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Abstract. In this paper I describe the way one might begin proving the
Borsuk-Ulam theorem using measure theory and what remains to be done for
such a proof. I then provide a proof of Borsuk-Ulam using graph theory and
use the Borsuk-Ulam theorem to prove the Ham Sandwich theorem.

1. Introduction 1
2. A Theorem of Many Monikers 1
3. A Measure Theoretic Approach to Lyustenik-Shnirel’man 4
4. Tucker’s Lemma: Detour De Force 6
5. The Ham Sandwich Theorem 13

1. Introduction
The Ham Sandwich Theorem states something like the following ordinary lan-
guage proposition: Take a sandwich made of a slice of ham and two slices of bread.
No matter where one places the pieces of the sandwich in the kitchen, or house, or
universe, so long as one’s knife is long enough one can cut all three pieces in half in
only one pass. The precise mathematical statement of the theorem, generalized to
n dimensions, is that given n compact sets in Rn there is a hyperplane which bisects
each compact set so that the two halves of both sets have equal measure. Somewhat
surprisingly, since the statement at first glance appears to result from a wedding
of plane geometry and measure theory, the proof is actually an easy consequence
of the Borsuk-Ulam Theorem, a theorem which imposes certain requirements upon
continuous maps from the n-sphere in Rn+1 to Rn . As it turns out, the Borsuk
Ulam theorem is a far deeper theorem in that it can be recapitulated in equivalent
form with rules governing various other types of continuous maps. In my paper
I will discuss two proofs of the theorem which draw on very different ideas, one
from combinatorics and the other from measure theory. Having established the
Borsuk-Ulam Theorem, I will prove the Ham Sandwich Theorem.

2. A Theorem of Many Monikers

One of the most common variations on the Borsuk-Ulam theorem is the claim
Theorem 2.1. Borsuk-Ulam. If n ≥ 0 then for any continuous mapping f : S n →
Rn there is a point x ∈ S n for which f (x) = f (−x)

Date: DEADLINE AUGUST 22, 2008.


But another common take on the theorem is as follows.

Theorem 2.2. Borsuk-Ulam. For every continuous mapping f : S n → Rn that is
antipodal there is a point x ∈ Sn for which f (x) = 0, where an antipodal map is
understood to be a map such that for all x ∈ S n , f (−x) = −f (x).
To show that these two are equivalent we present the following proof:
Proposition 2.3. Theorems 1.1 and 1.2 are equivalent.
Proof. Theorem 1.1 =⇒ Theorem 1.2. Pick a continuous antipodal function
f : S n → Rn . By the antipodality of f we have that for all x ∈ S n
f (−x) = −f (x).
Because f is continuous from the n + 1 sphere to Rn we can apply Theorem 1.1
and find that there is some x for which
f (x) = f (−x)
Combining these two expressions we get f (x) = −f (x) =⇒ 2f (x) = 0 =⇒
f (x) = 0.
Theorem 1.2 =⇒ Theorem 1.1. Take a continuous mapping f : S n → Rn .
Then let g(x) = f (x) − f (−x). This means that g(−x) = f (−x) − f (x) = −g(x).
Therefore g(x) is an antipodal mapping, and it is clearly continuous as it is the
sum of two continuous functions. We apply Theorem 1.2 and get that there is an
x for which g(x) = 0. But for that x, f (x) − f (−x) = 0 =⇒ f (x) = f (−x). 

Both these previous variants are statements regarding existence of certain kinds
of points for certain maps of the sphere. One can also show that the theorem is
equivalent to the non-existence of a certain kind of map between a sphere and a
sphere in a different dimension.
Theorem 2.4. Both versions of Borsuk-Ulam already mentioned are equivalent to
the statement that there is no continuous antipodal mapping f : S n → S n−1
Proof. Assume that there exists a continuous antipodal mapping f : S n → S n−1 .
By definition S n−1 = {x ∈ Rn |x21 + x22 + . . . x2n = 1}. Thus we can extend the
codomain of f to the whole of Rn and consider it as a function from S n → Rn .
As a function from S n → Rn we know that if f (x) is in the image of f then
f (x) ∈ S n−1 . Since 0 is not in S n−1 we have that f cannot equal zero anywhere.
But by the Borsuk-Ulam theorem, all antipodal continuous maps from S n → Rn
must equal zero somewhere. Therefore, we have a contradiction.
For the other direction, let’s say we know that there is no continuous antipodal
mapping from S n → S n−1 . Take any antipodal mapping from f : S n → Rn . If
f (x) 6= 0 for any x, then we can define g(x) = kff (x)
(x)k . Then this maps to S
is antipodal so we have a contradiction. 

The following variant on Borsuk Ulam, fairly similar to Theorem 2.4, will be
useful later.
Theorem 2.5. The versions of Borsuk-Ulam given before are equivalent to the
statement that there is no continuous mapping B n → S n−1 that is antipodal on the
boundary, i.e. for all x ∈ ∂B n = S n−1 we have f (−x) = −f (x).

