Why should K be closed to ensure X/K is complete?
$begingroup$
If $K$ is a closed subspace of Banach space $X$, then $X/K$ is complete. But I think the usual proof of this theorem doesn't make use of the fact that $K$ is closed. Would anyone explain it to me? Thanks a lot.
functional-analysis
asked Nov 11 '18 at 18:29
ZengZeng
524
$endgroup$
add a comment |
$begingroup$
If $K$ is a closed subspace of Banach space $X$, then $X/K$ is complete. But I think the usual proof of this theorem doesn't make use of the fact that $K$ is closed. Would anyone explain it to me? Thanks a lot.
functional-analysis
asked Nov 11 '18 at 18:29
ZengZeng
524
$endgroup$
add a comment |
$begingroup$
If $K$ is a closed subspace of Banach space $X$, then $X/K$ is complete. But I think the usual proof of this theorem doesn't make use of the fact that $K$ is closed. Would anyone explain it to me? Thanks a lot.
functional-analysis
asked Nov 11 '18 at 18:29
ZengZeng
524
$endgroup$
If $K$ is a closed subspace of Banach space $X$, then $X/K$ is complete. But I think the usual proof of this theorem doesn't make use of the fact that $K$ is closed. Would anyone explain it to me? Thanks a lot.
functional-analysis
functional-analysis
asked Nov 11 '18 at 18:29
ZengZeng
524
asked Nov 11 '18 at 18:29
ZengZeng
524
asked Nov 11 '18 at 18:29
ZengZeng
524
asked Nov 11 '18 at 18:29
ZengZeng
524
asked Nov 11 '18 at 18:29
ZengZeng
524
524
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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If the subspace $K$ is not closed, then the quotient $X/K$ is not even Hausdorff, so does not meet the usual requirements of a topological vector space at all!
(It's not about completeness or not, as $X/K$ is not metric in that case!)
EDIT: In more detail, for $K$ not closed, and for $x$ in the closure of $K$ but not in $K$ itself, every neighborhood of $0$ in the quotient (with the quotient topology) contains $x$, but $xnot=0$ in the quotient. So the quotient would be non-Hausdorff (since two distinct points, $0$ and $x$, do not have disjoint neighborhoods).
A Banach space, or even a normed space, is Hausdorff, as is every metric space.
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
$endgroup$
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
add a comment |
$begingroup$
If $K$ is not closed, then the function $lVert x+KrVert=inf_yin K lVert x-yrVert$ is not a norm on the quotient space, but just a seminorm, because $lVert x+KrVert=0$ for all $xinoverline K$.
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
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add a comment |
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You do need closure of $K$. In particular, you need it to show that $X/K$ is in fact a normed space: If $| x + K |=0$, closure of $K$ implies that $x+K=K=0_X/K$.
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
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add a comment |
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3 Answers
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3 Answers
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oldest
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active
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votes
$begingroup$
If the subspace $K$ is not closed, then the quotient $X/K$ is not even Hausdorff, so does not meet the usual requirements of a topological vector space at all!
(It's not about completeness or not, as $X/K$ is not metric in that case!)
EDIT: In more detail, for $K$ not closed, and for $x$ in the closure of $K$ but not in $K$ itself, every neighborhood of $0$ in the quotient (with the quotient topology) contains $x$, but $xnot=0$ in the quotient. So the quotient would be non-Hausdorff (since two distinct points, $0$ and $x$, do not have disjoint neighborhoods).
A Banach space, or even a normed space, is Hausdorff, as is every metric space.
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
$endgroup$
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
add a comment |
$begingroup$
If the subspace $K$ is not closed, then the quotient $X/K$ is not even Hausdorff, so does not meet the usual requirements of a topological vector space at all!
(It's not about completeness or not, as $X/K$ is not metric in that case!)
EDIT: In more detail, for $K$ not closed, and for $x$ in the closure of $K$ but not in $K$ itself, every neighborhood of $0$ in the quotient (with the quotient topology) contains $x$, but $xnot=0$ in the quotient. So the quotient would be non-Hausdorff (since two distinct points, $0$ and $x$, do not have disjoint neighborhoods).
