I realize an infinite set of Aleph-one cardinality is also known as uncountably infinite, but if you had a countably infinite set of counters to collectively assign to the task, could it be achieved? Would starting from the set for which its the powerset of help? Could I use the proof below on infinity ink at: http://www.ii.com/math/ch/#expCH
The set of reals between 0 and 1 can be represented by the set of all countably infinite sequences of 0's and 1's . Think of these as representing binary "decimals" between .000000… and .111111… . In this representation .1=1/2, .01=1/4, .11=3/4, etc.
The power set of the natural numbers, P(N), can also be represented by the set of all countably infinite sequences of 0's and 1's. Each sequence represents a subset of N by interpreting a 0 in position n to mean that the number n is not in the subset and a 1 in position n to mean the number n is in the subset. This way of specifying a set is called the "characteristic function" of the set.
One way to represent all countably infinite sequences of 0's and 1's is to use Cartesian product notation:
{0, 1} x {0, 1} x {0, 1} x … = {0, 1}^aleph0
Since in set theory {0, 1} = 2, we can also write this as:
2^aleph0
thanks in advance!
Best Answer
First of all, $2^{\aleph_0} = \mathfrak{c}$, the cardinality of the continuum. Now $\aleph_1 = \mathfrak{c}$ is exactly the Continuum Hypothesis.
Second, if by "an infinity of counters" you allow a non countable infinity, then yes :)
As for the question itself, since you are considering counters, and they can change their state at most a countable amount of times, then they won't be able to count to $\mathfrak{c}$.
Suppose they could, then we would have a uncountable amount of states our counters covered, which means at each step at least one of the counters changed its state, then by the Pigeon Hole Principle, since we have a countable set of counters and an uncountable amount of changes, so at least one of the counters must have chaged its state an uncountable amount of times. A contradiction.