1 Introduction
It has been shown recently that the sets of real numbers, irrational numbers, transcendental numbers, algebraic numbers, and rational numbers can all be partitioned into infinite collections of sets that are dense in the real line. Furthermore, each of the partitioning subsets can themselves be similarly partitioned indefinitely (see [Reference Ho and Zimmerman12, Reference Zimmerman and Ho17, Reference Zimmerman and Ho18]). In this article, we will show that the set of Liouville numbers shares this property.
Since their discovery, Liouville numbers have been studied intensively. They are not only the first examples of transcendental numbers but also the only irrational numbers that rational numbers can approximate more closely than they can approximate algebraic numbers (see Section 2). The set of Liouville numbers has been considered a “thin” set [Reference Chalebgwa and Morris4] since it is a zero-dimensional set whose Lebesgue measure and Hausdorff measure are both zero [Reference Chalebgwa and Morris4, Remark 4.3], also [Reference Priestley15, Theorem 1]. However, from a set-theoretical point of view, the set of Liouville numbers is rich and inexhaustible. As we will show in this article, the set of Liouville numbers can be partitioned into an uncountable collection of sets, each of which is uncountable and dense in the real line. Each of these sets can be similarly partitioned, and this can be repeated indefinitely.
In the next section, we will review some basic properties of Liouville numbers needed in Section 3. There we will describe in detail the repeated partition of the set of Liouville numbers.
2 Basic properties of the Liouville numbers
In his 1844 paper [Reference Liouville13] (see also [Reference Liouville14]), Liouville observed that when algebraic numbers are approximated by rational numbers, there is a limit to the accuracy of approximation: if a is an algebraic number of degree n then there exists a constant
$c=c(a)>0$
such that
$|a-\frac {p}{q}|>\frac {c}{q^n}$
for all rational numbers
$\frac {p}{q}$
. On the other hand, he constructed certain irrational numbers x which can be approximated by rational numbers without such limits of accuracy. These are the numbers x such that for each positive integer n, we can always find rational numbers
$\frac {p}{q}$
satisfying
$|x-\frac {p}{q}|<\frac {1}{q^n}$
. Thus, rational numbers can approximate these irrational numbers – later called Liouville numbers – more closely than they can approximate algebraic numbers. These were the first concrete examples of transcendental numbers, and as such, Alan Baker, himself a leading authority on transcendental number theory, is known to have regarded Liouville as the originator of the theory of transcendental numbers (see [Reference Baker2, p. 1]).
For the basic properties of Liouville numbers, see, e.g., [Reference Baker2–Reference Grant11]. It is known that Liouville numbers form an uncountable subset of transcendental numbers [Reference Chalebgwa and Morris6, p. 2], and that there are uncountably many transcendental numbers that are not Liouville numbers [Reference Priestley15, Corollary 2].
Lemma 2.1 For any integer
$b\geq 2$
and any sequence of integers
$a_1,a_2,\ldots ,a_k,\ldots $
such that for each
$k,\,0\leq a_k\leq b-1$
and
$a_k>0$
for infinitely many k, the number
$x=\sum _{k=1}^\infty \frac {a_k}{b^{k!}}$
, is a Liouville number.
This lemma is well known. For instance, the proofs given in [Reference Baker2, p. 2] or [Reference Stark16, Theorem 6.6] can be converted easily to the general number b.
Remark 2.2 The numbers in Lemma 2.1,
$\sum _{k=1}^\infty \frac {a_k}{b^{k!}}$
, are the numbers in base b whose digits are zero, except at the positions
$k!$
after the decimal point, where the digit is
$a_k$
.
When
$b=10$
, we will designate a sequence of integers
$s:a_1,a_2,\ldots $
to be an admissible sequence if for each
$k,\,0\leq a_k\leq 9$
and
$a_k>0$
for infinitely many k. A Liouville number of the form presented in Lemma 2.1 for an admissible sequence s will be called a primitive Liouville number for the sequence s. We may drop “for the sequence s” when it is unnecessary. As we will see later, many Liouville numbers are not primitive.
Lemma 2.3 There are uncountably many primitive Liouville numbers.
Proof Each admissible sequence of integers
$a_1,a_2,\ldots $
will give rise to a unique primitive Liouville number, and there are uncountably many such sequences, as can be shown by Cantor’s diagonalization argument.
