Example of a linear subset of Hilbert space that is not closed

[margaret37]

Homework Statement

Prove that for a linear set M a subset of Hilbert space, that the set perpendicular to the set perpendicular to M is equal to M iff M is closed.

The Attempt at a Solution

I already have my proof -- but what is an example of a linear subset of H that is not closed?

I think I understand that we need an infinite dimensional Hilbert space -- that in a finite space M would have to be closed.

Thanks

 

[Dick]

If H is a Hilbert space, and {e1,e2,...} is an orthonormal basis, then if you define S={a1*e1+a2*e2+...} where a finite number of the ai are nonzero, then S would be a subspace of H, right? But it wouldn't be closed.

 

[margaret37]

Thank you for answering, unfortunately I'm still confused.

Why wouldn't it be closed? If I set up a sequence within S, wouldn't it necessarily converge to something also inside S.

 

[Dick]

Thank you for answering, unfortunately I'm still confused.

Why wouldn't it be closed? If I set up a sequence within S, wouldn't it necessarily converge to something also inside S.

p1=e1, p2=e1+e2/2, p3=e1+e2/2+e3/4, p4=e1+e2/2+e3/4+e4/8. Do you see where this is going? Aren't all of the pi in S? Isn't the limit in the Hilbert space, since H is a complete metric space?