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Let $Y$ be a closed set of a Banach Space $X$. Is it true that the linear Span($Y$)is also closed?

For the examples I have tried, I see that the result holds true.

I understand that the linear span of any set is dense in the closed linear span of the same set. Now in my case, I am looking at the linear span of a closed set. However, I couldn't argue more over it to see as why the linear span of $Y$ should be closed in $X$.

G.Dinesh Nathan
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1 Answers1

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It's not true.

Take the set $Y\subset\ell_2$ consisting of the standard unit vectors. $Y$ is closed, since it has no accumulation points (any two distinct points in $Y$ are $\sqrt 2$ units apart).

Now, the linear span of $Y$ is dense in $\ell_2$ but not all of $\ell_2$. It follows from this that the linear span of $E$ is not closed.

David Mitra
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  • Thank you. As the standard unit vectors form a basis for $\ell_2$ why is that linear span of $Y$ is not all of $\ell_2$? – G.Dinesh Nathan Jun 03 '14 at 13:22
  • @G.DineshNathan The linear span is the set of finite linear combinations of elements of $Y$. Any element of the linear span of $Y$ has only finitely many non-zero coordinates. There are elements of $\ell_2$ with infinitely many non-zero coordinates. (Note when you say "basis", you mean "Schauder basis". A Schauder basis allows infinite sums.) – David Mitra Jun 03 '14 at 13:26