next up previous index
Next: 5.3 Monotonic Functions Up: 5. Area Previous: 5.1 Basic Assumptions about   Index

5.2 Further Assumptions About Area

In this section we will introduce some more assumptions about area. The assumptions in this section can actually be proved on the basis of the basic assumptions we have already made, and in fact the proofs are easy (the proofs are given in appendix B). The reason I have made assumptions out of them is that they are as intuitively plausible as the assumptions I have already made, and I do not have time to do everything I want to do. I am omitting the proofs with regret because I agree with Aristotle that

It is manifest that it is far better to make the principles finite in number. Nay, they should be the fewest possible provided they enable all the same results to be proved. This is what mathematicians insist upon; for they take as principles things finite either in kind or in number.[25, page 178]

5.14   Assumption (Addition rule for area.)

\put(1.5,1.3){$S\cap T$}
For any bounded sets $S$ and $T$ in $\mbox{{\bf R}}^2$
\alpha (S\cup T)=\alpha (S)+\alpha (T)-\alpha (S\cap T).
\end{displaymath} (5.15)

and consequently

\begin{displaymath}\alpha (S\cup T)\leq\alpha (S)+\alpha (T).\end{displaymath}

5.16   Assumption (Subadditivity of area.) Let $n \in \mbox{{\bf Z}}_{\geq 1}$, and let $A_1$, $A_2$, $\cdots$, $A_n$ be bounded sets in $\mbox{{\bf R}}^2$. Then
\alpha( \bigcup_{i=1}^n A_i) \leq \sum_{i=1}^n \alpha(A_i).
\end{displaymath} (5.17)

5.18   Assumption (Monotonicity of area.) Let $S,T$ be bounded sets such that $S \subset T$. Then $\alpha (S)\leq\alpha

5.19   Definition (Zero-area set.) We will call a set with zero area a zero-area set.

From the normalization property it follows that every horizontal or vertical segment has area equal to $0$. Thus every horizontal or vertical segment is a zero-area set.

5.20   Corollary (to assumption 5.18.) 5.1Every subset of a zero-area set is a zero-area set. In particular the empty set is a zero-area set.

5.21   Corollary (to assumption 5.16.) The union of a finite number of zero-area sets is a zero-area set.

5.22   Definition (Almost disjoint.) We will say that two bounded subsets $S,T$ of $\mbox{{\bf R}}^2$ are almost disjoint if $S\cap T$ is a zero-area set.

5.23   Examples. If $a,b,c$ are real numbers with $a<b<c$, then since

\begin{displaymath}B(a,b\colon p,q)\cap B(b,c\colon s,t)\subset B(b,b\colon p,q),\end{displaymath}

the boxes $B(a,b\colon p,q)$ and $B(b,c\colon s,t)$ are almost disjoint.

Any zero-area set is almost disjoint from every set - including itself.

5.24   Assumption (Additivity for almost disjoint sets.) Let $\{R_1,\cdots ,R_n\}$ be a finite set of bounded sets such that $R_i$ and $R_j$ are almost disjoint whenever $i\neq j$. Then
\alpha (\bigcup_{i=1}^n R_i)=\sum_{i=1}^n \alpha (R_i).
\end{displaymath} (5.25)

5.26   Notation (Area functions $\alpha$ or area) Any real valued function $\alpha$, whose domain is the family of bounded subsets of $\mbox{{\bf R}}^2$, and which satisfies all of the assumptions listed in sections 5.1 and 5.2 will be called an area function. In these notes I will use the names ``$\alpha$'' and ``area'' to denote area function. Thus

\begin{displaymath}\alpha(B(a,b:c,d)) = \mbox{\rm area}(B(a,b:c,d)) = (b-a)(d-c). \end{displaymath}

next up previous index
Next: 5.3 Monotonic Functions Up: 5. Area Previous: 5.1 Basic Assumptions about   Index
Ray Mayer 2007-09-07