Lemma 2.4 For any Liouville number x and rational number r, the numbers
$x+r$
and
$xr$
are also Liouville numbers.
Proof See [Reference Chalebgwa and Morris6, Proposition 2] and the comments following the proposition there.
For ease of description, we will first restrict our attention to the numbers in the interval
$(0,1)$
, unless stated otherwise.
Remark 2.5 Let x be a primitive Liouville number for the sequence
$s:a_1,a_2,\ldots $
; by Lemma 2.4, the number
$x/10$
is a Liouville number. This Liouville number is not primitive, since the position for each
$a_i$
is shifted one position after each
$i!$
, and all the digits at the positions
$i!$
, for
$i>1$
, are zero.
We will now give a definition of certain Liouville numbers that will be used in the next section. Let
$p_1,p_2,\ldots $
be a sequence of distinct primes and let
$s:a_1,a_2,\ldots $
be an admissible sequence. We will construct certain primitive Liouville numbers
$l(s)$
as any real number satisfying the following property: The digits of the number are all zero except at positions
$k!$
for some integer k. While at the positions of
$k!$
, the digits of
$l(s)$
are arbitrary, i.e., they can be any digit from 0 to 9, except when
$k=(p_1)^n$
, a power of the prime
$p_1$
. In this case, the digit of
$l(s)$
at the position
$k!$
is
$a_n$
, the
$n^{\text {th}}$
term of the admissible sequence s. Since at the positions
$(p_1)^n!$
for some positive integers n, the digits are not all zero,
$l(s)$
is a primitive Liouville number.
Let l be a Liouville number. We will call a real number x eventually l if there is an integer k (different k for different x) such that the first k digits of x after the decimal point are arbitrary, but the
$m^{\text {th}}$
digit, for
$m>k$
, is the
$m^{\text {th}}$
digit of the number l. We will call a number x eventually primitive, if it is eventually l for some primitive Liouville number l. If x is eventually a Liouville number l, the number x itself will also be a Liouville number because we may obtain x by adding an appropriate rational number (with a finite decimal expansion) to the Liouville number l. We will call the first k digits of the number x, which are arbitrary, the rational part of the number x. Clearly, most of the eventually primitive Liouville numbers are not primitive Liouville numbers.
3 Repeated partitions of Liouville numbers
We first note that by [Reference Chalebgwa and Morris6, Corollary 27], the set of Liouville numbers, with the subspace topology inherited from the real line, is homeomorphic to the space of the irrational numbers. This homeomorphism can translate any repeated partitions of irrational numbers into repeated partitions of the Liouville numbers. However, our partitions will be explicit, not merely existential. First some preliminaries.
Definition 3.1 Two admissible sequences
$s:a_1,a_2,\ldots $
and
$t:b_1,b_2,\ldots $
are said to be eventually identical if there exists a positive integer k such that
$a_i$
=
$b_i$
for all
$i\geq k$
.
The relation “eventually identical” is an equivalence relation.
Lemma 3.1 There are uncountably many equivalence classes for the eventually identical relation in the set of admissible sequences.
Proof The difference between any two sequences in each equivalence class must be a terminating decimal. The set of terminating decimals has cardinality
$\aleph _0$
, and thus the cardinality of each equivalence class is
$\aleph _0$
. Equivalence classes are disjoint, and their union is the set of all sequences, which has cardinality
$\mathbf {c}$
. Thus, there must be
$\mathbf {c}$
equivalence classes.
We will now describe a method of partitioning the Liouville numbers. In the following, let Y be a complete set of representatives of this equivalence relation. Any two different sequences
$s_1:a_1,a_2,\ldots $
and
$s_2:b_1,b_2,\ldots $
of Y are not eventually identical, i.e., for each positive integer
$k,$
there exists an integer
$i>k$
such that
$a_i\neq b_i$
. For each sequence
$s_1$
in this set Y of representatives, we will first let
$L_0(s_1)$
be the set of all the numbers in
$(0,1)$
which are eventually the primitive Liouville number
$l(s_1)$
, as defined above in the second paragraph following Remark 2.5. We will then let
$L(s_1)$
be the set of all the real numbers x such that
$x-\lfloor x\rfloor $
belongs to
$L_0(s_1)$
, where
$\lfloor x\rfloor $
is the greatest integer less than or equal to x. Finally, let H be the set consisting of all the Liouville numbers that do not belong to
$L(s)$
for any
$s\in Y$
.
