# How Do We Know the Annihilator Operation Is Associative?

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In summary: This shows that $(A^\bot)^{\bot\bot} = (A^{\bot\bot})^\bot$, and thus $(A^\bot)^{\bot\bot} = A^{\bot\bot\bot}$. Therefore, the assumption made in Garling's proof is true. In summary, the conversation discusses understanding Garling's proof of Proposition 11.3.5, specifically the assumption that $(A^\bot)^{\bot\bot} = (A^{\bot\bot})^\bot = A^{\bot\bot\bot}$. This assumption is shown to be true by observing the associativity of the annihilator operation.
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MHB
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 understand Garling's proof of Proposition 11.3.5 - 5 ...

Garling's statement and proof of Proposition 11.3.5 reads as follows:View attachment 8965I can follow Garling's proof of Proposition 11.3.5 - 5 (except that when he refers to (iii) and (iv) ... he means 3 and 4 ...)

... ... BUT ... the proof assumes $$\displaystyle (A^{ \bot } )^{ \bot \bot } = (A^{ \bot \bot } )^{ \bot } = A^{ \bot \bot \bot }$$ ... ...How do we know that this is the case ... that is, true ... ?

Peter

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• Garling - Proposition 11.3.5 ... Annihilator ... .,.png
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Peter said:
... ... BUT ... the proof assumes $$\displaystyle (A^{ \bot } )^{ \bot \bot } = (A^{ \bot \bot } )^{ \bot } = A^{ \bot \bot \bot }$$ ... ...

$A^{\bot\bot\bot}$ is just a shorthand way of writing $(A^{ \bot } )^{ \bot \bot }$ or $(A^{ \bot \bot } )^{ \bot}$ – if both are equal. And they are, being equal to $A^\bot$.

Last edited:
Peter said:
the proof assumes $$\displaystyle (A^{ \bot } )^{ \bot \bot } = (A^{ \bot \bot } )^{ \bot } = A^{ \bot \bot \bot }$$ ... ...How do we know that this is the case ... that is, true ... ?
To see that the annihilator operation is associative, notice that the condition for $x\in (A^\bot)^{\bot\bot}$ is the same as the condition for $x\in(A^{\bot\bot})^\bot$. In both cases, the condition is that $\langle x,y\rangle = 0$ for all $y$ in $A^{\bot\bot}$.

## 1. What is the Annihilator of a Set?

The Annihilator of a Set is a mathematical concept used in abstract algebra. It is defined as the set of all elements in a given algebraic structure that multiply to zero when paired with any element in the original set.

## 2. How is the Annihilator of a Set calculated?

The Annihilator of a Set is calculated by finding all elements in the original set that, when multiplied with any element in the given algebraic structure, result in zero. These elements are then combined to form the Annihilator of the Set.

## 3. What is the significance of the Annihilator of a Set?

The Annihilator of a Set is used to study the structure of algebraic systems. It helps in identifying elements that have a special property, such as being a zero divisor, and can provide insights into the properties of the original set.

## 4. Can the Annihilator of a Set be empty?

Yes, the Annihilator of a Set can be empty. This happens when the original set contains only elements that are not zero divisors, or when the algebraic structure is a division ring.

## 5. How is the Annihilator of a Set related to Proposition 11.3.5 - 5 in Garling's book?

Proposition 11.3.5 - 5 in Garling's book states that the Annihilator of a Set is an ideal in the given algebraic structure. This means that it satisfies certain properties, such as being closed under addition and multiplication by elements in the algebraic structure. This proposition helps in understanding the properties of the Annihilator of a Set in a more general context.

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