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A projection operator on a Hilbert space $H$ is defined as operator that projects a vector $x$ of $H$ onto an closed subspace $S$ of $H$. Why the subspace $S$ has to be closed?

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It depends: exactly what properties would you like your projection operator to have? – Nate Eldredge Nov 30 '11 at 23:55

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$S = \{x: P(x) = x\}$, so if the projection $P$ is continuous $S$ must be closed.

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Can we say that if $S$ wasn't closed, the projection $P$ would be discontinuous function at some $x$'s so we couldn't project this $x$'s onto $S$? – Andyk Dec 1 '11 at 16:36
We might project it onto $S$, but not with a continuous projection. – Robert Israel Dec 2 '11 at 0:21
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Your definition of projection could be regarded redundant. However something interesting is given by the case when your subspace is not closed, but a non-closed dense subset. To which vector do project your $x \notin S$ then?

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Can you give an example of such an $S$? – Srivatsan Dec 1 '11 at 2:58

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