proof of ham sandwich theorem

proof of ham sandwich theorem

This proof uses the

Borsuk-Ulam theorem

, which states that any

continuous function

from

S

n

to

ℝ

n

maps some pair of

antipodal points

to the same point.

Let

A

be a

measurable

bounded

subset of

ℝ

n

. Given any

unit vector

n

^

∈

S

n

-

1

and

s

∈

ℝ

, there is a unique

n

-

1

dimensional

hyperplane

normal to

n

^

and containing

s

⁢

n

^

.

Define

f

:

S

n

-

1

×

ℝ

→

[

0

,

∞

)

by sending

(

n

^

,

s

)

to the measure of the subset of

A

lying on

the side of the plane corresponding to

(

n

^

,

s

)

in the

direction

in which

n

^

points. Note that

(

n

^

,

s

)

and

(

-

n

^

,

-

s

)

correspond to the same plane, but to different sides of the plane, so that

f

⁢

(

n

^

,

s

)

+

f

⁢

(

-

n

^

,

-

s

)

=

m

⁢

(

A

)

.

Since

A

is bounded, there is an

r

>

0

such that

A

is contained in

B

r

¯

, the

closed ball

of radius

r

centered at the origin. For sufficiently small changes in

(

n

^

,

s

)

, the measure of the portion of

B

r

¯

between the different corresponding planes can be made arbitrarily small, and this bounds the change in

f

⁢

(

n

^

,

s

)

, so that

f

is a continuous function.

Finally, it’s easy to see that, for fixed

n

^

,

f

⁢

(

n

^

,

s

)

is monotonically decreasing in

s

, with

f

⁢

(

n

^

,

-

s

)

=

m

⁢

(

A

)

and

f

⁢

(

n

^

,

s

)

=

0

for

s

sufficiently large.

Given these properties of

f

, we see by the

intermediate value theorem

that, for fixed

n

^

, there is an

interval

[

a

,

b

]

such that the set of

s

with

f

⁢

(

n

^

,

s

)

=

m

⁢

(

A

)

/

2

is

[

a

,

b

]

. If we define

g

⁢

(

n

^

)

to be the

midpoint

of this interval, then, since

f

is

continuous

, we see

g

is a continuous function from

S

n

-

1

to

ℝ

. Also, since

f

⁢

(

n

^

,

s

)

+

f

⁢

(

-

n

^

,

-

s

)

=

m

⁢

(

A

)

, if

[

a

,

b

]

is the interval corresponding to

n

^

, then

[

-

b

,

-

a

]

is the interval corresponding to

-

n

^

, and so

g

⁢

(

n

^

)

=

-

g

⁢

(

-

n

^

)

.

Now let

A

1

,

A

2

,

…

,

A

n

be measurable bounded subsets of

ℝ

n

, and let

f

i

,

g

i

be the maps constructed above for

A

i

. Then we can define

h

:

S

n

-

1

→

R

n

-

1

by:

h

⁢

(

n

^

)

=

(

f

1

⁢

(

n

^

,

g

n

⁢

(

n

^

)

)

,

f

2

⁢

(

n

^

,

g

n

⁢

(

n

^

)

)

,

…

⁢

f

n

-

1

⁢

(

n

^

,

g

n

⁢

(

n

^

)

)

)

This is continuous, since each

coordinate

function is the composition of continuous functions. Thus we can apply the Borsuk-Ulam theorem to see there is some

n

^

∈

S

n

-

1

with

h

⁢

(

n

^

)

=

h

⁢

(

-

n

^

)

, ie, with:

f

i

⁢

(

n

^

,

g

n

⁢

(

n

^

)

)

=

f

i

⁢

(

-

n

^

,

g

n

⁢

(

-

n

^

)

)

=

f

i

⁢

(

-

n

^

,

-

g

n

⁢

(

n

^

)

)

where we’ve used the property of

g

mentioned above. But this just means that for each

A

i

with

1

≤

i

≤

n

-

1

, the measure of the subset of

A

i

lying on one side of the plane corresponding to

(

n

^

,

g

n

⁢

(

n

^

)

)

, which is

f

i

⁢

(

n

^

,

g

n

⁢

(

n

^

)

)

, is the same as the measure of the subset of

A

i

lying on the other side of the plane, which is

f

i

⁢

(

-

n

^

,

-

g

n

⁢

(

n

^

)

)

. In other words, the plane corresponding to

(

n

^

,

g

n

⁢

(

n

^

)

)

bisects each

A

i

with

1

≤

i

≤

n

-

1

. Finally, by the definition of

g

n

, this plane also bisects

A

n

, and so it bisects each of the

A

i

as claimed.

Title

proof of ham sandwich theorem

Canonical name

ProofOfHamSandwichTheorem

Date of creation

2013-03-22 16:40:29

Last modified on

2013-03-22 16:40:29

Owner

Statusx (15142)

Last modified by

Statusx (15142)

Numerical id

5

Author

Statusx (15142)

Entry type

Proof

Classification

msc 54C99

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