Abstract

In (3), some of us proved that Brownian paths in n-space have double points with probability 1 if n = 2 or 3; but, for n ≥ 4, there are no double points with probability 1. The question naturally arises as to whether or not Brownian paths in n-space (n = 2 or 3) have triple points. The case of paths in the plane is settled by (4), where it is shown that, with probability 1, Brownian paths in the plane have points of multiplicity k (k = 2, 3,4, …). The purpose of the present paper is to settle the remaining case, n = 3. We prove that, with probability 1, Brownian paths in 3-dimensional space have no triple points. The general idea behind our proof is to show that there are not too many double points. That is, we show that the set of double points has sigma-fmite linear measure, and therefore zero capacity, with probability 1.

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