Uniqueness of dimension in the infinite-dimensional case.
$begingroup$
Say $V$ is a vector space over some field.
Lemma Suppose $A,B,Csubset V$, $Acup B$ spans $V$ and $Bcup C$ is independent. For every $cin C$ there exists $ain A$ such that if $A'=Asetminus{a}$ and $B'=Bcup{c}$ then $A'cup B'$ spans $V$.
(Write $c$ as a linear combination of elements of $Acup B$ and note that some element of $A$ must appear with a non-zero coefficient, since $Bcup C$ is independent...)
Cor Suppose $A$ spans $V$ and $C$ is independent. If $A$ is finite then $|A|ge|C|$.
Proof: Let $A_0=A$, $B_0=emptyset$, $C_0=C$. Apply the lemma repeatedly, setting $A_{n+1}=A_n'$, $B_{n+1}=B_n'$, $C_{n+1}=C_n'=C_nsetminus {c}$. If $|A|<|C|$ then there exists $n$ so that $A_n=emptyset$ and $C_nneemptyset$; hence $B_n$ is a proper subset of the independent set $C$ and $B_n$ spans $V$, contradiction.
Question: Is there a way to get the Corollary from the Lemma if $A$ is infinite?
"My work so far:" The obvious transfinite recursion. The problem is if $alpha$ is a limit ordinal: I can't imagine what' $A_alpha$ and $B_alpha$ could be other than $bigcap_{beta<alpha}A_beta$ and $bigcup_{beta<alpha}B_beta$, but in that case I see no reason that $A_alphacup B_alpha$ should span $V$.
The reason I suspect I may just be missing something is that in Finite Dimensional Vector Spaces Halmos comments that he tends to give proofs that can be generalized to the infinite-dimensional case.
Edit: It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia; see the Answer below for a paraphrase.
For anyone missing the point:
Exercise. (i) Show that "maximal independent set" is equivalent to "minimal spanning set". (ii) Explain why nonetheless the first phrase always comes up in a proof that every vector space has a basis, never the second..
linear-algebra transfinite-induction
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
|
show 2 more comments
$begingroup$
Say $V$ is a vector space over some field.
Lemma Suppose $A,B,Csubset V$, $Acup B$ spans $V$ and $Bcup C$ is independent. For every $cin C$ there exists $ain A$ such that if $A'=Asetminus{a}$ and $B'=Bcup{c}$ then $A'cup B'$ spans $V$.
(Write $c$ as a linear combination of elements of $Acup B$ and note that some element of $A$ must appear with a non-zero coefficient, since $Bcup C$ is independent...)
Cor Suppose $A$ spans $V$ and $C$ is independent. If $A$ is finite then $|A|ge|C|$.
Proof: Let $A_0=A$, $B_0=emptyset$, $C_0=C$. Apply the lemma repeatedly, setting $A_{n+1}=A_n'$, $B_{n+1}=B_n'$, $C_{n+1}=C_n'=C_nsetminus {c}$. If $|A|<|C|$ then there exists $n$ so that $A_n=emptyset$ and $C_nneemptyset$; hence $B_n$ is a proper subset of the independent set $C$ and $B_n$ spans $V$, contradiction.
Question: Is there a way to get the Corollary from the Lemma if $A$ is infinite?
"My work so far:" The obvious transfinite recursion. The problem is if $alpha$ is a limit ordinal: I can't imagine what' $A_alpha$ and $B_alpha$ could be other than $bigcap_{beta<alpha}A_beta$ and $bigcup_{beta<alpha}B_beta$, but in that case I see no reason that $A_alphacup B_alpha$ should span $V$.
The reason I suspect I may just be missing something is that in Finite Dimensional Vector Spaces Halmos comments that he tends to give proofs that can be generalized to the infinite-dimensional case.
Edit: It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia; see the Answer below for a paraphrase.
For anyone missing the point:
Exercise. (i) Show that "maximal independent set" is equivalent to "minimal spanning set". (ii) Explain why nonetheless the first phrase always comes up in a proof that every vector space has a basis, never the second..
linear-algebra transfinite-induction
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
I believe you want to do this with Zorn's lemma, instead of transfinite induction. If you want clarification I can write a complete answer.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:15
$begingroup$
@SmileyCraft Please do write that complete answer! The problem with Zorn's lemma is the same, as far as I can see: The intersection of a (decreasing) chain of spanning sets need not be a spanning set. (Hence part (ii) of the exercise...)
$endgroup$
– David C. Ullrich
Dec 14 '18 at 15:21
$begingroup$
I'll admit that I underestimated the difficulty here. Do you know an example of a decreasing chain of spanning sets of which the intersection is not a spanning set?
