A rather annoying fact about Hausdorff dimension is that, if one has a family of sets parametrized by
, then it is typically not true, or at least not easy to show, that the Hausdorff dimension of
is
, as one would expect from, say, algebraic geometry. For example, if
are Borel sets, then it is not necessarily true that
. When one goes a bit beyond the Borel sets (and does not assume the existence of large cardinals), things get much more ridiculous, and in fact there can exist a coanalytic set of full dimension whose closed subsets are all countable.
I ran into this issue recently, when I needed the following theorem to be true:
Theorem. Let
be an asymmetric norm on
, let
be a subset of the unit sphere of the dual norm of
, and let
.
Then
.
Here, is the dual flat of
.
The techniques used to prove this theorem have nothing to do with the problem I am working on, and I think it is an instructive example of how they are used; so I record the theorem here, rather than in the paper I am writing.
Specifically, the theorem shall be proven using algorithmic randomness, following the framework of Algorithmic information, plane Kakeya, and conditional dimension. We recall this framework here.
Let be the set of all finite binary strings. If
, we write
for their concatenation. We write
for the length of
.
A set is prefix-free if for every
and
, if
, then
. By standard coding techniques one can represent points in
or
as finite binary strings, in such a way that the set of binary strings which are used to code points in
or
is prefix-free; we shall exploit this fact without comment.
A function is partial computable if
and there exists a Turing machine
such that for every binary string
,
iff
halts with input
, and if
, then
prints
. A partial computable function
is prefix-free if
is prefix-free. A prefix-free function
is universal if for every prefix-free partial computable function
, there exists
such that
is prefix-free, and for every
,
. Thus
is playing the role of a “compiler”. There exists a universal prefix-free function
; we fix one, once and for all. If
is prefix-free and
is chosen so that
, then
is a program which computes
.
The Kolmogorov complexity of is
.
Since we can code rational points, or integers, as binary strings, it makes sense to take their Kolmogorov complexities. For any , the program which simply decodes the binary expansion of
provides an upper bound on
, namely
(where logarithms are base-
). Given
and
, the Kolmogorov complexity of
to
bits is
.
The effective Hausdorff dimension of is
.
It can be shown that does not depend on the choice of the universal function
.
The above concepts can be relativized. Let , and in the above definitions, replace “Turing machine” with “Turing machine with oracle
.” Let
and
denote the Kolmogorov complexity and effective Hausdorff dimension of
measured with oracle
.
The key theorem, which interprets Hausdorff dimension as a measure of indescribability, and is proven in Plane Kakeya… is,
Point-to-set principle. For every set
,
.
In particular, for each set , there exists an oracle
such that if
then
. The oracle
is called a Hausdorff oracle for
.
Lemma 1. Let
be a compact convex subset of
, and assume
.
There existssuch that for every
, with
,
.
To prove the lemma, observe that since ,
has empty interior. Since
is convex, it follows that
is contained in an
-plane
. If
, then it is enough to prove the theorem with
replaced by
, so we may use induction on
to reduce to the case that
.
If , then
has nonempty interior in
, so
contains a ball
in
, and (viewing
as a subset of
)
is a cylinder of length
, so
. Conversely, since
is compact,
is contained in a ball
in
, and the converse inequality holds, proving the lemma.
Lemma 2. Let
be an asymmetric norm on
. Then there exists
such that for each point
on the unit sphere of the dual norm of
, and each
,
.
This is just some linear algebra, and comparability of norms, so I’ll leave out the proof of this lemma.
Now to prove the main theorem. By the point-to-set principle, we may choose a Hausdorff oracle for
. By rounding off
to a suitable rational number for each
, we may find a function
such that for each
and
,
. Then there is an oracle
which computes
; let
.
For each and
, choose
such that
and
. By comparability of norms, there exists
such that for every
, if
, then
. So
is a finite set and therefore the function
which returns the lexicographically
th
such that
, and returns
if no such point exists, is well-defined.
Fix and
, and define the following algorithm which takes input
, and uses
as an oracle:
- Let
.
- For each
in lexicographic order:
- If
, increment
.
- If
, return
.
- If
- Return
.
Thus this algorithm computes , so for any
such that
,
.
For each , choose
such that
. Then, if
, one can check that
,
which establishes that there exists such that
. In particular, with
,
.
It remains to estimate , but by Lemma 2, there exists
which only depends on
, such that for every
such that
,
. Let
,
the largest
such that
, and
, so by Lemma 1,
; for any
,
; and for any
,
. Combining these facts, we see that
and therefore, since each ball has volume ,
. From this, one can show that
.
The theorem now follows from the point-to-set principle.