Proof. First we note that there is a homeomorphism π : (x1 , . . . , xn+1 ) → (x1 , x2 , . . . , xn )

between the upper hemisphere of the n sphere and the ball B n . For an idea of what
this function looks like, take the crust of the earth on the northern hemisphere to be
the upper hemisphere of a realization of S 2 . Then the B 2 ball could be considered
as that portion of the earth (including inside the crust) that is on the longitude
of the equator. Each point on the crust of the Northern-Hemisphere sits above a
point on the disc of the equator, and so our function associates these points. In
any event, if there existed an antipodal mapping f : S n → S n−1 we could define
f ◦ π −1 a function from B n → S n−1 that would be antipodal on the boundary of
B n . Therefore we have proved by contrapositive that if our version of Borsuk Ulam
in Theorem 2.5 is true then the other variants of Borsuk-Ulam introduced so far
also hold.
The other direction is equally simple. Assume that there is a function g that
goes from B n → S n−1 that is antipodal on the boundary. Then f (x) = g(π(x)) is
defined on the upper hemisphere of S n . Further we can extend f to the whole sphere
by letting f (−x) = −f (x). The fact that B n is antipodal on the boundary assures
us that these associations are consistent and that our function is well-defined. This
function is continuous and is an antipodal mapping of S n → S n−1 . 
Another variant of Borsuk Ulam, which goes in a somewhat different direction,
is a statement congenial to elementary point-set topology.
Theorem 2.6. Lyusternik-Shnirel’man. If U1 , U2 , . . . , Un+1 is a cover of S n with
with each Ui open then for some i, Ui contains a pair of antipodal points.
Remark 2.7. Lyusternik-Shnirel’man is also equivalent to the same statement with
all the Ui taken to be closed sets. Formalizing the argument with full rigor would
be a substantial digression, but seeing why it should be true is not difficult.
Assume that the theorem holds for closed covers. Take any {Ui }n+1 i=1 that is
an open cover of S n . One can shave off a tiny bit from each element in this set
and get a collection of closed sets which covers S n . Then the antipodal points
are guaranteed to be in one of these sets by Lyusternik Shnirel’man for closed
sets. But whichever set has antipodal points is a subset of one of the open sets so
Lyusternik-Shnirel’man is true for open sets.
Assume that the theorem holds for open covers. Take a closed cover {Fi }n+1 i=1 .
Let Ui = {x ∈ S n |dist(x, Fi ) < n1 . So there is for each  an open cover of S n and
so for each  we have an x and −x contained in the same set. It is possible to find
a sequence x1 , x2 , . . . such that limj→∞ dist(x, Fi ) = limj→∞ dist(−x, Fi ) = 0 for
a fixed i. Because S n is compact we know that there is a convergent subsequence
and we take the limit of this sequence to be x. Then this x must have −x in Fi as
Proposition 2.8. The Lyusternik-Shnirel’man theorem is equivalent to the Borsuk-
Ulam theorem.
Proof. Lyusternik-Shnirel’man =⇒ Borsuk-Ulam. It is a useful fact for this theo-
rem that S n−1 can be covered using n + 1 open sets in such a way as to have none
of the open sets contain antipodal points. Let us denote this cover {Ui }n+1 i=1 .
If we assume Borsuk-Ulam is false and that there is a continuous antipodal map
f : S n → S n−1 then by the definition of continuity f −1 (Ui ) is open for all Ui . Fur-
ther, it covers S n . Therefore by Lyusternik-Shnirel’man we have x, −x ∈ f −1 (Ui )

for some i. But this is impossible. Since f is an antipodal map f (−x) = −f (x) so
f (x) and f (−x) are antipodes on S n−1 . But they are both in Ui by definition of
f −1 (Ui ). But by our construction we have that no Ui contains antipodes and so we
have reached a contradiction.
Borsuk-Ulam =⇒ Lyusternik-Shnirel’man. It is convenient for the proof to use
closed covers instead of open covers. Take {Ui }n+1 n
i=1 to be a closed cover of S .
Now we define a function f : S → R coordinate-wise. For all x ∈ S n let
n n

fi (x) = inf y∈Fi |y − x|. Then by Borsuk-Ulam we have that f (x) = f (−x) for some
x ∈ S n . If the ith coordinate is 0 at the point f (x) then x, −x ∈ Fi . If none of
the coordinates are zero, then it means that neither x nor −x are in any of the
sets Fi . But because all the Fi are a cover we have that these two must both be in
Fn+1 . 

3. A Measure Theoretic Approach to Lyustenik-Shnirel’man

Although the last section established that there was a large family of mutually
equivalent statements it did little to show how one might get started actually prov-
ing any of them. As a first example, when things are very concrete and easily visu-
alizable, we now set out to prove the case of the Borsuk-Ulam theorem when n = 1
which concerns maps of the circle into the real line. The Lyusternik-Shnirel’man
variant of Borsuk Ulam can be proved quite easily.
Theorem 3.1. If S 1 is covered by two open sets U1 and U2 then one of the two
sets contains a pair of antipodal points.
Proof. First, it is a basic fact that S 1 is not a separable space, so U1 ∩ U2 6= ∅. The
proof is immediate. Pick x ∈ U1 ∩ U2 . Then the antipodal point −x ∈ S 1 ⊂ U1 ∪ U2
and therefore −x ∈ U1 or −x ∈ U2 . If −x ∈ U1 , U1 contains x and −x. If −x ∈ U2
then U2 is the set which contains antipodal points. 
This proof is startlingly simple, but it is apparent that this strategy would not
work for higher dimensions. Consider, for example, the Earth as an example of
S 2 . Then when we take the Northern Hemisphere without the equator as one open
set, the Southern Hemisphere without the equator as another, and an open set that
looks like a slim belt around the equator, we get an open cover of S 2 but the mutual
intersection of all three sets is empty.
So let’s return to the drawing board and see if we can come up with another way
to prove Lyusternik-Shnirel’man in S 1 .
Proof. Let’s say that we are given two open sets U1 and U2 which cover S 1 . The
naive guess we might have is that it will that the ”‘larger”’ of the two covers which
will contain antipodal points. But what does ”large” mean for us in this situation?
One good measure of the size of these sets is the arc length of that portion of the
circle covered by each i.e. the Lebesgue measure on S 1 . As it turns out, it is
possible to show that the larger of the two covers contains
We let µ denote the arc-length of a set and say that µ(S 1 ) = 2π which fits
with our standard notion of arc-length. Because U1 , U2 are open we know they
are measurable and since they cover S 1 we have that µ(U1 ) + µ(U2 ) > 2π. This
means that the average value of the two is greater than π, which implies that one
or the other has value greater than π. Therefore without loss of generality we take
µ(U1 ) > π.