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:43
$begingroup$
Nevermind, I found one myself: $A_n:={1/k:k>n}subsetmathbb{R}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:46
$begingroup$
I think that the question exactly as it is stated would be a definite "no". You an't get the infinite-dimensional case from the lemma. That lemma will not get you past a limit ordinal as you've noticed. Of course, that doesn't mean there aren't other ways to handle it.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:42
|
show 2 more comments
$begingroup$
Say $V$ is a vector space over some field.
Lemma Suppose $A,B,Csubset V$, $Acup B$ spans $V$ and $Bcup C$ is independent. For every $cin C$ there exists $ain A$ such that if $A'=Asetminus{a}$ and $B'=Bcup{c}$ then $A'cup B'$ spans $V$.
(Write $c$ as a linear combination of elements of $Acup B$ and note that some element of $A$ must appear with a non-zero coefficient, since $Bcup C$ is independent...)
Cor Suppose $A$ spans $V$ and $C$ is independent. If $A$ is finite then $|A|ge|C|$.
Proof: Let $A_0=A$, $B_0=emptyset$, $C_0=C$. Apply the lemma repeatedly, setting $A_{n+1}=A_n'$, $B_{n+1}=B_n'$, $C_{n+1}=C_n'=C_nsetminus {c}$. If $|A|<|C|$ then there exists $n$ so that $A_n=emptyset$ and $C_nneemptyset$; hence $B_n$ is a proper subset of the independent set $C$ and $B_n$ spans $V$, contradiction.
Question: Is there a way to get the Corollary from the Lemma if $A$ is infinite?
"My work so far:" The obvious transfinite recursion. The problem is if $alpha$ is a limit ordinal: I can't imagine what' $A_alpha$ and $B_alpha$ could be other than $bigcap_{beta<alpha}A_beta$ and $bigcup_{beta<alpha}B_beta$, but in that case I see no reason that $A_alphacup B_alpha$ should span $V$.
The reason I suspect I may just be missing something is that in Finite Dimensional Vector Spaces Halmos comments that he tends to give proofs that can be generalized to the infinite-dimensional case.
Edit: It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia; see the Answer below for a paraphrase.
For anyone missing the point:
Exercise. (i) Show that "maximal independent set" is equivalent to "minimal spanning set". (ii) Explain why nonetheless the first phrase always comes up in a proof that every vector space has a basis, never the second..
linear-algebra transfinite-induction
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
Say $V$ is a vector space over some field.
Lemma Suppose $A,B,Csubset V$, $Acup B$ spans $V$ and $Bcup C$ is independent. For every $cin C$ there exists $ain A$ such that if $A'=Asetminus{a}$ and $B'=Bcup{c}$ then $A'cup B'$ spans $V$.
(Write $c$ as a linear combination of elements of $Acup B$ and note that some element of $A$ must appear with a non-zero coefficient, since $Bcup C$ is independent...)
Cor Suppose $A$ spans $V$ and $C$ is independent. If $A$ is finite then $|A|ge|C|$.
Proof: Let $A_0=A$, $B_0=emptyset$, $C_0=C$. Apply the lemma repeatedly, setting $A_{n+1}=A_n'$, $B_{n+1}=B_n'$, $C_{n+1}=C_n'=C_nsetminus {c}$. If $|A|<|C|$ then there exists $n$ so that $A_n=emptyset$ and $C_nneemptyset$; hence $B_n$ is a proper subset of the independent set $C$ and $B_n$ spans $V$, contradiction.
Question: Is there a way to get the Corollary from the Lemma if $A$ is infinite?
"My work so far:" The obvious transfinite recursion. The problem is if $alpha$ is a limit ordinal: I can't imagine what' $A_alpha$ and $B_alpha$ could be other than $bigcap_{beta<alpha}A_beta$ and $bigcup_{beta<alpha}B_beta$, but in that case I see no reason that $A_alphacup B_alpha$ should span $V$.
The reason I suspect I may just be missing something is that in Finite Dimensional Vector Spaces Halmos comments that he tends to give proofs that can be generalized to the infinite-dimensional case.
Edit: It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia; see the Answer below for a paraphrase.
For anyone missing the point:
Exercise. (i) Show that "maximal independent set" is equivalent to "minimal spanning set". (ii) Explain why nonetheless the first phrase always comes up in a proof that every vector space has a basis, never the second..
linear-algebra transfinite-induction
linear-algebra transfinite-induction
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
edited Dec 15 '18 at 14:59
David C. Ullrich
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
asked Dec 14 '18 at 15:07
David C. UllrichDavid C. Ullrich
61.2k43994
61.2k43994
$begingroup$
I believe you want to do this with Zorn's lemma, instead of transfinite induction. If you want clarification I can write a complete answer.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:15
$begingroup$
@SmileyCraft Please do write that complete answer! The problem with Zorn's lemma is the same, as far as I can see: The intersection of a (decreasing) chain of spanning sets need not be a spanning set. (Hence part (ii) of the exercise...)
