Mixing
Intersection of totally ordered set of spanning sets is still spanning.
$begingroup$
Suppose $Ssubseteq V$ is a spanning set(possibly infinite) of a non-zero vector space $V$.
Let $s'={ssubseteq S|$$s$ also spans V$}$. Suppose $ l subseteq s'$ is a totally ordered non-empty subset of $s'$.
How do I show that span$(cap l)$ =V.
It looks true to me. If we take a $cin l$. It would contain all the vectors of those sets(whose spans are also V) that are subsets of it.
And all supersets would also contain vectors of $c$. But I just can't seem to show vigorously that the intersection spans V. (i.e. can't find least element.)
Any ideas on how to prove it or is it false?
P.S. I'm trying to prove that every spanning set has a basis. i.e. Zorn's Lemma, Dual order etc
linear-algebra vector-spaces
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
$endgroup$
add a comment |
$begingroup$
Suppose $Ssubseteq V$ is a spanning set(possibly infinite) of a non-zero vector space $V$.
Let $s'={ssubseteq S|$$s$ also spans V$}$. Suppose $ l subseteq s'$ is a totally ordered non-empty subset of $s'$.
How do I show that span$(cap l)$ =V.
It looks true to me. If we take a $cin l$. It would contain all the vectors of those sets(whose spans are also V) that are subsets of it.
And all supersets would also contain vectors of $c$. But I just can't seem to show vigorously that the intersection spans V. (i.e. can't find least element.)
Any ideas on how to prove it or is it false?
P.S. I'm trying to prove that every spanning set has a basis. i.e. Zorn's Lemma, Dual order etc
linear-algebra vector-spaces
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
$endgroup$
1
$begingroup$
As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span.
$endgroup$
– Arturo Magidin
Jan 19 at 3:32
$begingroup$
@ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true?
$endgroup$
– Jhon Doe
Jan 19 at 3:38
$begingroup$
I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $Lsubseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $Lsubseteq Bsubseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=varnothing$, and the existence of bases by taking $L=varnothing$ and $S=V$.
$endgroup$
– Arturo Magidin
Jan 19 at 4:13
1
$begingroup$
Actually, I think the statement you are trying to prove is false.
$endgroup$
– Arturo Magidin
Jan 19 at 4:36
add a comment |
$begingroup$
Suppose $Ssubseteq V$ is a spanning set(possibly infinite) of a non-zero vector space $V$.
Let $s'={ssubseteq S|$$s$ also spans V$}$. Suppose $ l subseteq s'$ is a totally ordered non-empty subset of $s'$.
How do I show that span$(cap l)$ =V.
It looks true to me. If we take a $cin l$. It would contain all the vectors of those sets(whose spans are also V) that are subsets of it.
And all supersets would also contain vectors of $c$. But I just can't seem to show vigorously that the intersection spans V. (i.e. can't find least element.)
Any ideas on how to prove it or is it false?
P.S. I'm trying to prove that every spanning set has a basis. i.e. Zorn's Lemma, Dual order etc
linear-algebra vector-spaces
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
$endgroup$
Suppose $Ssubseteq V$ is a spanning set(possibly infinite) of a non-zero vector space $V$.
Let $s'={ssubseteq S|$$s$ also spans V$}$. Suppose $ l subseteq s'$ is a totally ordered non-empty subset of $s'$.
How do I show that span$(cap l)$ =V.
It looks true to me. If we take a $cin l$. It would contain all the vectors of those sets(whose spans are also V) that are subsets of it.
And all supersets would also contain vectors of $c$. But I just can't seem to show vigorously that the intersection spans V. (i.e. can't find least element.)
Any ideas on how to prove it or is it false?
P.S. I'm trying to prove that every spanning set has a basis. i.e. Zorn's Lemma, Dual order etc
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
edited Jan 19 at 3:24
Jhon Doe
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
asked Jan 19 at 2:57
Jhon DoeJhon Doe
614314
614314
1
$begingroup$
As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span.
$endgroup$
– Arturo Magidin
Jan 19 at 3:32
$begingroup$
@ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true?
$endgroup$
– Jhon Doe
Jan 19 at 3:38
$begingroup$
I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $Lsubseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $Lsubseteq Bsubseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=varnothing$, and the existence of bases by taking $L=varnothing$ and $S=V$.
$endgroup$
– Arturo Magidin
Jan 19 at 4:13
1
$begingroup$
Actually, I think the statement you are trying to prove is false.
$endgroup$
– Arturo Magidin
Jan 19 at 4:36
add a comment |
1
$begingroup$
As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span.
$endgroup$
– Arturo Magidin
Jan 19 at 3:32
$begingroup$
@ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true?
$endgroup$
– Jhon Doe
Jan 19 at 3:38
$begingroup$
I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $Lsubseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $Lsubseteq Bsubseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=varnothing$, and the existence of bases by taking $L=varnothing$ and $S=V$.