Theorem 3.2 Let L be the set of Liouville numbers in the real line, and let Y be a complete set of representatives of the eventually identical relation on the set of all the admissible sequences. Then L=
$H\cup (\cup _{s\in Y}L(s))$
is a partition of L into an uncountable collection of dense subsets.
Proof We first observe that the indexing set Y in the conclusion of the theorem is uncountable. This is because by Lemma 3.1 there are uncountably many equivalence classes for the eventually identical relation, and hence there are uncountably many different sequences s in the set Y.
We then claim that the sets in the union are mutually disjoint: H is disjoint from each of the sets
$L(s)$
by definition. To show that two sets
$L(s_1)$
and
$L(s_2)$
are disjoint, it is sufficient to show that the sets
$L_0(s_1)$
and
$L_0(s_2)$
are disjoint. Consider any two elements
$a\in L_0(s_1)$
and
$b\in L_0(s_2)$
for two different
$s_1$
and
$s_2$
in the set Y. Since
$s_1$
and
$s_2$
are representatives of two different equivalence classes, they are not eventually identical. Hence, the decimal representations of a and b will eventually differ at many positions
$k=(p_1)^n$
for the powers of
$p_1$
.
We next claim that any of these sets in the union is dense. Again, it is sufficient to show that each set
$L_0(s)$
is dense in the open interval
$(0,1)$
. Observe that the set
$L_0(s)$
contains Liouville numbers with arbitrarily long rational parts, which give rise to rational numbers of a finite decimal expansion. It is well known that a rational number has a finite decimal expansion if and only if it can be represented as a quotient of two integers with the denominator a product of a power of 2 and a power of 5 (see, e.g., [Reference Apostol1, Exercises 1and 2, p. 19]). Such a denominator can be arbitrarily large. For any nonempty open interval,
$(a,b)$
, choose a denominator
$q=2^n5^m$
large enough so that
$\frac {2}{q}<b-a$
; then some rational numbers
$\frac {p}{q}$
and
$\frac {p+1}{q}$
will both be in the interval
$(a,b)$
and so will any eventually primitive number with
$\frac {p}{q}$
as its rational part.
Finally, we need to prove that the set H is dense. First note that the set
$H_0$
, the set of all the elements of H that belong to the open interval
$(0,1)$
, is dense in this interval. This is because, as noted in Remark 2.3, for any primitive Liouville number x, the number
$x_0=x/10$
is not a primitive Liouville number. Consider the Liouville numbers which are eventually
$x_0$
. Since
$x_0$
is not a primitive Liouville number, any eventually
$x_0$
number is an element of
$H_0$
. The set
$H_0$
is then dense in the open interval
$(0,1)$
since the eventually
$x_0$
numbers contain arbitrarily long rational parts. Now, consider any open interval
$(n,n+1)$
for a positive integer n; for each eventually
$x_0$
number l in the open interval
$(0,1)$
, the number
$n+l$
is an element of H that belongs to the open interval
$(n,n+1)$
. Thus, H is also dense in
$(n,n+1)$
. A set which is dense in every interval
$(n,n+1)$
must also be dense in the real line.
Theorem 3.3 Each of the partitioning subsets of
$L=H\cup (\cup _{s\in Y}L(s))$
can itself be partitioned into an uncountable family of dense subsets. Furthermore, this process can be repeated indefinitely, while still achieving the same result, each time applied to any member of the previous partition.