A Banach space, or even a normed space, is Hausdorff, as is every metric space.
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
$endgroup$
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
add a comment |
$begingroup$
If the subspace $K$ is not closed, then the quotient $X/K$ is not even Hausdorff, so does not meet the usual requirements of a topological vector space at all!
(It's not about completeness or not, as $X/K$ is not metric in that case!)
EDIT: In more detail, for $K$ not closed, and for $x$ in the closure of $K$ but not in $K$ itself, every neighborhood of $0$ in the quotient (with the quotient topology) contains $x$, but $xnot=0$ in the quotient. So the quotient would be non-Hausdorff (since two distinct points, $0$ and $x$, do not have disjoint neighborhoods).
A Banach space, or even a normed space, is Hausdorff, as is every metric space.
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
$endgroup$
If the subspace $K$ is not closed, then the quotient $X/K$ is not even Hausdorff, so does not meet the usual requirements of a topological vector space at all!
(It's not about completeness or not, as $X/K$ is not metric in that case!)
EDIT: In more detail, for $K$ not closed, and for $x$ in the closure of $K$ but not in $K$ itself, every neighborhood of $0$ in the quotient (with the quotient topology) contains $x$, but $xnot=0$ in the quotient. So the quotient would be non-Hausdorff (since two distinct points, $0$ and $x$, do not have disjoint neighborhoods).
A Banach space, or even a normed space, is Hausdorff, as is every metric space.
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
edited Nov 11 '18 at 18:55
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
answered Nov 11 '18 at 18:31
paul garrettpaul garrett
31.9k362118
31.9k362118
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
add a comment |
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
$begingroup$
Thanks! But it has been a long time since my last topology class, so would you give me some hint?
$endgroup$
– Zeng
Nov 11 '18 at 18:39
add a comment |
$begingroup$
If $K$ is not closed, then the function $lVert x+KrVert=inf_yin K lVert x-yrVert$ is not a norm on the quotient space, but just a seminorm, because $lVert x+KrVert=0$ for all $xinoverline K$.
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
$endgroup$
add a comment |
$begingroup$
If $K$ is not closed, then the function $lVert x+KrVert=inf_yin K lVert x-yrVert$ is not a norm on the quotient space, but just a seminorm, because $lVert x+KrVert=0$ for all $xinoverline K$.
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
$endgroup$
add a comment |
$begingroup$
If $K$ is not closed, then the function $lVert x+KrVert=inf_yin K lVert x-yrVert$ is not a norm on the quotient space, but just a seminorm, because $lVert x+KrVert=0$ for all $xinoverline K$.
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
$endgroup$
If $K$ is not closed, then the function $lVert x+KrVert=inf_yin K lVert x-yrVert$ is not a norm on the quotient space, but just a seminorm, because $lVert x+KrVert=0$ for all $xinoverline K$.
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
answered Nov 11 '18 at 18:32
Saucy O'PathSaucy O'Path
5,9591627
5,9591627
add a comment |
add a comment |
$begingroup$
You do need closure of $K$. In particular, you need it to show that $X/K$ is in fact a normed space: If $| x + K |=0$, closure of $K$ implies that $x+K=K=0_X/K$.
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
$endgroup$
add a comment |
$begingroup$
You do need closure of $K$. In particular, you need it to show that $X/K$ is in fact a normed space: If $| x + K |=0$, closure of $K$ implies that $x+K=K=0_X/K$.
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
$endgroup$
add a comment |
$begingroup$
You do need closure of $K$. In particular, you need it to show that $X/K$ is in fact a normed space: If $| x + K |=0$, closure of $K$ implies that $x+K=K=0_X/K$.
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
$endgroup$
You do need closure of $K$. In particular, you need it to show that $X/K$ is in fact a normed space: If $| x + K |=0$, closure of $K$ implies that $x+K=K=0_X/K$.
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
answered Nov 11 '18 at 18:36
Alonso DelfínAlonso Delfín
3,79411132
3,79411132
add a comment |
add a comment |
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