Because µ is invariant under rotation we know that if −U1 = {x ∈ S 1 | −x ∈ U1 },

then µ(−U1 ) = µ(U1 ) > π.
Since µ(U1 ) + µ(S 1 \ U1 ) = 2π we have that µ(S 1 \ U1 ) < π. If U1 ∩ −U1 = ∅
then −U1 ⊂ S 1 \ U1 . But this implies that µ(−U1 ) ≤ µ(S 1 \ U1 ). Therefore we
have that π < µ(−U ) ≤ µ(S 1 \ U ) < π. This is a contradiction.
Actually, we can make this argument a little better without the assumption that
U1 ∩ −U1 = ∅. Note that

µ(−U1 ) = µ(U1 ∩ −U1 ) + µ(−U1 \ U1 )

Further, we observe that
−U1 \ U1 ⊂ S 1 \ U1
Combining these observations with what was already stated we get following

π < µ(−U ) = µ(−U1 ∩U1 )+µ(−U1 \U1 ) < µ(−U1 ∩U1 )+µ(S 1 \U1 ) < µ(−U1 ∩U1 )+π
Clearly whenever µ(−U1 ∩ U1 ) = 0 we get a contradiction, meaning that our set
of intersection has positive measure. 

Comparing the two proofs of Lyusternik-Shnirel’man in S 1 that have been given

so far, it appears that the measure theoretic proof has at least two comparative
advantages over the set theoretic proof. One potential comparative advantage is
that the first proof generated only a single antipodal point, while this second one
gives us an uncountable number of antipodal points. But one could easily reach
the same conclusion from the first proof with only a little more work. Here’s
how. The map which sends each point of the sphere to its antipode is a bijective
homeomorphism. Thus if ν(x) = −x denotes this bijective homeomorphism then
when we take ν(U1 ∩ U2 ) we know that we get an open set. ν(U1 ∩ U2 ) ∩ Ui is open
for i = 1 and i = 2 and for one of these numbers is not equal to the empty set.
The only open set containing a countable number of elements is the empty set so
ν(U1 ∩ U2 ) ∩ Ui is uncountable for some i.
The real comparative advantage of the second proof is how it uses the fact that
U1 ∩ U2 is not empty. The first proof uses it as the entire proof, while the second
proof only uses the fact to show that U1 ∩ U2 has positive measure and therefore
that µ(U1 ) + µ(U2 ) > 2π. The reason this is a real comparative advantage is that
what we need in the first proof is that all the sets in the two element cover intersect.
As mentioned before in the earth with belt example, this does not hold up in higher
dimensions. In the second proof all we need is that the arc length covered by the
elements in the cover is greater than the arc-length of the entire circle. Analogues
of this fact do hold up in higher dimensions, each element in an open cover of
the sphere must overlap with its adjacent neighbors, meaning that the sum of all
the surface areas must be greater than the surface area total. In fact, it yields
immediately, the following Borsuk-Ulam-type theorem.
Theorem 3.2. If we take an open cover of S n and any one element U in that
cover has surface area greater than half the surface area of the entire sphere then
that element U contains a non-trivial subset that is invariant under the antipodal
mapping v(x) = −x. Further, that subset has positive Lebesgue measure.

Proof. Let µ denote the Lebesgue measure on a sphere S n . Further let Ui be an

element of an open cover of S n with µ(Ui ) > µ(S2 ) . Then because surface area of
Ui is equal to the surface area of µ(−Ui ) we have that µ(−Ui ) > µ(S2 ) . Further,
similar to the argument in the proof above we have the following inequality.

µ(S n ) µ(S n )
< µ(−Ui ) = µ(−Ui \Ui )+µ(−Ui ∩Ui ) ≤ µ(S n \Ui )+µ(−Ui ∩Ui ) < +µ(−Ui ∩Ui )
2 2
Whenever µ(−Ui ∩ Ui ) = 0 we have a contradiction, yielding the theorem. 
The simplicity of this application of measure theory may belie the non-triviality
of its result. This theorem says that so long as an open set on the sphere is
sufficiently big it will have a large number of points with antipodes still in the set.
Additionally, it actually reduces what needs to be proved for Lyusternik Shnirel’man
in S 2 to the following.
Conjecture 3.3. If U1 , U2 , U3 is an open cover of S 2 and for each Ui the surface
area measure µ(Ui ) ≤ 2π then one of the elements contains a pair of antipodal
While this initially does not seem like we have moved ahead that far, after some
thought it is apparent that we can use the fact that Lyusternik-Shnirel’man holds
for S 1 . That is to say, if the following conjecture were true:
Conjecture 3.4. If U1 , U2 , U3 is an open cover of S 2 and for each Ui the surface
area measure µ(Ui ) ≤ 2π then for some i and j the union Ui ∪ Uj must contain an
Then one could say two open sets contain an equator and so must contain a set
of positive measure invariant under the map which exchanges antipodes. Thus one
might get the somewhat nice result that if the cover contains an element sufficiently
big then that element has antipodes, and if none of the elements in the cover are
sufficiently big then none is big enough to stop the other two from containing
an equator and thus proving our theorem. After substantial thought about lots of
messy covers, the conjecture still seems true. Sadly, this problem is a difficult one to
solve and the author was unable to come up with any techniques to do it. Indeed,
this last piece were it found might really yield a complete solution and it might
generalize to higher dimensions. Nevertheless, there are proofs of Borsuk-Ulam
already well-known from a variety of different angles, one of which we approach