$endgroup$
– David C. Ullrich
Dec 14 '18 at 15:21
$begingroup$
I'll admit that I underestimated the difficulty here. Do you know an example of a decreasing chain of spanning sets of which the intersection is not a spanning set?
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:43
$begingroup$
Nevermind, I found one myself: $A_n:={1/k:k>n}subsetmathbb{R}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:46
$begingroup$
I think that the question exactly as it is stated would be a definite "no". You an't get the infinite-dimensional case from the lemma. That lemma will not get you past a limit ordinal as you've noticed. Of course, that doesn't mean there aren't other ways to handle it.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:42
|
show 2 more comments
$begingroup$
I believe you want to do this with Zorn's lemma, instead of transfinite induction. If you want clarification I can write a complete answer.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:15
$begingroup$
@SmileyCraft Please do write that complete answer! The problem with Zorn's lemma is the same, as far as I can see: The intersection of a (decreasing) chain of spanning sets need not be a spanning set. (Hence part (ii) of the exercise...)
$endgroup$
– David C. Ullrich
Dec 14 '18 at 15:21
$begingroup$
I'll admit that I underestimated the difficulty here. Do you know an example of a decreasing chain of spanning sets of which the intersection is not a spanning set?
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:43
$begingroup$
Nevermind, I found one myself: $A_n:={1/k:k>n}subsetmathbb{R}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:46
$begingroup$
I think that the question exactly as it is stated would be a definite "no". You an't get the infinite-dimensional case from the lemma. That lemma will not get you past a limit ordinal as you've noticed. Of course, that doesn't mean there aren't other ways to handle it.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:42
$begingroup$
I believe you want to do this with Zorn's lemma, instead of transfinite induction. If you want clarification I can write a complete answer.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:15
$begingroup$
I believe you want to do this with Zorn's lemma, instead of transfinite induction. If you want clarification I can write a complete answer.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:15
$begingroup$
@SmileyCraft Please do write that complete answer! The problem with Zorn's lemma is the same, as far as I can see: The intersection of a (decreasing) chain of spanning sets need not be a spanning set. (Hence part (ii) of the exercise...)
$endgroup$
– David C. Ullrich
Dec 14 '18 at 15:21
$begingroup$
@SmileyCraft Please do write that complete answer! The problem with Zorn's lemma is the same, as far as I can see: The intersection of a (decreasing) chain of spanning sets need not be a spanning set. (Hence part (ii) of the exercise...)
$endgroup$
– David C. Ullrich
Dec 14 '18 at 15:21
$begingroup$
I'll admit that I underestimated the difficulty here. Do you know an example of a decreasing chain of spanning sets of which the intersection is not a spanning set?
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:43
$begingroup$
I'll admit that I underestimated the difficulty here. Do you know an example of a decreasing chain of spanning sets of which the intersection is not a spanning set?
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:43
$begingroup$
Nevermind, I found one myself: $A_n:={1/k:k>n}subsetmathbb{R}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:46
$begingroup$
Nevermind, I found one myself: $A_n:={1/k:k>n}subsetmathbb{R}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:46
$begingroup$
I think that the question exactly as it is stated would be a definite "no". You an't get the infinite-dimensional case from the lemma. That lemma will not get you past a limit ordinal as you've noticed. Of course, that doesn't mean there aren't other ways to handle it.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:42
$begingroup$
I think that the question exactly as it is stated would be a definite "no". You an't get the infinite-dimensional case from the lemma. That lemma will not get you past a limit ordinal as you've noticed. Of course, that doesn't mean there aren't other ways to handle it.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:42
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia;. Here's the same proof, in a version that seems somewhat cleaner to me:
Suppose $A$ is infinite. Since any independent set is contained in a maximal independent set, wlog $C$ is actually a basis. For $cin C$ let $pi_c$ be the corresponding coordinate functional, so that $$x=sum_{cin C}pi_c(x)c$$for every $xin V$.
Let $S(x)$ be the support of $xin V$: $$S(x)={cin C:pi_c(x)ne0}.$$Since $A$ spans $V$ it's clear that $$S(x)subset S(A):=bigcup_{ain A}S(a)$$for every $xin V$. We certainly have $$S(c)={c}quad(cin C),$$ so $$Csubset S(A).$$And since $A$ is infinite and each $S(a)$ is finite it follows that $$|S(A)|le|A|.$$
I feel so silly not having come up with that on my own.
