My Math Forum circle area

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April 10th, 2018, 05:37 AM   #11
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Quote:
 Originally Posted by v8archie We can define area perfectly well without calculus.
But can we though? Sure, you can axiomatically define the area of a square and a rectangle. That's it. Is that enough to say that we can define area without calculus. Shouldn't there be minimum requirements before we can say we have an area?
I am thinking of the following requirements at least for $A$ to be an area function
1) $A(\emptyset) = 0$
2) $A$ must be defined on at least all the compact sets of $\mathbb{R}^2$
3) If $F$ and $F'$ are disjoint and $A$ is defined on them, then $A(F\cup F') = A(F) + A(F')$
4) The area must remain stable under rigid motions (=rotations, translations, reflections)
5) If $F_1\subseteq F_2\subseteq F_3 \subseteq ...$, then $A\left(\bigcup_n F_n\right) = \sup_n A(F_n)$
6) If $D=[a,b]\times [c,d]$ is a rectangle with sides parallel to the x and y axes, then $A(D) = (b-a)(d-c)$

Is an elementary definition of this possible? I would say measure theory provides us with such, but that might not fit the bill of the OP. An elementary definition of length is even harder.

 April 10th, 2018, 06:48 AM #12 Senior Member   Joined: Jun 2015 From: England Posts: 905 Thanks: 271 Micromass, Whilst I understand your list, is area not a functional defined on $\displaystyle R \times R \to R$ since it needs must have a two parameter input in the domain and a single real output in the co-domain?
April 10th, 2018, 06:53 AM   #13
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Quote:
 Originally Posted by studiot Micromass, Whilst I understand your list, is area not a functional defined on $\displaystyle R \times R \to R$ since it needs must have a two parameter input in the domain and a single real output in the co-domain?
My area is a function taking as input a subset of $\mathbb{R}^2$ and outputs a number. Thus it would be a function
$$A:\mathcal{P}(\mathbb{R}^2)\rightarrow \mathbb{R}$$
with as domain the power set of $\mathbb{R}^2$. However, as is known, not all subsets of the plane have an area, so $A$ is not a function rather than a partial function that is not everywhere defined.

April 10th, 2018, 07:07 AM   #14
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Quote:
 $\displaystyle R \times R \to R$

Either that or there is a kernal which maps to zero for lines and such?

April 10th, 2018, 11:50 PM   #15
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Quote:
 Originally Posted by studiot Either that or there is a kernal which maps to zero for lines and such?
Right, ok, that works too!

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