded but not sequentially compact. It also shows (3) does not imply (2).
$endgroup$
– Henno Brandsma
Jan 5 at 22:30
add a comment |
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$begingroup$
You arguments and examples are correct, but you did not answer the question. $X= (0,1)$ is not a normed vector space. But you may consider $X = mathbb{R}$ to show (3) does not imply (1) nor (2).
$endgroup$
– Paul Frost
Jan 5 at 17:48
$begingroup$
But then what should I need to choose as the complete subset of X? can I take $[0,infty)$
$endgroup$
– Charith Jeewantha
Jan 5 at 18:48
1
$begingroup$
Just $S=X=mathbb{R}$ itself is complete. And not bounded nor sequentially compact.
$endgroup$
– Henno Brandsma
Jan 5 at 22:29
1
$begingroup$
As for (1) not implying (2) : take $S$ as the closed unit ball in say $X=ell_2$ (or any inifnite-dimensional Banach space), closed and bounded but not sequentially compact. It also shows (3) does not imply (2).
$endgroup$
– Henno Brandsma
Jan 5 at 22:30