Does completeness of $X/Y$ imply completeness of $X$?

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Does completeness of $X/Y$ imply completeness of $X$ ? Where $X$ is normed space and $Y$ closed inifinty dimensional subspace of $X$. I tried take $c_00$ (space of the finite sequence with supremum norm) and for the subspace every sequence that have $0$ on the first coordinate but I can't prove that this quotient space is complete.

linear-algebra functional-analysis

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edited Jul 16 at 17:52

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up vote
3
down vote

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Does completeness of $X/Y$ imply completeness of $X$ ? Where $X$ is normed space and $Y$ closed inifinty dimensional subspace of $X$. I tried take $c_00$ (space of the finite sequence with supremum norm) and for the subspace every sequence that have $0$ on the first coordinate but I can't prove that this quotient space is complete.

linear-algebra functional-analysis

share|cite|improve this question

edited Jul 16 at 17:52

mechanodroid

22.3k52041

asked Jul 16 at 17:10

user577360

331

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Does completeness of $X/Y$ imply completeness of $X$ ? Where $X$ is normed space and $Y$ closed inifinty dimensional subspace of $X$. I tried take $c_00$ (space of the finite sequence with supremum norm) and for the subspace every sequence that have $0$ on the first coordinate but I can't prove that this quotient space is complete.

linear-algebra functional-analysis

share|cite|improve this question

edited Jul 16 at 17:52

mechanodroid

22.3k52041

asked Jul 16 at 17:10

user577360

331

Does completeness of $X/Y$ imply completeness of $X$ ? Where $X$ is normed space and $Y$ closed inifinty dimensional subspace of $X$. I tried take $c_00$ (space of the finite sequence with supremum norm) and for the subspace every sequence that have $0$ on the first coordinate but I can't prove that this quotient space is complete.

linear-algebra functional-analysis

share|cite|improve this question

edited Jul 16 at 17:52

mechanodroid

22.3k52041

asked Jul 16 at 17:10

user577360

331

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 17:52

mechanodroid

22.3k52041

edited Jul 16 at 17:52

mechanodroid

22.3k52041

edited Jul 16 at 17:52

mechanodroid

22.3k52041

22.3k52041

asked Jul 16 at 17:10

user577360

331

asked Jul 16 at 17:10

user577360

331

asked Jul 16 at 17:10

user577360

331

331

add a comment | 

add a comment | 

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Indeed, you can construct a counterexample with $c_00$.

Equip $c_00$ with any $|cdot|_p$ norm and let $Y = (x_n)_n in c_00 : x_1 = 0$. Then $Y$ is the kernel of the bounded linear functional $f : c_00 to mathbbC$, given by $f(x_n)_n = x_1$. Hence $Y$ is closed in $c_00$ and the induced map $tildef : c_00/Y to mathbbC$ given by $tildef((x_n)_n + Y) = x_1$ is an isometric isomorphism.

So $c_00/Y cong mathbbC$, which is finite-dimensional and hence Banach.

---

The statement is true if $Y$ itself assumed to be complete.

Let $(x_n)_n$ be a Cauchy sequence in $X$. Then

$$|(x_m+Y) - (x_n+Y)| = |(x_m - x_n) + Y| le |x_m-x_n|$$

so $(x_n + M)_n$ is a Cauchy sequence in $X/Y$ so it converges to an element $x + M$.

We have $|(x_n -x) + M| = |(x_n+M) - (x + M)| to 0$ so for every $n in mathbbN$ there exists $p(n) in mathbbN$ and $y_n in Y$ such that

$$|x_p(n) + y_n - x| < frac1n$$

Notice that $(y_n)_n$ is Cauchy in $Y$

$$|y_m-y_n| le |y_m - x_p(m) - x| + |x-x_p(n) - y_n| + |x_p(m) - x_p(n)| xrightarrowm,ntoinfty 0$$

so it converges to an element $y in Y$.

$$|x_p(m) + y - x| le |x_p(m) + y_n - x| + |y - y_n| xrightarrowntoinfty 0$$

so $(x_p(n))_n$ converges to $x-y$. Since $(x_n)_n$ is Cauchy, it also converges to $x-y$.

