In mathematics, the Borsuk–Ulam theorem states that every continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point. Here, two points on a sphere are called antipodal if they are in exactly opposite directions from the sphere's center. Formally: if is continuous then there exists an such that: . The case can be illustrated by saying that there always exist a pair of opposite points on the Earth's equator with the same temperature. The same is true for any circle. This assumes the temperature varies continuously. The case is often illustrated by saying that at any moment, there is always a pair of antipodal points on the Earth's surface with equal temperatures and equal barometric pressures. The Borsuk–Ulam theorem has several equivalent statements in terms of odd functions. Recall that is the n-sphere and is the n-ball:
If is a continuous odd function, then there exists an such that: .
If is a continuous function which is odd on , then there exists an such that: .
History
According to, the first historical mention of the statement of the Borsuk–Ulam theorem appears in. The first proof was given by, where the formulation of the problem was attributed to Stanislaw Ulam. Since then, many alternative proofs have been found by various authors, as collected by.
Equivalent statements
The following statements are equivalent to the Borsuk–Ulam theorem.
With odd functions
A function is called odd if for every :. The Borsuk–Ulam theorem is equivalent to the following statement: A continuous odd function from an n-sphere into Euclidean n-space has a zero. PROOF:
If the theorem is correct, then it is specifically correct for odd functions, and for an odd function, iff. Hence every odd continuous function has a zero.
For every continuous function, the following function is continuous and odd:. If every odd continuous function has a zero, then has a zero, and therefore,. Hence the theorem is correct.
With retractions
Define a retraction as a function The Borsuk–Ulam theorem is equivalent to the following claim: there is no continuous odd retraction. Proof: If the theorem is correct, then every continuous odd function from must include 0 in its range. However, so there cannot be a continuous odd function whose range is. Conversely, if it is incorrect, then there is a continuous odd function with no zeroes. Then we can construct another odd function by: since has no zeroes, is well-defined and continuous. Thus we have a continuous odd retraction.
Proofs
1-dimensional case
The 1-dimensional case can easily be proved using the intermediate value theorem. Let be an odd real-valued continuous function on a circle. Pick an arbitrary. If then we are done. Otherwise, without loss of generality, But Hence, by the IVT, there is a point between and at which.
Assume that is an odd continuous function with . By passing to orbits under the antipodal action, we then get an induced continuous function between real projective spaces, which induces an isomorphism on fundamental groups. By the Hurewicz theorem, the induced ring homomorphism on cohomology with coefficients , sends to. But then we get that is sent to, a contradiction. One can also show the stronger statement that any odd map has odd degree and then deduce the theorem from this result.
The Borsuk-Ulam theorem can be proved from Tucker's lemma. Let be a continuous odd function. Because g is continuous on a compact domain, it is uniformly continuous. Therefore, for every, there is a such that, for every two points of which are within of each other, their images under g are within of each other. Define a triangulation of with edges of length at most. Label each vertex of the triangulation with a label in the following way:
Because g is odd, the labeling is also odd:. Hence, by Tucker's lemma, there are two adjacent vertices with opposite labels. Assume w.l.o.g. that the labels are. By the definition of l, this means that in both and, coordinate #1 is the largest coordinate: in this coordinate is positive while in it is negative. By the construction of the triangulation, the distance between and is at most, so in particular and so. But since the largest coordinate of is coordinate #1, this means that for each . So, where is some constant depending on and the norm which you have chosen. The above is true for every ; since is compact there must hence be a point u in which.
The ham sandwich theorem: For any compact sets A1,..., An in we can always find a hyperplane dividing each of them into two subsets of equal measure.
Equivalent results
Above we showed how to prove the Borsuk–Ulam theorem from Tucker's lemma. The converse is also true: it is possible to prove Tucker's lemma from the Borsuk–Ulam theorem. Therefore, these two theorems are equivalent.
Generalizations
In the original theorem, the domain of the function f is the unitn-sphere. In general, it is true also when the domain of f is the boundary of any open bounded symmetric subset of containing the origin.
Consider the function A which maps a point to its antipodal point: Note that The original theorem claims that there is a point x in which In general, this is true also for every function A for which However, in general this is not true for other functions A.