Proof As shown in the proof of Theorem 3.2, it is sufficient to prove the case when all the sets are subsets of the open interval
$(0,1)$
. To show that the partitioning subsets can be partitioned repeatedly, we need to consider a general situation. For a set of m representatives
$s_1,s_2,\ldots ,s_m$
, chosen from the set Y of representatives, let
$l(s_1,s_2,\ldots ,s_m)$
be any number satisfying the following properties: The digits of
$l(s_1,s_2,\ldots ,s_m)$
are all zero except for those at the positions
$k!$
after the decimal point, for a positive integer k. When
$k=(p_1)^n$
,
$\,(p_2)^n, \ldots $
, or
$(p_m)^n$
, a power of any of the first m primes in our list of primes, this digit of
$l(s_1,s_2,\ldots ,s_m)$
at the position
$(p_i)^n!$
is
$s_{i_n}$
, the
$n^{\text {th}}$
term of the admissible sequence
$s_i$
for each
$i=1,\,2,\,\ldots m$
. If k is not one of these m primes, the digit of
$l(s_1,s_2,\ldots ,s_m)$
is arbitrary. Each
$l(s_1,s_2,\ldots ,s_m)$
is a primitive Liouville number. Now, let
$L(s_1,s_2,\ldots ,s_m)$
be the set of all the eventually primitive numbers
$l(s_1,s_2,\ldots ,s_m)$
. Note that when
$m=1$
and
$s_1=s$
for some representative
$s\in Y,\,L(s_1,s_2,\ldots ,s_m)$
becomes one of the partitioning subsets
$L(s)$
in Theorem 3.2.
We observe that for a fixed set of representatives
$s_1,s_2,\ldots ,s_{m-1}$
there are uncountably many ways to choose an
$m^{\text {th}}$
representative
$s_m$
, because the complete set of representatives, Y, is uncountable. Furthermore, if two representatives at m are different, say
$s_m\neq s^{\prime }_m$
, the sets
$L(s_1,s_2,\ldots ,s_m)$
and
$L(s_1,s_2,\ldots ,s^{\prime }_m)$
have no element in common. This is because
$s_m$
and
$s^{\prime }_m$
are two admissible sequences which are not eventually identical, and any two elements from
$L(s_1,s_2,\ldots ,s_m)$
and
$L(s_1,s_2,\ldots ,s^{\prime }_m)$
, respectively, will have a different digit at the position
$(p_m)^n!$
for some positive integer n. Finally, each of the sets
$L(s_1,s_2,\ldots ,s_m)$
is dense in the real line. This is because these sets contain Liouville numbers of arbitrarily long rational parts. Now, for a fixed set of representatives
$s_1,s_2,\ldots ,s_{m-1}$
, let
$H(s_1,s_2,\ldots ,s_{m-1})$
be the set consisting of all the elements of
$L(s_1,s_2,\ldots ,s_{m-1})$
that do not belong to any
$L(s_1,s_2,\ldots ,s_m)$
for any
$s_m\in Y$
.
$$ \begin{align} L(s_1,s_2,\ldots,s_{m-1})=H(s_1,s_2,\ldots,s_{m-1})\cup(\cup_{s_m\in Y}L(s_1,s_2,\ldots,s_{m-1},s_m)) \end{align} $$
is then a partition of the set
$L(s_1,s_2,\ldots ,s_{m-1})$
into an uncountable collection of sets, where each of the sets
$L(s_1,s_2,\ldots ,s_{m-1},s_m)$
is a dense subset of the real line.
Partition (3.1) shows that the sets
$L(s_1,s_2,\ldots ,s_{m-1})$
can be partitioned repeatedly. However, we still need to show that the set
$H(s_1,s_2,\ldots ,s_{m-1})$
is dense and can also be partitioned repeatedly. Since it is a subset of
$L(s_1,s_2,\ldots ,s_{m-1})$
, for each of its elements, a digit at the positions
$(p_m)^n!$
, for
$1\leq i\leq m-1$
, must eventually be the
$n^{\text {th}}$
digit of the representative
$s_i$
. The condition that
$H(s_1,s_2,\ldots ,s_{m-1})$
does not contain any element of
$L(s_1,s_2,\ldots ,s_m)$
for any
$s_m\in Y$
means that the digit at the position
$(p_m)^n$
! will be zero for sufficiently large n. However, if
$k=(p_j)^n$
, for
$j>m$
and any positive integer n, the digit of any element of
$H(s_1,s_2,\ldots ,s_{m-1})$
at the position
$k!$
is still arbitrary. There are infinitely many such primes,
$p_j$
’s, and the set
$H(s_1,s_2,\ldots ,s_{m-1})$
will be uncountable. Furthermore, the set also contains Liouville numbers with arbitrarily long rational parts, and hence, the set is dense.
