Orthogonal complements- properties

Question

Show that for a subspace E we have (E⊥)⊥ = E.

For this, I am assuming one would need to start with the decomposition that is x= x_1 + x_2. So we would have x_1 in E, and x_2 in E⊥, or x_2 perpendicular to x_1. But where do I go from there?

Answer 1

Let $x \in E$. Then, by definition of $\perp$, $x \perp y$ for all $y \in E_\perp$, since elements of $E_\perp$ are orthogonal to every element of $E$. Again, by definition of $\perp$, $x$ is perpendincular to every element of $E_\perp$, so $x \in (E_\perp)_\perp$. One way is done.

The other way proceeds as follows: We know that $E\oplus E_\perp = X$, the whole space. Now, take a further orthogonal complement, and see that $E_\perp \oplus (E_\perp)_\perp = X$. Hence, the dimension of $S$ and $(S_\perp)_\perp$ are the same, hence if one is contained in the other, then they are equal since both are subspaces. But we already proved one inclusion, hence both are the same.

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In general, in the infinite dimensional setting , not all subspaces are closed. Therefore, it is not true anymore, that $E = (E^{\perp})^\perp$. It turns out that this is true if $E$ is closed. More precisely, $(E^{\perp})^{\perp} = \overline E$, the closure of $E$. Finite dimensional subspaces are closed even in the infinite dimensional setting, therefore if $E$ is finite dimensional even inside an infinite dimensional space, this holds true.

An example of this is to take a separable Hilbert space and take the span of any countable basis. This is a subspace that is dense, but is not the whole space (can you find an element of the Hilbert space not inside this span? The hint is that it must be an infinite linear combination , with coefficients having finite sum), and is therefore not closed. In fact, if $E$ is this space, then its orthogonal complement is the zero vector space, by the Hahn-Banach theorem.