cl$\left(span(S)\right)$=cl$\left(span(span(S))\right)$?

Question

Ive managed to confuse myself regarding the cl$\left(span(S)\right)$=cl$\left(span(span(S))\right)$? relation

Is any element in the closed linear span is series of scalar multiplied elements of the span?

We have that every element of cl$(span(S))$ cannot always be written as a series of scalar multiplies of elements of $S$ unless $S$ is linear and then it can. Furthermore, every element of cl$\left(span(span(S))\right)$ can be written as a series of scalar multiples of elements of $span(S)$ which is also a series of scalar multiples of elements of $S$. I.e

$\sum^{\infty}_{1} a_{n}( \sum^{n}_{1}b_{n}s_{n}) = \sum^{\infty}_{1}c_{n}s_{n}$

Hence cl$\left(span(S)\right)$=cl$\left(span(span(S))\right)$? , but this cant be since every element of cl$(span(S))$ was not a series of scalar multiplied elements of $S$ while the RHS is.

Use the characterisation "$\operatorname{span} S$ is the smallest linear subspace that contains $S$". — Daniel Fischer, Oct 12 '15 at 10:16
@DanielFischer Im sorry but that just gives me span(S) $ \subseteq $ span(span(S)). not that span(span(S)) is striclty larger accodrind to me atleast — user123124, Oct 12 '15 at 11:01
$\operatorname{span} S$ is a linear subspace that contains $\operatorname{span} S$. Clearly, it is the smallest such. Therefore $\operatorname{span} (\operatorname{span} S) = \operatorname{span} S$. Perhaps it's easier to see more abstractly, if $L$ is a linear subspace, then $L$ is the smallest linear subspace containing $L$. Therefore $\operatorname{span} L = L$. Take $L = \operatorname{span} S$. — Daniel Fischer, Oct 12 '15 at 11:11
@DanielFischer This is what I get out of the linked post. That if we take closed linear span of a set which is a subspace then we get the set of norm converget series but we cant garantee this unless we take a subspace as Yukis counterexample shows. — user123124, Oct 12 '15 at 11:21
@HenningMakholm that for some norm induced topology we dont have the same sets when taking closure or RHS and LHS — user123124, Oct 12 '15 at 11:22
The linked post isn't very clear. It confounds several issues. Let's first get our definitions straight. Our universe is a topological vector space $E$. a) The (linear) span of a subset $S\subset E$ is the smallest linear subspace of $E$ that contains $S$. b) The closed (linear) span of a subset $S\subset E$ is the closure of the linear span of $S$, or equivalently, the smallest closed linear subspace of $E$ containing $S$. If we denote the closed linear span operator by $\overline{\operatorname{span}}$, then both $\operatorname{span}$ and $\overline{\operatorname{span}}$ are idempotent … — Daniel Fischer, Oct 12 '15 at 11:36
… operators, i.e. $\operatorname{span} (\operatorname{span} S) = \operatorname{span} S$ and $\overline{\operatorname{span}} (\overline{\operatorname{span}} S) = \overline{\operatorname{span}} S$ for all $S\subset E$. It is an entirely different question whether each element of $\overline{\operatorname{span}} S$ can be written as a series of elements of $\operatorname{span} S$ (that is always the case if $E$ is metrisable) or a series of elements of $S$ (that is rarely the case). — Daniel Fischer, Oct 12 '15 at 11:36
@DanielFischer Every element of cl(span(S)) cannot always be written as a series of elements of S but if S is linear then it can. I.e every element of cl(span(span(S)) can be written as a series of elements of span(S) which are sums of elements of S .But if span(span(S))=span(S) then so can the elements of cl(span(S)). If you are right which im sure you are then somthing where donst add up. — user123124, Oct 12 '15 at 11:58
elements of span(S) which are elements of S No. Unless $S$ was already a linear subspace, elements of $\operatorname{span} S$ are usually not elements of $S$. Let $L = \operatorname{span} S$, and assume that the ambient topological vector space is metrisable, perhaps even a normed space if you're more comfortable with that. Then every element of $\overline{\operatorname{span} S}$ can be written as a series of elements of $L$. — Daniel Fischer, Oct 12 '15 at 12:04
@DanielFischer Im sorry I edited to "sums of elemets of S" if that changes anything — user123124, Oct 12 '15 at 12:06
They're not "sums of elements of $S$" generally, they're linear combinations of elements of $S$. And that makes a difference. — Daniel Fischer, Oct 12 '15 at 12:13
@DanielFischer I didnt see the difference I figured a series of elements of L is also a series of elemtes of S since the sequence of partial sums are sums in elemets of S, just factor out $s_{i}$ but this probobly donst work in the limit right? — user123124, Oct 12 '15 at 12:18
No. It doesn't even work for finite sums. The elements of $L$ are linear combinations of elements of $S$, that is, you have coeffcients. If we assume that we're dealing with real or complex (or rational) vector spaces, the only coefficients you can get from sums of elements of $S$ are non-negative integers (aka natural numbers, $0,1,2,3,4,\dotsc$). But in a linear combination, you can have negative coefficients and non-integer coefficients. That gets you tons and tons of more elements. — Daniel Fischer, Oct 12 '15 at 12:23
@DanielFischer They way I looked at it i just factored out the vector to the right and summed the "coefficents" using distubutivity of scalars. Hence i did mean combinations above — user123124, Oct 12 '15 at 12:28
But you must not confuse a "sum of (linear combinations of elements of $S$)" and a "sum of elements of $S$". If $S$ happens to be a linear subspace, the two things coincide, but in general the set described by the first sentence is much larger than the set described by the second. — Daniel Fischer, Oct 12 '15 at 12:38
@DanielFischer I think this is true even if we replace sum in your last comment by "linear combination", right? sums seem unimportant in this context. — user123124, Oct 12 '15 at 12:45
No, linear combinations behave differently. A "linear combination of (linear combinations of elements of $S$)" is a "linear combination of elements of $S$". The distinction between linear combinations and sums/series is the point. — Daniel Fischer, Oct 12 '15 at 12:53
@DanielFischer Im very sorry, sums was a poor choice of word. If we consider linear combinations all the way then I think we will understand eachother. — user123124, Oct 12 '15 at 13:58