We may use a similar method to show that each of the sets
$H(s_1,s_2,\ldots ,s_{m-1})$
can be partitioned into an uncountable collection of dense subsets. Consider a particular set
$H(s_1,s_2,\ldots ,s_{m-1})$
. To simplify the presentation, let
$H_m=H(s_1,s_2,\ldots ,s_{m-1})$
. We first try to partition this set. As noted above, the digits at the positions
$(p_i)^n!$
for each element of
$H_m$
, are eventually the
$n^{\text {th}}$
digit of the representative
$s_i$
for
$1\leq i\leq m-1$
, but the digits at the positions
$(p_{\text {m}})^n!$
will eventually be zero for sufficiently large n. However, if
$k=(p_j)^n$
, for
$j>m$
and any positive integer n, the digit of any element of
$H_m$
at the position k! is arbitrary. We can let
$H_m(t)$
, for each representative t of Y, be the set of those elements of
$H_m$
whose digit at the positions
$(p_{m+1})^n$
! is eventually the
$n^{\text {th}}$
digit of the sequence t.
$H_m$
=
$M_m\cup (\cup _{t\in Y}$
$H_m(t))$
is a partition of
$H_m$
into an uncountable collection of dense subsets, where
$M_m$
is the set consisting of all the elements of
$H_m$
that do not belong to any
$H_m(t)$
. We know that this is a partition because for different
$t_1$
and
$t_2$
in Y, different elements in
$H_m(t)$
and
$H_m(t_2)$
, respectively, will eventually have different digits in the position
$(p_{m+1})^n!$
for some large enough n, and thus,
$H_m(t)$
and
$H_m(t_2)$
will have no common element. The partition is into an uncountable collection of subsets, since the set of representatives Y is uncountable. The subsets
$H_m(t)$
are all dense subsets since
$H_m(t)$
contains elements of arbitrarily long rational parts.
We now show that each of the partitioning sets of
$H_m$
can itself be partitioned repeatedly in a similar way. Let
$t_1,\,t_2,\,\ldots ,\,t_r$
be any sequence of representatives chosen from the set Y, and assume the set
$H_m(t_1,\,t_2,\,\ldots ,\,t_{r-1})$
has been defined. We may let
$H_m(t_1,\,t_2,\,\ldots ,\,t_{r-1}, \, t_r)$
for a fixed r, be the set of all the numbers of
$H_m(t_1,\,t_2,\,\ldots ,\,t_{r-1})$
that, at the location
$(p_{\text {m}+r})^n!$
, are eventually the digits of
$t_{r_n}$
, the
$n^{\text {th}}$
term of
$t_r$
. In addition, let
$M_m(t_1,\,t_2,\,\ldots ,\,t_{r-1})$
be the set consisting of all the elements of
$H_m(t_1,\,t_2,\,\ldots ,\,t_{r-1})$
that do not belong to any
$H_m(t_1,\,t_2,\,\ldots ,\,t_{r-1},t_r)$
. We then have a repeated partition
$$ \begin{align*} H_m(t_1,\,t_2,\,\ldots,\,t_{r-1})=M_m(t_1,\,t_2,\,\ldots,\,t_{r-1})\cup( \cup_{t_r\in Y}H_m(t_1,\,t_2,\,\ldots,\,t_{r-1},t_r)) \end{align*} $$
The proof that this is a partition into an uncountable collection of dense sets, and that the sets
$M_m$
and
$M_m(t_1,\,t_2,\,\ldots ,\,t_{r-1})$
can also be partitioned repeatedly, is similar to what we did for the sets
$H_m$
and
$H_m(t_1,\,t_2,\,\ldots ,\,t_{r-1})$
, respectively. Since the method is so similar, we will omit the proof.
4 Concluding remarks
We now look at some properties of the partitioning subsets and will then pose some questions as possible future work. One natural question is whether any of the partitioning subsets possess any additional arithmetic or algebraic structures. In particular, are any of them closed under addition or multiplication? We first note that none of the sets used in the partition (3.1)
$$ \begin{align*} L(s_1,s_2,\ldots,s_{m-1})=H(s_1,s_2,\ldots,s_{m-1})\cup(\cup_{s_m\in Y}L(s_1,s_2,\ldots,s_{m-1},s_m)) \end{align*} $$
is closed under either addition or multiplication if
$m>1$
. First consider the sets
$L(s_1,s_2,\ldots ,s_{m-1})$
. These sets consist of points whose digits at the places
$(p_i)^n!$
will eventually be
$s_{i_n}$
, the
$n^{\text {th}}$
term of the admissible sequence
$s_i$
, for each
$i=1,\,2,\,\ldots m-1$
. The sum or product of two such elements will not be such an element. The sets
$H(s_1,s_2,\ldots ,s_{m-1})$
will not be closed under addition or multiplication either, since these are subsets of
$L(s_1,s_2,\ldots ,s_{m-1})$
and we have just noted that the sum or product of two elements of this set would not be an element of this set.