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
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$begingroup$
It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia;. Here's the same proof, in a version that seems somewhat cleaner to me:
Suppose $A$ is infinite. Since any independent set is contained in a maximal independent set, wlog $C$ is actually a basis. For $cin C$ let $pi_c$ be the corresponding coordinate functional, so that $$x=sum_{cin C}pi_c(x)c$$for every $xin V$.
Let $S(x)$ be the support of $xin V$: $$S(x)={cin C:pi_c(x)ne0}.$$Since $A$ spans $V$ it's clear that $$S(x)subset S(A):=bigcup_{ain A}S(a)$$for every $xin V$. We certainly have $$S(c)={c}quad(cin C),$$ so $$Csubset S(A).$$And since $A$ is infinite and each $S(a)$ is finite it follows that $$|S(A)|le|A|.$$
I feel so silly not having come up with that on my own.
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia;. Here's the same proof, in a version that seems somewhat cleaner to me:
Suppose $A$ is infinite. Since any independent set is contained in a maximal independent set, wlog $C$ is actually a basis. For $cin C$ let $pi_c$ be the corresponding coordinate functional, so that $$x=sum_{cin C}pi_c(x)c$$for every $xin V$.
Let $S(x)$ be the support of $xin V$: $$S(x)={cin C:pi_c(x)ne0}.$$Since $A$ spans $V$ it's clear that $$S(x)subset S(A):=bigcup_{ain A}S(a)$$for every $xin V$. We certainly have $$S(c)={c}quad(cin C),$$ so $$Csubset S(A).$$And since $A$ is infinite and each $S(a)$ is finite it follows that $$|S(A)|le|A|.$$
I feel so silly not having come up with that on my own.
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
add a comment |
$begingroup$
It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia;. Here's the same proof, in a version that seems somewhat cleaner to me:
Suppose $A$ is infinite. Since any independent set is contained in a maximal independent set, wlog $C$ is actually a basis. For $cin C$ let $pi_c$ be the corresponding coordinate functional, so that $$x=sum_{cin C}pi_c(x)c$$for every $xin V$.
Let $S(x)$ be the support of $xin V$: $$S(x)={cin C:pi_c(x)ne0}.$$Since $A$ spans $V$ it's clear that $$S(x)subset S(A):=bigcup_{ain A}S(a)$$for every $xin V$. We certainly have $$S(c)={c}quad(cin C),$$ so $$Csubset S(A).$$And since $A$ is infinite and each $S(a)$ is finite it follows that $$|S(A)|le|A|.$$
I feel so silly not having come up with that on my own.
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
It seems likely the answer is no, since nobody's come up with a yes. This raises the question "ok, how do you prove the corollary for infinite $A$?" In fact there's a very simple proof of the corollary for infinite $A$ on Wikipedia;. Here's the same proof, in a version that seems somewhat cleaner to me:
Suppose $A$ is infinite. Since any independent set is contained in a maximal independent set, wlog $C$ is actually a basis. For $cin C$ let $pi_c$ be the corresponding coordinate functional, so that $$x=sum_{cin C}pi_c(x)c$$for every $xin V$.
Let $S(x)$ be the support of $xin V$: $$S(x)={cin C:pi_c(x)ne0}.$$Since $A$ spans $V$ it's clear that $$S(x)subset S(A):=bigcup_{ain A}S(a)$$for every $xin V$. We certainly have $$S(c)={c}quad(cin C),$$ so $$Csubset S(A).$$And since $A$ is infinite and each $S(a)$ is finite it follows that $$|S(A)|le|A|.$$
I feel so silly not having come up with that on my own.
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
answered Dec 15 '18 at 14:57
David C. UllrichDavid C. Ullrich
61.2k43994
61.2k43994
add a comment |
add a comment |
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$begingroup$
I believe you want to do this with Zorn's lemma, instead of transfinite induction. If you want clarification I can write a complete answer.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:15
$begingroup$
@SmileyCraft Please do write that complete answer! The problem with Zorn's lemma is the same, as far as I can see: The intersection of a (decreasing) chain of spanning sets need not be a spanning set. (Hence part (ii) of the exercise...)
$endgroup$
– David C. Ullrich
Dec 14 '18 at 15:21
$begingroup$
I'll admit that I underestimated the difficulty here. Do you know an example of a decreasing chain of spanning sets of which the intersection is not a spanning set?
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:43
$begingroup$
Nevermind, I found one myself: $A_n:={1/k:k>n}subsetmathbb{R}$.
$endgroup$
– SmileyCraft
Dec 14 '18 at 15:46
$begingroup$
I think that the question exactly as it is stated would be a definite "no". You an't get the infinite-dimensional case from the lemma. That lemma will not get you past a limit ordinal as you've noticed. Of course, that doesn't mean there aren't other ways to handle it.
$endgroup$
– Paul Sinclair
Dec 14 '18 at 23:42