4. Tucker’s Lemma: Detour De Force

Many of the most commonly known proofs of Borsuk-Ulam theorem are rela-
tively simple consequences of quite sophisticated machinery. The following sketch
should give a reader un-initiated in this machinery a feel for what’s missing and
the advanced reader an idea of how it might work.
Proposition 4.1. The Borsuk-Ulam theorem is true.
Proof. First it is a fact, in itself something to prove, that antipodal maps of the
sphere must have something called an odd degree, where by degree we mean a spe-
cialized notion attached to Homology theory. Now let’s say there were an existent

continuous antipodal map f of S n to S n−1 . Then the S n−1 we are mapping to is the
boundary of a disc in Rn . Consider the restriction of f to an equator of S n−1 ⊂ S n .
We know that it must have odd degree. But if we look at f restricted to the upper
hemisphere of S n we know that f must be nullhomotopic when restricted to S n−1 .
This means that our restriction of f to S n−1 must have degree zero. 

In contrast with the above homological proof, which require a lot of prerequisite
knowledge in homology theory, there is a proof using graph theory which can be
found in Jiri Matousek’s book on the Borsuk-Ulam theorem. As it turns out,
one can prove a weak version of Tucker’s Lemma, derive from it the Borsuk-Ulam
theorem, and then double back and use Borsuk-Ulam to derive Tucker’s Lemma in
its full glory. We intend to recount this proof in this paper.
But how does graph theory get introduced to the problem at all? The Borsuk
Ulam theorem typically lays down rules governing continuous maps of the sphere.
Continuous maps have a certain liquidity about them, however, and it does not
take much effort to see that the Borsuk-Ulam theorem is saying something about
a larger class of shapes than might be obvious at first. After all, if one takes a
shape different from the sphere but that nevertheless has a continuous map onto
the sphere, one can compose the map from the shape to the sphere with a map
from the sphere to say another sphere and see that the composed function certainly
is governed in some way by the Borsuk-Ulam theorem. The hope is that if one can
prove a statement about a shape that is similar enough to the sphere than what
goes for that shape will hold for the sphere as well.
This provides some motivation for the notion of triangulations, but to understand
what exactly triangulations are we need a little background in simplices.
Definition 4.2. If v0 , v1 , . . . , vk are points in Rn then we say that these points are
affinely independent if v1 − v0 , v2 − v0 , . . . , vk − v0 are linearly independent.
Which is why affinely independent sets look like this (the lines are added to
make the linear independence more clear, the affinely independent set is just the

The convex hull of a set of points is the intersection of all convex sets which
contain those points. In other words the convex hull of any of the points in the
figure above is the same sets but instead of containing nothing in the interior we
have that the interior is filled in. This suggests the following definition.
Definition 4.3. A simplex σ is the convex hull of a finite collection of affinely
independent points in Rn . The elements in the finite collection are called the

vertices of σ and we denote them V (σ). The dimension of a simplex is 1 less than
the number of vertices it has.
Definition 4.4. If A is a subset of V (σ) then we call the convex hull of A a face
of σ. From this it readily follows that the face of a simplex is also a simplex.
The requirements for something to be a simplex are rather rigid and sometimes
we want to use the more loose concept of simplicial complexes.
Definition 4.5. A nonempty family ∆ of simplices is called a simplicial complex
so long as the following two conditions are met.
(1) Each face of a simplex σ ∈ ∆ is also a simplex in ∆
(2) The intersection of two simplices in ∆ is a face of each simplex.
Further if one takes the union of all simplices in a simplicial complex ∆ then
this union is called the polyhedron of ∆ and is denoted by k∆k
Roughly speaking a major difference between the two is that simplexes are con-
vex, while simplicial complexes are only cobbled together from things which are
convex and are not necessarily convex themselves. Further, we can easily go back
and forth between simplicial complexes geometrically realized in the plane and the
so called “abstract” simplicial complexes.
Definition 4.6. An abstract simplicial complex is a pair (V, K) where V is a set
of ”vertices” and K ⊂ P (V ) the power set of V . We require that if F ∈ K and
G ⊂ F then G ∈ K. Dimension is defined similarly as one less than the dimension
of the vertex set.
The idea of triangulation is to analyze these ”triangular” looking sets instead of
analyzing sets with any other shape. From the standpoint of topology, looking at
these shapes comes cost free so long as the two spaces are homeomorphic, that is to
say if there is a continuous bijection with continuous inverse between two spaces.
Definition 4.7. If X is a topological space and ∆ is a simplicial complex such that
X is homeomorphic to k∆k then we call ∆ a triangulation of X.
To really be able to switch over from talking about topological spaces to simplicial
complexes we need one more piece of the puzzle. The functions which go from the
topology of one space to the topology of another are called continuous. Likewise
we can talk about functions which map simplices from one complex to another.
Definition 4.8. For two abstract simplicial complexes K and L if f is a function
from V (K) to V (L) and has the property that f (F ) ∈ L for all F ∈ K then we say
that f is a simplicial mapping.
Such a mapping on abstract simplicial complex induces a mapping between their
geometric realizations.
Definition 4.9. Let K1 and K2 be abstract simplicial complexes. Let f be a
simplicial mapping from the first complex to the second. If ∆1 and ∆2 are the
aforementioned geometric realization then define kf k : k∆1 k → k∆2 k as follows. If
x is in the relative interior of a face σ ∈ ∆1 then by convexity of σ we have that
Pd Pd
x =
Pd i=1 ai vi where vi ∈ V (σ) and i=1 ai = 1. Therefore we say kf k (x) =
i=1 ai f (vi ). We call kf k the affine extension of f .