Now, consider the sets in the partition
$L=H\cup (\cup _{s\in Y}L(s))$
of Theorem 3.2. These are the same sets as in the partition (3.1) for the cases
$m=1$
which we have not considered yet. The sets
$L(s)$
are not closed under either addition or multiplication for the reason discussed above. However, the set H will be closed under both addition and multiplication. This is because H consists of elements whose digit at the positions
$(p_1)^n!$
will eventually be zero for sufficiently large n. The sum or product of any two such elements will again be one of such elements.
In fact, there is a more interesting algebraic structure among sets
$L(s)$
and H. Consider any two elements
$x_1 \in L(s_1)$
and
$x_2 \in L(s_2)$
. If we define
$x_1 \oplus x_2$
, or
$x_1\otimes x_2$
, by adding, or multiplying, modulo 10, the corresponding digits of
$x_1$
and
$x_2$
, respectively, we will see that the digits of
$x_1 \oplus x_2$
, or
$x_1\otimes x_2$
, at the place
$(p_1)^n!$
will eventually either be some admissible sequence
$s_0$
or be a sequence of zeros from a certain point on (the results will be different for
$x_1 \oplus x_2$
and
$x_1\otimes x_2$
in general). In the first place
$x_1 \oplus x_2$
, or
$x_1\otimes x_2$
, will be an element
$L(s_0)$
and in the second place,
$x_1 \oplus x_2$
, or
$x_1\otimes x_2$
, will be an element of H. Furthermore, this operation is independent of the choices of elements
$x_1$
and
$x_2$
from these two sets. In fact, we may define the operations
$L(s_1) \oplus L(s_2)$
, and
$L(s_1)\otimes L(s_2)$
, and the results of these operations are either a set
$L(s_0)$
for some admissible sequence
$s_0$
, or the set H. (Again the results for
$L(s_1) \oplus L(s_2)$
and
$L(s_1)\otimes L(s_2)$
may be different in general.) Furthermore, if one of the elements
$x_i$
belongs to the set H, it is straightforward to show that
$$ \begin{align*} H \oplus L(s)=L(s) \oplus H=L(s) \,\,\,\, \text{and} \,\,\,\, H \otimes L(s)=L(s) \otimes H=H \end{align*} $$
for all the sets
$L(s)$
. With the operations
$ \oplus $
and
$ \otimes $
, the sets
$L(s)$
and H form a ring with H the additive identity. We will leave the reader to explore whether this would lead to some interesting algebraic properties.
We have shown that all the partitioning subsets of the Liouville numbers used in this article are dense. Are any of them a
$G_\delta $
set? If so, it would imply that any real number can be expressed as a sum (and as a product) of two elements from the set. This would be a nice homage to the beautiful paper [Reference Erdös8] of Erdös, to which, at least in part, this line of pursuit can be attributed.
Another question is that “the infinite repetition” phenomenon characterized in the article hints at the existence of some kind of underlying “fractal” or “self-similarity” with the set L, at least when viewed from some specific perspective. Is it possible to elaborate on this structure?
Finally, as shown in [Reference Ho and Zimmerman12, Reference Zimmerman and Ho17], and [Reference Zimmerman and Ho18], the sets of real numbers, irrational numbers, transcendental numbers, algebraic numbers, and rational numbers can all be partitioned into infinite collections of sets that are dense in the real line. We have now shown that the set of Liouville numbers shares this property. This prompts us to conjecture that any dense subset of real numbers can be similarly partitioned.
Acknowledgements
The authors would like to thank the reviewers, whose comments and suggestions greatly improved the article’s presentation. This is especially true of the final section, much of whose work was inspired by the reviewers’ remarks.
