Ham-sandwich cut of red points and blue points in the plane.

Question

For a finite set of points in the plane, each colored "red" or "blue", there is a line that simultaneously bisects the red points and bisects the blue points, that is, the number of red points on either side of the line is equal and the number of blue points on either side of the line is equal. Can somebody tell me a proof of this result using the ham-sandwich theorem or Borsul-Ulam Theorem or some algebraic topology tool? Wikipedia page of Ham Sandwich Theorem (http://en.wikipedia.org/wiki/Ham_sandwich_theorem) suggests there is such a prooof! Thanks in advance!

Answer 1

It follows by applying $n=2$, and using the discrete counting measure.

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The Ham-Sandwich Theorem states that given $n$ measurable "objects" in $n-$dimensional space, it is possible to divide all of them in half (with respect to their measure, i.e. volume) with a single $(n − 1)-$dimensional hyperplane.

Take $n=2$, with the discrete counting measure. Then, if we are given $k$ red points and $l$ blue points, it is possible to divide all of them in half (with respect to the discrete counting measure), with a single $1-$dimensional hyperplane, aka a line.

With this line, the numb of red points are equal, and the number of blue points are equal. Note that if $k$ (or $l$) is odd, then this implies that there is a red (or blue) point on the line.