Showing that kf k is well-defined is not challenging, if the reader is interested

in doing so on their own it is helpful to first show that x must be in the relative
interior of only one face of ∆1 . Further, the reader may try to verify the also true
statement that kf k is continuous.
Now we can understand the statement of Tucker’s Lemma.
Theorem 4.10. Let T be a triangulation of B n with a finite number of vertices
and which is antipodally symmetric on the boundary. By antipodally symmetric on
the boundary, we mean that the set of all simplices of T contained in ∂B n = S n−1
is a triangulation of S n−1 and further that if σ ⊂ ∂B n then −σ ∈ T . Given such a
triangulation if there is a map

λ : V (T ) → {+1, −1, +2, −2, . . . , +n, −n}

for which λ(−v) = −λ(v) for all vertices v ∈ ∂B n then there is a 1-dimensional
simplices, or an edge, contained in T such that if v1 and v2 are its vertices λ(v1 ) =
−λ(v2 ).
One can think of λ as labeling the points and Tucker’s lemma as stating that
somewhere there is an edge connecting two vertices which have been given opposite
Before we go forward, however, it is helpful to also recast Tucker’s lemma as
saying something not just about labellings but about simplicial maps. Recall that
analogous to the conceptualization of continuity as a description of the way topolo-
gies go to other topologies, we have that simplicial maps are the way that we
describe simplices as going into simplices. Indeed, if the point at the end is to
say something about continuous maps by using simplicial maps as a way of get-
ting a handle on the problem then we had better cast Tucker’s Lemma in terms of
simplicial maps.
But how to do that? Thankfully the machinery we developed before in terms
of abstract simplicial complices allows us to quickly make a link between the set
{±1, ±2, . . . , ±n} and a simplicial complex. Let us consider the abstract simplicial
complex Λ formed by this set of labels. Tucker’s Lemma gives us the fact that if
there is a simplicial mapping of the vertex set of our triangulation T into Λ then
there must be a complementary edge. In other words, there must be a simplex that
looks like an edge connecting the vertices +i and −i for some i. Further, we could
form a new simplicial complex from Λ by cutting out of Λ all edges connecting
opposite labels (we will also have to cut out additional simplices from Λ that had
these edges as their faces or else our result after our little simplicial surgery would
have ceased to be a simplicial complex at all). For this new simplex, we would find
that there could not be a simplicial map from V (T ) to this complex.
This discussion suggests we define n−1 to be the abstract simplicial complex
formed by the vertex set V (n−1 ) = {±1, ±2, . . . , ±n} and under the requirement
that a set F ⊂ V (n−1 ) is a simplex if and only if there is no i such that +i
and −i are both in F . If one thinks about the geometric realization of n−1 it is
most natural to consider all the vertices as a unit direction along the corresponding
coordinate axes because this restriction agrees with our requirement for simplices
to be affinely independent. That is to say V (n−1 ) = ±e1 , ±e2 , . . . , ±en . At this
point it is helpful to look at the picture.
The only requirement we put on n−1 is that it cannot have a face which contains
two opposite vertices, i and −i,which means that there is no simplex in n−1 that

