Explain why, if $x \in F$, then $x \in {F}^{\perp\perp}$

Question

As far as the definition of orthogonality ${F}^{\perp} \!$ Is ${y \in E, \forall f \in F, (f|y)=0}.$

Let be $x \in F$,

We have by definition of ${F}^{\perp} \!: \forall y \in {F}^{\perp} \!, (x|y)=0$.

And then by definition of ${{F}^{\perp\perp} \! }: x \in {{F}^{\perp\perp} \! }$. But I don't understand why.

This lead to say that $F \subset {{F}^{\perp\perp} \! }$

Answer 1

By definition, we have $$ F^{\perp} = \{y \in X: \forall x \in F, (x|y) = 0\}\\ F^{\perp \perp} = \{y \in X: \forall x \in F^\perp, (x|y) = 0 \} $$ Now, if $x \in F$, then by definition of $F^\perp$: $(x|y)= (y|x) = 0$ for all $y$ in $F^\perp$. So, if $x \in F$, then it must be that $x \in F^{\perp \perp}$.