Show that the annihilator (M perpendicular) of a non-empty set M in an inner product space X, is a closed subspace of X.

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- May 31st 2009, 07:47 AMfrater_cpAnnihilator of a set M in inner product space X
Show that the annihilator (M perpendicular) of a non-empty set M in an inner product space X, is a closed subspace of X.

- May 31st 2009, 09:32 AMTheEmptySet
- May 31st 2009, 11:02 AMRebesques
I think the question demands "show it is topologically closed".

To that end, define for fixed the function . This function is continuous (a fundamental property of the inner product), so is a closed set. This means

is closed as the intersection of a family of closed sets. - May 31st 2009, 11:38 AMfrater_cp
Thank you for your replies.