Prop 11.3.5-4: Peter's Help with Garling's The Annihilator of a Set

[Math Amateur]

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help in order to formulate a proof of Proposition 11.3.5 - 4 ...

Garling's statement and proof of Proposition 11.3.5 reads as follows:
View attachment 8964Can someone please help me formulate a formal and rigorous proof that $$A \subseteq A^{ \bot \bot }$$ ... ... ?Help will be much appreciated ...

Peter

 

[Math Amateur]

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help in order to formulate a proof of Proposition 11.3.5 - 4 ...

Garling's statement and proof of Proposition 11.3.5 reads as follows:
Can someone please help me formulate a formal and rigorous proof that $$A \subseteq A^{ \bot \bot }$$ ... ... ?Help will be much appreciated ...

Peter

I have been reflecting on my post above ...

I think I have a proof ... but not sure ... we proceed as follows ...We have $$A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0$$ for all $$a \in A \}$$

and

$$A^{ \bot \bot } = \{ y \in V \ : \ \langle b, y \rangle = 0$$ for all $$ b \in A^{ \bot } \}$$Now ... ... let $$u \in A$$ ...

then $$u \in A^{ \bot \bot }$$ if $$ \langle b, u \rangle = 0 \ \forall \ b \in A^{ \bot }$$But $$b \in A^{ \bot } \Longrightarrow \langle a, b \rangle = 0 \ \forall \ a \in A$$

$$\Longrightarrow \langle u, b \rangle = 0$$

$$\Longrightarrow \overline{ \langle b, u \rangle} = 0$$

$$\Longrightarrow \langle b, u \rangle = 0$$

$$\Longrightarrow u \in A^{ \bot \bot }$$... so that $$u \in A \Longrightarrow u \in A^{ \bot \bot }$$

Hence $$A \subseteq A^{ \bot \bot }$$
Can someone please confirm that the above proof is correct and/or point out errors and give a correct version of the proof ...

Peter

 

[Opalg]

Yes, that's completely correct. When dealing with annihilators and second annihilators, you need to get used to the fact that $\langle x,y\rangle = 0 \Longleftrightarrow \langle y,x\rangle = 0$.

 

[Olinguito]

Hi Peter.

Yep, the proof looks correct to me. (Yes)