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

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Homework Help Overview

The discussion revolves around identifying an example of a linear subset of a Hilbert space that is not closed. The original poster seeks clarification on the properties of such subsets, particularly in the context of infinite dimensional Hilbert spaces.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the nature of a specific subset defined by a finite combination of basis elements and question its closure properties. There is an exploration of sequences within this subset and their convergence behavior.

Discussion Status

The conversation includes attempts to clarify why certain subsets are not closed, with participants expressing confusion about convergence and limits within the context of Hilbert spaces. There is an ongoing exploration of the implications of completeness in these spaces.

Contextual Notes

Participants note that the discussion is focused on infinite dimensional Hilbert spaces, suggesting that closure properties differ from those in finite dimensional spaces.

margaret37
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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
 
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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.
 
Last edited:
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.
 
margaret37 said:
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?
 

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