1
There are uncountably many binary sequences whose limiting
relative frequency is x.
Mariusz Popieluch, April 12, 2015
The proof will proceed with reference to subsets of and their
natural densities, but it should be noted that each result about
subsets of can be transformed to a corresponding result about
binary sequences that are characteristic functions of the sets in
question. I.e.,
for each , the characteristic function of is defined as the
function : {,} , such that = if , else = . Such binary sequences,
in this context, have the intended interpretation of Bernoulli
processes.
SOME USEFUL DEFINITIONS AND LEMMAS
(i) Let and for any let = {, , } and = ||. Natural (or
asymptotic) density of is defined as the limit on the left, below
(if the limit exists): = lim = lrf = lim =
To that value, there corresponds the limit on the right, which
is the limiting relative
frequency , denoted lrf, of the binary sequence (derived from ).
I just presetig this correspondence here, to stress that it exists,
but the discussion will make reference
only to subsets of and their natural densities, as I believe
this approach contributes to the clarity of the argument. So a
result about the cardinality of families of subsets of with some
fixed natural density, corresponds to a result about the
cardinality of families of binary sequences with some fixed
limiting relative frequency.
(ii) For each [,] there exists a set such that = . Proof: Take
some equidistributed sequence {} in [,] and define a monotone
family {}[,] of sets: = { < }. Then, by definition = , for all .
I only need one such sequence {} and its above property for the
main proof. The existence of such a sequence is guaranteed by the
equidistribution theorem.
(iii) The set of all the infinite subsets of the set of all
primes , is uncountable.
(iv) That = is a well-known fact. But also, for each = . Proof:
this is an immediate consequence of the definition of natural
density: if is the set of primes, and , then | []| | []| for each ,
where [] ={, , }. In this case the limits exist by the squeeze
theorem, since zero is a lower bound for | []|/, and the upper
bound of | []|/ tends to zero.
(v) If = > and = , then = , where denotes the symmetric
difference of sets and , defined as = . Moreover = , since = and
the fact that has density zero.
Proof on next page.
2
PROOF
Take an element from the family {}[,] given in (ii). It follows
that = . Next, the idea is to rid of all of its prime elements,
without altering its natural density. This follos directly from .
Lets denote, with , and note that = = , by (v). Now, since for each
= , then it follows that = for each , by (iv) and (v). This holds
for each , by (ii). Hence |{ }| = || for each . Hence |{ }| is
uncountable, by (ii), for each [,], as required.
However, the above cardinal equality rests on the key fact that
for all , if , then , for each . In other words, the above result
is conditioned on thus generated sets maintaining the natural
density of , whilst remaining distinct sets. But this holds by
virtue of the construction of for each . For suppose = , then = by
definition of symmetric difference, and given that = for all , by
the careful construction of , it follows that = . That implies that
= , as required.