$endgroup$
– Arturo Magidin
Jan 19 at 4:13
1
$begingroup$
Actually, I think the statement you are trying to prove is false.
$endgroup$
– Arturo Magidin
Jan 19 at 4:36
1
1
$begingroup$
As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span.
$endgroup$
– Arturo Magidin
Jan 19 at 3:32
$begingroup$
As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span.
$endgroup$
– Arturo Magidin
Jan 19 at 3:32
$begingroup$
@ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true?
$endgroup$
– Jhon Doe
Jan 19 at 3:38
$begingroup$
@ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true?
$endgroup$
– Jhon Doe
Jan 19 at 3:38
$begingroup$
I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $Lsubseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $Lsubseteq Bsubseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=varnothing$, and the existence of bases by taking $L=varnothing$ and $S=V$.
$endgroup$
– Arturo Magidin
Jan 19 at 4:13
$begingroup$
I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $Lsubseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $Lsubseteq Bsubseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=varnothing$, and the existence of bases by taking $L=varnothing$ and $S=V$.
$endgroup$
– Arturo Magidin
Jan 19 at 4:13
1
1
$begingroup$
Actually, I think the statement you are trying to prove is false.
$endgroup$
– Arturo Magidin
Jan 19 at 4:36
$begingroup$
Actually, I think the statement you are trying to prove is false.
$endgroup$
– Arturo Magidin
Jan 19 at 4:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I believe the statement you are trying to prove is false.
Consider $V=mathbb{R}$ as a vector space over $mathbb{R}$. Let $S$ be the interval $(-1,1)$ (which is certainly a spanning set; note that a subset of $V$ is a spanning set if and only if it contains at least one nonzero number). Now let $l={(-frac{1}{n},frac{1}{n})mid ninmathbb{N}_{gt 0}}$. This is a totally ordered subset of the collection of all subsets of $S$ that span $V$. However, $cap l = {0}$, which is not a spanning set.
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
add a comment |
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$begingroup$
I believe the statement you are trying to prove is false.
Consider $V=mathbb{R}$ as a vector space over $mathbb{R}$. Let $S$ be the interval $(-1,1)$ (which is certainly a spanning set; note that a subset of $V$ is a spanning set if and only if it contains at least one nonzero number). Now let $l={(-frac{1}{n},frac{1}{n})mid ninmathbb{N}_{gt 0}}$. This is a totally ordered subset of the collection of all subsets of $S$ that span $V$. However, $cap l = {0}$, which is not a spanning set.
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
add a comment |
$begingroup$
I believe the statement you are trying to prove is false.
Consider $V=mathbb{R}$ as a vector space over $mathbb{R}$. Let $S$ be the interval $(-1,1)$ (which is certainly a spanning set; note that a subset of $V$ is a spanning set if and only if it contains at least one nonzero number). Now let $l={(-frac{1}{n},frac{1}{n})mid ninmathbb{N}_{gt 0}}$. This is a totally ordered subset of the collection of all subsets of $S$ that span $V$. However, $cap l = {0}$, which is not a spanning set.
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
add a comment |
$begingroup$
I believe the statement you are trying to prove is false.
Consider $V=mathbb{R}$ as a vector space over $mathbb{R}$. Let $S$ be the interval $(-1,1)$ (which is certainly a spanning set; note that a subset of $V$ is a spanning set if and only if it contains at least one nonzero number). Now let $l={(-frac{1}{n},frac{1}{n})mid ninmathbb{N}_{gt 0}}$. This is a totally ordered subset of the collection of all subsets of $S$ that span $V$. However, $cap l = {0}$, which is not a spanning set.
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
$endgroup$
I believe the statement you are trying to prove is false.
Consider $V=mathbb{R}$ as a vector space over $mathbb{R}$. Let $S$ be the interval $(-1,1)$ (which is certainly a spanning set; note that a subset of $V$ is a spanning set if and only if it contains at least one nonzero number). Now let $l={(-frac{1}{n},frac{1}{n})mid ninmathbb{N}_{gt 0}}$. This is a totally ordered subset of the collection of all subsets of $S$ that span $V$. However, $cap l = {0}$, which is not a spanning set.
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
answered Jan 19 at 4:32
Arturo MagidinArturo Magidin
264k34588915
264k34588915
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
add a comment |
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
$begingroup$
Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"...
$endgroup$
– David C. Ullrich
Jan 19 at 14:17
add a comment |
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$begingroup$
As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span.
$endgroup$
– Arturo Magidin
Jan 19 at 3:32
$begingroup$
@ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true?
$endgroup$
– Jhon Doe
Jan 19 at 3:38
$begingroup$
I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $Lsubseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $Lsubseteq Bsubseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=varnothing$, and the existence of bases by taking $L=varnothing$ and $S=V$.
$endgroup$
– Arturo Magidin
Jan 19 at 4:13
1
$begingroup$
Actually, I think the statement you are trying to prove is false.
$endgroup$
– Arturo Magidin
Jan 19 at 4:36