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It is well known that:

A closed subspace $S\subseteq H$ and $H$ is Hilbert space, then $H = S\oplus S^{\perp}$ and $ S^{\perp}$ is also closed.

I'm thinking that since $S^{\perp} = H\setminus S$ and $S$ is closed, so $S^{\perp}$ is also open?


hmm, sure.$S^{\perp}\cap S = {0}\neq \emptyset$, the previous equality is invalid.

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    If $S$ is a subspace it has to contain $0$, and since $H^{\vert}$ also contains $0$, we can't have the equality you wrote. If a subspace $F$ is open, it must contain a ball $B(0,r)$. Hence for all $x\in H$ we must have $\frac{r x}{2\lVert x\rVert}\in F$, hence this subspace has to be the whole space.2011-11-19

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I think you're interpreting the direct sum as a union, which is incorrect. $S\oplus S^\perp$ is not the same as $S\cup S^\perp$. Writing $H=S\oplus S^\perp$ means that every member of $H$ can be written as a sum of vectors, one from $H$ and the other from $H^\perp$. As mentioned by Daniel, this can happen without $S\cup S^\perp$ being the whole space.

As another example, consider $\Bbb R^2$ and the coordinate axes...

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Since $H$ is convex, hence connected, this never happens. Take for example $S$ a line through origin in $\mathrm R^3$, then $S^\perp$ is a plane, which is not open.

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    The equality $S^\perp = H \setminus S$ is not true, it is actually $S^\perp = H/S$ (considering isomorphisms as equalities).2011-11-19