is on the interior of the convex hull of V (n−1 ). Thus we get from Tucker’s Lemma
the following theorem.
Theorem 4.11. Let T be a triangulation of B n that is antipodally symmetric on
the boundary. Then there is no map λ : V (T ) → V (n−1 ) that is a simplicial map
of T into n−1 and is antipodal on the boundary.
There is a real impact to this theorem which needs to be teased out. n−1 is
the set of faces of the boundary complex formed by a set of coordinate vectors,
and so is a triangulation of S n−1 . Tucker’s Lemma thus appears to have as a
consequence a statement stating that it is impossible to have certain kinds of maps
from triangulations of the n − ball to triangulations of the n − 1 sphere. Tucker’s
Lemma is starting to look a lot more like Borsuk-Ulam.
Further, it is apparent that if one were to have a labeling of the triangulation
and that there were no complementary edge, or edge with opposite labels, then one
could reason similarly as I have done above and find that there was a simplicial map
from the triangulation into n−1 . The non-existence of such a simplicial map thus
implies that there is no complementary edge. Thus we have that these theorems
are both equivalent and henceforth we refer to both as Tucker’s Lemma.
The statement of the theorem should now be somewhat understood and we can
now talk about proving the statement. As mentioned at the outset of this section,
the theorem is difficult to approach directly. Instead of doing it for all triangulations
we do it for a special class of triangulations. What exactly is this special class? We
define it below.
Pn 4.12. The ball B n is homeomorphic to the set B̂ n = {x = x1 , x2 , . . . , xn |
i=1 xi = 1} which is the unit ball with the `1 norm. Let 4 be the ”natural” tri-
angulation of B̂ n induced by the coordinate hyperplanes. Explicitly that means
that σ ∈ 4 means that σ ∈ n−1 or σ = τ ∪ {0} where τ ∈ n−1 . Then we say that
T is a special triangulation of B̂ n if
(1) T is antipodally symmetric on the boundary, i.e. is contained in S n−1 and
if σ ∈ T and is a subset of S n−1 then −σ is also a simplex of T .
(2) For each σ ∈ T we have that there is a τ ∈ 4 such that σ ⊂ τ .
Proposition 4.13. Tucker’s Lemma is true for special triangulations.
Proof. Let T be a special triangulation of B̂ n and λ : V (T ) → {±1, ±2, ±3, . . . , ±n}.
For a simplex σ ∈ T we consider the set λ(σ) = {λ(v) : v is a vertex of σ}. λ(σ)
can be thought of as the collection of all labels which have been proscribed by the
mapping λ to that simplex.
What we do next is compare this list to a different list of labels that one could
ascribe to the simplex based on its location in Rd . Pick an element x in the relative
interior of σ. Then we say that S(σ) = {+i | xi > 0, i = 1, 2, . . . n} ∪ {−i : xi <
0, i = 1, 2, ..., n}. It follows from the second requirement of special triangulations
that no matter where we pick x so long as it is in the relative interior we will get the
same value of S(σ) (the reason we say the relative interior is that there are many
simplices in a special triangulation which have one or more edge on the coordinate
axes, which, were it picked instead, would certainly give a different set).
We will call σ a happy simplex if S(σ) ⊂ λ(σ). In other words, happy simplices
have been proscribed labels which fit with the ones we’d naturally imagine them
being given.

Observation 4.14. Happy simplices have the following properties.

(1) The dimension of a happy simplex is equal to the number of elements in
S(σ) or it is one less than that number
The reason for this fact is that σ lies inside the linear subspace spanned
by the coordinate axes xi such that i ∈ S(σ) or −i ∈ S(σ). Therefore it
must have dimension less than or equal to that subspace. At the same time,
however, there must be at least as many vertices being considered in the
set λ(σ) as there are elements of S(σ) for it to be happy. Therefore since
the dimension of a simplex is defined to be one less than the number of
vertices, σ must be no more than one less than the number of elements in
This suggests that we give the name tight to those happy simplices for
which dim S(σ) = k − 1 and loose for those happy simplices for which
dim S(σ) = k.
(2) A boundary simplex, if it is happy, is necessarily tight. Non-boundary
simplices are either tight or loose.
The reason is that when a simplex is on the boundary it has as many
vertices as there are prescribed labels.
(3) {0} is a loose happy simplex
It must always be happy because S({0}) = ∅, and so the number of
elements in S({0}) is zero. This also implies that it must always be loose
because {0} has one vertex and dimension zero.

For brevity of expression, let us agree to call τ a facet of a simplex σ if τ is a

face of σ and τ has dimension only one less than that of σ. We will now define a
graph G with vertices all happy simplices. We say that two vertices σ, τ ∈ T are
connected by an edge if σ and τ are antipodal boundary simplices or σ is a facet
of τ and the labels of σ alone already make τ happy, i.e. S(τ ) ⊂ λ(σ).
{0} has degree 1 since it is connected to the edge of the triangulation that
is made happy by the label λ(0). What we want to show is that if there is no
complementary edge, then any other vertex σ of the graph G has degree 2. This
will yield a contradiction since a finite graph cannot contain only one vertex of odd

Observation 4.15. (1) Suppose σ is a tight happy simplex. Then if τ is a

neighbor of σ then either τ is −σ or has σ as a facet.
(a) If σ is on the boundary ∂ B̂ n then −σ is one of its neighbors. Any other
neighbor τ has σ as a facet and is made happy by its labels. That is
to say, it is a loose happy simplex. Thus it has to lie in the subspace
Lσ mentioned above. Then Lσ ∩ B̂ n is a k-dimensional crosspolytope
and since σ is in the boundary of a (k − 1)-dimensional simplex it is a
facet of only one k-simplex.
(b) σ does not lie on the boundary means that σ is a facet of only two
simplices made happy by its labels, and these are the two neighbors.
(2) Suppose σ is a loose happy simplex.
(a) We have that S(σ) = λ(σ) and so one of the labels occurs twice on σ.
Then σ is adjacent to exactly two of its facet.
(b) There is an extra label i ∈ λ(σ) \ S(σ). It follows that −i 6∈ S(σ) for
otherwise we would have a complementary edge and Tucker’s Lemma

would be true. One of the neighbors of σ is a facet of σ not containing

the vertex with the extra label i. Further σ is a facet of exactly one
loose simplex σ 0 made happy by the labels of σ. Namely, one with
S(σ 0 ) = λ(σ) = S(σ) ∪ {i} We enter that σ 0 if we go from an interior
point of σ in the direction of the x|i| − axis in the positive direction
for i > 0 and in the negative direction for i < 0.
Thus for each possibility we have two neighbors, and so {0} is the only vertex
where we can only go out one direction, a contradiction. 

