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 \(\displaystyle A \subseteq A^{ \bot \bot }\) ... ... ?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 \(\displaystyle A \subseteq A^{ \bot \bot }\) ... ... ?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 \(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0\) for all \(\displaystyle a \in A \}\)

and

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

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

\(\displaystyle \Longrightarrow \langle u, b \rangle = 0\)

\(\displaystyle \Longrightarrow \overline{ \langle b, u \rangle} = 0\)

\(\displaystyle \Longrightarrow \langle b, u \rangle = 0\)

\(\displaystyle \Longrightarrow u \in A^{ \bot \bot }\)... so that \(\displaystyle u \in A \Longrightarrow u \in A^{ \bot \bot }\)

Hence \(\displaystyle A \subseteq A^{ \bot \bot }\)
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)