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Theorem of the month

Counter-directional theorem

If the path [math]\displaystyle{ \gamma: [a, b[ \, \cap \, C \rightarrow V }[/math] with [math]\displaystyle{ C \subseteq \mathbb{R} }[/math] passes the edges of every [math]\displaystyle{ n }[/math]-cube of side length d0 in the [math]\displaystyle{ n }[/math]-volume [math]\displaystyle{ V \subseteq {}^{(\omega)}\mathbb{R}^{n} }[/math] with [math]\displaystyle{ n \in \mathbb{N}_{\ge 2} }[/math] exactly once, where the opposite edges in all two-dimensional faces of every [math]\displaystyle{ n }[/math]-cube are traversed in reverse direction, but uniformly, then, for [math]\displaystyle{ D \subseteq \mathbb{R}^{2}, B \subseteq {V}^{2}, f = ({f}_{1}, ..., {f}_{n}): V \rightarrow {}^{(\omega)}\mathbb{R}^{n}, \gamma(t) = x, \gamma(\curvearrowright D t) = \curvearrowright B x }[/math] and [math]\displaystyle{ {V}_{\curvearrowright } := \{\curvearrowright B x \in V: x \in V, \curvearrowright B x \ne \curvearrowleft B x\} }[/math], it holds that

[math]\displaystyle{ \int\limits_{t \in [a,b[ \, \cap \, C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}=\int\limits_{\begin{smallmatrix} (x,\curvearrowright B\,x) \\ \in V\times {{V}_{\curvearrowright}} \end{smallmatrix}}{f(x)dBx}=\int\limits_{\begin{smallmatrix} t \in [a,b[ \, \cap \, C, \\ \gamma | {\partial{}^{\acute{n}}} V \end{smallmatrix}}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}. }[/math]

Proof:

If two arbitrary squares are considered with common edge of length d0 included in one plane, then only the edges of [math]\displaystyle{ V\times{V}_{\curvearrowright} }[/math] are not passed in both directions for the same function value. They all, and thus the path to be passed, are exactly contained in [math]\displaystyle{ {\partial}^{\acute{n}}V.\square }[/math]

Recommended reading

Nonstandard Mathematics