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answered Jul 16 at 17:51

mechanodroid

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1 Answer
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1 Answer
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up vote
2
down vote

Indeed, you can construct a counterexample with $c_00$.

Equip $c_00$ with any $|cdot|_p$ norm and let $Y = (x_n)_n in c_00 : x_1 = 0$. Then $Y$ is the kernel of the bounded linear functional $f : c_00 to mathbbC$, given by $f(x_n)_n = x_1$. Hence $Y$ is closed in $c_00$ and the induced map $tildef : c_00/Y to mathbbC$ given by $tildef((x_n)_n + Y) = x_1$ is an isometric isomorphism.

So $c_00/Y cong mathbbC$, which is finite-dimensional and hence Banach.

---

The statement is true if $Y$ itself assumed to be complete.

Let $(x_n)_n$ be a Cauchy sequence in $X$. Then

$$|(x_m+Y) - (x_n+Y)| = |(x_m - x_n) + Y| le |x_m-x_n|$$

so $(x_n + M)_n$ is a Cauchy sequence in $X/Y$ so it converges to an element $x + M$.

We have $|(x_n -x) + M| = |(x_n+M) - (x + M)| to 0$ so for every $n in mathbbN$ there exists $p(n) in mathbbN$ and $y_n in Y$ such that

$$|x_p(n) + y_n - x| < frac1n$$

Notice that $(y_n)_n$ is Cauchy in $Y$

$$|y_m-y_n| le |y_m - x_p(m) - x| + |x-x_p(n) - y_n| + |x_p(m) - x_p(n)| xrightarrowm,ntoinfty 0$$

so it converges to an element $y in Y$.

$$|x_p(m) + y - x| le |x_p(m) + y_n - x| + |y - y_n| xrightarrowntoinfty 0$$

so $(x_p(n))_n$ converges to $x-y$. Since $(x_n)_n$ is Cauchy, it also converges to $x-y$.

share|cite|improve this answer

answered Jul 16 at 17:51

mechanodroid

22.3k52041

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up vote
2
down vote

Indeed, you can construct a counterexample with $c_00$.

Equip $c_00$ with any $|cdot|_p$ norm and let $Y = (x_n)_n in c_00 : x_1 = 0$. Then $Y$ is the kernel of the bounded linear functional $f : c_00 to mathbbC$, given by $f(x_n)_n = x_1$. Hence $Y$ is closed in $c_00$ and the induced map $tildef : c_00/Y to mathbbC$ given by $tildef((x_n)_n + Y) = x_1$ is an isometric isomorphism.

So $c_00/Y cong mathbbC$, which is finite-dimensional and hence Banach.

---

The statement is true if $Y$ itself assumed to be complete.

Let $(x_n)_n$ be a Cauchy sequence in $X$. Then

$$|(x_m+Y) - (x_n+Y)| = |(x_m - x_n) + Y| le |x_m-x_n|$$

so $(x_n + M)_n$ is a Cauchy sequence in $X/Y$ so it converges to an element $x + M$.

We have $|(x_n -x) + M| = |(x_n+M) - (x + M)| to 0$ so for every $n in mathbbN$ there exists $p(n) in mathbbN$ and $y_n in Y$ such that

$$|x_p(n) + y_n - x| < frac1n$$

Notice that $(y_n)_n$ is Cauchy in $Y$

$$|y_m-y_n| le |y_m - x_p(m) - x| + |x-x_p(n) - y_n| + |x_p(m) - x_p(n)| xrightarrowm,ntoinfty 0$$

so it converges to an element $y in Y$.

$$|x_p(m) + y - x| le |x_p(m) + y_n - x| + |y - y_n| xrightarrowntoinfty 0$$

so $(x_p(n))_n$ converges to $x-y$. Since $(x_n)_n$ is Cauchy, it also converges to $x-y$.

share|cite|improve this answer

answered Jul 16 at 17:51

mechanodroid

22.3k52041

add a comment | 

up vote
2
down vote

up vote
2
down vote

Indeed, you can construct a counterexample with $c_00$.