Thus we have established Tucker’s Lemma for special triangulations.

Proposition 4.16. If Tucker’s Lemma is true for special triangulations then Borsuk-
Ulam is also true.
Proof. Assume Borsuk-Ulam is false. Specifically, we assume that f : B n → S n−1
is a continuous map and that it is antipodal on the boundary. We will show this
implies that there is a map λ and a special triangulation T such that λ : V (T ) →
V (n−1 ) that is a simplicial map and is antipodal on the boundary.
First of all it is apparent that the special triangulations can be made with arbi-
trarily small simplex diameter. We will use this fact. For the sake of convenience
we will use the ||`p notation,
Pnwhere |y|`∞ is equal to the largest absolute value of yi .
If y ∈ S n−1 then because i=1 yi2 = 1 we know that the absolute value of at least
one of the yi is greater than √1n . That is, |y|`∞ ≥ √1n . Because f is a continuous
function on a compact set we know that there is a δ such that |x − x0 | < δ implies
that |f (x) − f (x0 )|`∞ < √2n . We then take T to be a triangulation with simplex
diameter less than δ.
Now we define our labeling map which will lead to contradiction. First let
k(v) = min{i : |f (v)| ≥ √1n }. We define λ : V (T ) → {±1, ±2, . . . , ±n} as follows.

λ(v) = +k(v)iff (v)k(v) > 0, −k(v)iff (v)k(v) < 0

Since f is antipodal on ∂B n we have that λ(−v) = −λ(v). Therefore we know
that Tucker’s Lemma applies and that there is a pair v, v 0 that are connected by a
complementary edge. But by definition of λ this can only happen if f (v)i ≥ √1n and
f (v 0 )i ≤ √ n
or vice-versa, but we assume that these are the inequalities without
loss of generality. But then we have that |f (v) − f (v)|`∞ ≥ |f (v)i − f (v 0 )i | ≥ √2n .
This contradicts the uniform continuity of f . Contradiction. 

Thus Borsuk-Ulam theorem has been established. The full-fledged Tucker’s

Lemma, for all triangulation not just special ones, is low hanging fruit.
Proposition 4.17. Borsuk-Ulam implies Tucker’s Lemma
Proof. Assume there is an antipodal simplicial map λ of T into n−1 , where T
n n−1
is any Triangulation of B . But then kλk : kT k →  is a continuous map
antipodal on the boundary. Then there is a continuous map f : B n → T antipodal
on the boundary and a continuous map g : n−1 → S n−1 that is antipodal on the
boundary. Therefore g ◦ kλk ◦ f is a continuous map from B n → S n−1 which is
antipodal on the boundary. But this contradicts Borsuk-Ulam. 

5. The Ham Sandwich Theorem

At the beginning of the paper we stated that the Ham-Sandwich theorem is a
result about hyperplanes bisecting a collection of compact sets. All the discussion of
the Borsuk-Ulam theorem would be for naught if there was not some link between
the two ideas. As it turns out, it is possible to think of hyperplanes in Rn being
uniquely determined by points in S n , and from there it is possible to design a
function which measures how well the hyperplane defined by the point cuts the sets
in Rn . But first it is essential to see how one associates for each point in S n a
hyperplane in Rn and the related notion of Half Space
If one takes a point u = (u0 , u1 , . . . , un ) ∈ S n then each oriented hyperplane
going through the origin is associated uniquely to a point on S n by the dot product.
That is to say, the set of all points x = (x0 , x1 , . . . , xn ) for which u · x = 0 is the
hyperplane going through the origin orthogonal to the vector going from the origin
to u. When we take the set of all points with u · x ≤ 0 we get a half space,
or the half of Rn+1 below the hyperplane (although what exactly below means
depends on how our axes are set up, so we should just ensure that we are consistent
in our algebraic definition). But what we wanted was a hyperplane that divided
Rn . In order to get that hyperplane what we can do is we intersect that with the
hyperplane defined by all the points < x0 , x1 , . . . , xn > where x0 = −1, that is we
force the planes to have less freedom where they could be. The choice x0 = −1 is
somewhat arbitrary, in that we would get unique associations for every half-space in
Rn if x0 = −2, for example. Further we could have locked any single xi at −1 and
gotten rather different looking associations for all the hyperplanes in Rn but every
oriented hyperplane would still be covered uniquely. Nevertheless picking x0 = −1
will reveal the simplest form for our unique associations. That is to say, we have
two requirements on x for it to be in our set.
• x · u ≤ 0.
• x0 = −1
We can rewrite the first equation with the dot product as i=1 xi ui ≤ u0 . Hence,
we have the motivation to make the following definition.
Definition 5.1. Half Space. Given a point u ∈ S n we define P the half space at u,
denoted as h+ (u), as follows: h+ (u) = {(x1 , x2 , . . . , xn ) ∈ Rn | i=1 xi ui < u0 }.
This definition is essential to the proof I will present of Ham Sandwich, and
before jumping into the proof a few more words about this definition. As written
in n-dimensional space the equation looks bewildering but really this function is
not difficult to imagine even in three-dimensions. What the definition is saying, for
the case of S 2 in three dimensions, is that when one picks a point on the sphere one
draws a vector connecting that sphere to the origin. This vector in turn uniquely
determines a plane going through the origin, and in fact every plane through the
origin is determined by some vector. Additionally, antipodal points on the sphere
generate planes with same slope but pointing in opposite directions, an important
property true for all n-dimensions that we will use later. Once one has a given plane
through the origin then one takes everything below it, where one has to be careful
about what below means, and also takes the intersection of that with the plane
which runs parallel to the yz-plane through the point (−1, 0, 0). All that remains
after this intersection is an infinite portion of the yz-plane through (−1, 0, 0). This
infinite-portion, deceptively called a half space because it divides the yz-plane into

