arclength as filtered limit

The length (http://planetmath.org/Rectifiable) of a rectifiable curve may be phrased as a filtered limit.To do this, we will define a filter of partitions of an interval $[a,b]$ . Let ${\\bf P}$ be the set of all ordered tuplets of distinctelements of $[a,b]$ whose entries are increasing:

{\\bf P}=\\{(t_{1},\\ldots t_{n})\\mid(a\\leq t_{1}<t_{2}<\\cdots<t_{n}\\leq b)\\wedge%(n\\in\\mathbb{Z})\\wedge(n>0)\\}

We shall refer to elements of ${\\bf P}$ as partitions of the interval $[a,b]$ . We shall say that $(t_{1},\\ldots,t_{n})$ is a refinement of apartition $(s_{1},\\ldots,s_{m})$ if $\\{t_{1},\\ldots,t_{n}\\}\\supset\\{s_{1},\\ldots,s_{m}\\}$ . Let ${\\bf F}\\subset\\mathcal{P}({\\bf P})$ be theset of all subsets of ${\\bf P}$ such that, if a certain partitionbelongs to ${\\bf F}$ then so do all refinements of that partition.

Let us see that ${\\bf F}$ is a filter basis. Suppose that $A$ and $B$ are elements of ${\\bf F}$ . If a partition belongs to both $A$ and $B$ then every one of its refinements will also belong to both $A$ and $B$ , hence $A\\cap B\\in{\\bf F}$ .

Next, note that, if a partition of $B$ is a refinement of a partitionof $A$ then, by the triangle inequality, the length of $\\Pi(B)$ isgreater than the length of $\\Pi(A)$ . By definition, for every $\\epsilon>0$ , we can pick a partition $A$ such that the length of $\\Pi(A)$ differs from the length of the curve by at most $\\epsilon$ .Since the length of $\\Pi(B)$ for any partition $B$ refining $A$ liesbetween the length of $\\Pi(A)$ and the length of the curve, we seethat the length of $\\Pi(B)$ will also differ by at most $\\epsilon$ , sothe length of the curve is the limit of the length of polygonal linesaccording to the filter generated by ${\\bf F}$ .