Equip $c_00$ with any $|cdot|_p$ norm and let $Y = (x_n)_n in c_00 : x_1 = 0$. Then $Y$ is the kernel of the bounded linear functional $f : c_00 to mathbbC$, given by $f(x_n)_n = x_1$. Hence $Y$ is closed in $c_00$ and the induced map $tildef : c_00/Y to mathbbC$ given by $tildef((x_n)_n + Y) = x_1$ is an isometric isomorphism.

So $c_00/Y cong mathbbC$, which is finite-dimensional and hence Banach.

---

The statement is true if $Y$ itself assumed to be complete.

Let $(x_n)_n$ be a Cauchy sequence in $X$. Then

$$|(x_m+Y) - (x_n+Y)| = |(x_m - x_n) + Y| le |x_m-x_n|$$

so $(x_n + M)_n$ is a Cauchy sequence in $X/Y$ so it converges to an element $x + M$.

We have $|(x_n -x) + M| = |(x_n+M) - (x + M)| to 0$ so for every $n in mathbbN$ there exists $p(n) in mathbbN$ and $y_n in Y$ such that

$$|x_p(n) + y_n - x| < frac1n$$

Notice that $(y_n)_n$ is Cauchy in $Y$

$$|y_m-y_n| le |y_m - x_p(m) - x| + |x-x_p(n) - y_n| + |x_p(m) - x_p(n)| xrightarrowm,ntoinfty 0$$

so it converges to an element $y in Y$.

$$|x_p(m) + y - x| le |x_p(m) + y_n - x| + |y - y_n| xrightarrowntoinfty 0$$

so $(x_p(n))_n$ converges to $x-y$. Since $(x_n)_n$ is Cauchy, it also converges to $x-y$.

share|cite|improve this answer

answered Jul 16 at 17:51

mechanodroid

22.3k52041

Indeed, you can construct a counterexample with $c_00$.

Equip $c_00$ with any $|cdot|_p$ norm and let $Y = (x_n)_n in c_00 : x_1 = 0$. Then $Y$ is the kernel of the bounded linear functional $f : c_00 to mathbbC$, given by $f(x_n)_n = x_1$. Hence $Y$ is closed in $c_00$ and the induced map $tildef : c_00/Y to mathbbC$ given by $tildef((x_n)_n + Y) = x_1$ is an isometric isomorphism.

So $c_00/Y cong mathbbC$, which is finite-dimensional and hence Banach.

---

The statement is true if $Y$ itself assumed to be complete.

Let $(x_n)_n$ be a Cauchy sequence in $X$. Then

$$|(x_m+Y) - (x_n+Y)| = |(x_m - x_n) + Y| le |x_m-x_n|$$

so $(x_n + M)_n$ is a Cauchy sequence in $X/Y$ so it converges to an element $x + M$.

We have $|(x_n -x) + M| = |(x_n+M) - (x + M)| to 0$ so for every $n in mathbbN$ there exists $p(n) in mathbbN$ and $y_n in Y$ such that

$$|x_p(n) + y_n - x| < frac1n$$

Notice that $(y_n)_n$ is Cauchy in $Y$

$$|y_m-y_n| le |y_m - x_p(m) - x| + |x-x_p(n) - y_n| + |x_p(m) - x_p(n)| xrightarrowm,ntoinfty 0$$

so it converges to an element $y in Y$.

$$|x_p(m) + y - x| le |x_p(m) + y_n - x| + |y - y_n| xrightarrowntoinfty 0$$

so $(x_p(n))_n$ converges to $x-y$. Since $(x_n)_n$ is Cauchy, it also converges to $x-y$.

share|cite|improve this answer

answered Jul 16 at 17:51

mechanodroid

22.3k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 17:51

mechanodroid

22.3k52041

answered Jul 16 at 17:51

mechanodroid

22.3k52041

answered Jul 16 at 17:51

mechanodroid

22.3k52041

22.3k52041

add a comment | 

add a comment | 

 

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