two equal but infinite chunks, is what we want. All the x coordinates in this half-
space are -1, so we can safely collapse all our coordinates, which are in R3 , to
merely (y, z) coordinates in R2 . In n-dimensions, the process is similar although
more difficult to visualize.
Theorem 5.2. Ham Sandwich. Given n compact sets in Rn there is a hyperplane
which bisects each set into two sets of equal measure.
Proof. Let K1 , . . . , Kn be our collection of compact sets. Let µi (A) = µ(A ∩ Ki )
for all sets K. It is a fact that all µi are still Radon measures. It is a property
of Radon measures that the measure of compact sets is always finite, therefore
µi (R) = µ(Ki ) < ∞.
The salient feature of the Half Space construction is that we get a series of
planes defined by each point on S n with points near each other leading to similar
hyperplanes in Rn . What we do next is define a function for each point on S n
based on the measure of the half space each point generates. For all u ∈ S n let f
be a function from S n → Rn defined coordinate-wise as fi (u) = µi (h+ (u)). This
function going from S n → R is well-defined for each u because the measure of a
half space is finite.
It should be clear that this construction is a map from S n → Rn and that as
a result the Borsuk-Ulam Theorem becomes a useful tool. In fact, its application
proves the theorem. If one can apply Borsuk-Ulam to this function then one gets
that there is a point u on the sphere for which f (u) = f (−u). In other words,
for all i we have µi (h+ (u)) = µi (h+ (−u)). But as mentioned before, antipodal
points on the half plane produce opposite half-spaces. Therefore the space below
the hyperplane generated by u has equal measure to the space above it. But the
µi measure of the whole space is only the measure of Ki . Therefore we have that
the measure of the portion of Ki above the plane is equal to the measure of that
below it. Therefore if we can use Borsuk-Ulam we will have established the fact for
all Ki and be done.
Nevertheless, to use Borsuk-Ulam we need to establish continuity. Pick u ∈
S n and a sequence x(k) ∈ S n so that limk→∞ x(k) = u. We want to show that
limk→∞ f (x(k) ) = f (u). Because we have only a finite number of coordinates, we
can do that by showing that the limit holds termwise.
Therefore we only need to show for all i that:
lim µi (h+ (x( k))) = µi (h+ (u))

Another way to state this fact is in terms of integrals. That is to say, if we set
gk (y) as the characteristic function of h+ (xk ) and gu as the characteristic function
of h+ (u) then what we want to show is that:
lim g(y)dµi = gu (y)dµi

An analyst will immediately recognize that the best way to show this fact is
Lebesgue’s Dominated Convergence Theorem. Lebesgue’s Theorem will show that
this equality is true if we can bound all the gk and gu by another function which
has finite integral and show that limk→∞ gk (y) = gu (y) almost everywhere with
respect to the measure µi . That is to say, we need almost everywhere point-wise
convergence. The bounding part is easy, the characteristic function is always less

than equal to 1, and 1µi is finite for all i. Let y 6∈ δh+ (u), the boundary of h+ (u).

Then we know y =< y1 , y2 , . . . , yn >∈ h+ (u) if and only if:

y1 u 1 + y 2 u 2 + . . . y n u n < u 0
(k) (k) (k)
Let δ = u0 − y1 u1 + y2 u2 + . . . yn un > 0. Then we say x(k) = x0 , x1 , . . . , xn .
limk→∞ x(k) = u means that there is for each coordinate an Ni such that for all
(k) δ (k) δ
k > Ni we have xi < ui + 2ny i
and xi > ui − 2ny i
, except in the case of y0
which we have not defined yet but for convenience we simply let equal 1. Therefore
there is a largest N so that we have for all i the same property when k > N . For
k > N we thus get that

X (k) δ δ δ
yi xi < y1 (u1 + ) + y2 (ui + ) + . . . + yn (un + )=
2ny1 2ny2 2nyn
n n
X δ X δ δ (k)
=( yi u i ) + = ( yi ui ) + δ − < u0 − < x0
2 i=1
2 2
Therefore we have that for all y ∈ h+ (u) that are not in the boundary, that
limk→∞ gk (y) = gu (y). Now we need to check that for all y 6∈ h+ (u) that gk (y) also
+ (k)
converges on gy (u). That is to say, for large enough k, y willPnnot be in h (x ).
ButPthis is similar to the argument before. We know that i=1 yi ui > u0 . If we
let i=1 yi ui − u0 = δ > 0 then we know that we can find an N such that for all
(k) δ (k) δ
k > N we have xi < ui + 2ny i
and xi > ui − 2ny i
, except in the case of y0 which
we again set to be 1. Then we have that
n n n
(k) δ X δ X δ X (k)
x0 < + u0 < ( yi u i ) − = ( yi (ui − )< yi xi
2 i=1
2 i=1
2ny i i=1
But this means that y not in h+ (x(k) ) for all sufficiently large k, so again gk
converges point wise. Thus we have shown that we can use Lebesgue Dominated
Convergence and have established continuity of our function f , which by using
Borsuk-Ulam finishes the proof of Ham Sandwich. 