score 1 · Answer 1 · answered Oct 12 '15 at 11:01

1

Following the comment of @Daniel Fischer,Here is a solution:

Recall that span($S$) is the smallest subspace of vector space containing $S$ and elements of span($S$) are finite linear combination of elements of $S$.

As, $ S \subset$ span($S$) therefore clearly span($S$)$\subset$ span(span($S$)).Also as span(span(S)) contains $S$ and is a subspace of $V$ hence must contain span($S$),because span($S$) is smallest subspace containing $S$.Therefore span(span($S$)=span($S$)

answered Oct 12 '15 at 11:01

Arpit Kansal

10,268

But then thier closure should be equal aswell right? which its not by http://math.stackexchange.com/questions/1475138/is-any-element-in-the-closed-linear-span-is-series-of-elements-of-the-span – user123124 Oct 12 '15 at 11:06
This seems unnecessarily convoluted. Instead say: Because the span of A is the smallest linear subspace that contains A, every linear subspace is its own span! In particular, since span(S) is a subspace, it is its own span, so span(span(S))=span(S). – hmakholm left over Monica Oct 12 '15 at 11:13
@HenningMakholm agree,but is there any probelm is my solution? – Arpit Kansal Oct 12 '15 at 11:18
@ArpitKansal I also have this..but I think its contradicting the linked post and the question is why it dosnt – user123124 Oct 12 '15 at 11:34

user123124 · Accepted Answer · 2015-10-13T06:59:50.697

The confusion seems to been arising from the following conclusion ; "Furthermore, every element of cl(span(span(S)) can be written as a series of scalar multiples of elements of span(S) which is aslo a series of scalar multiples of elements of S. But

$\sum^{\infty}_{1} a_{n} \sum^{n}_{1}b_{n}s_{n} \neq \sum^{\infty}_{1}c_{n}s_{n}$ unless S is a vector space. Since $b_{n}s_{n}$ might not be in $S$

Its not even true for finite sums.

cl$\left(span(S)\right)$=cl$\left(span(span(S))\right)$?

2 Answers2