Bijection between $[0,1)$ and the space of binary sequences

Question

My question deals with the problem of showing that the set $$ \Omega = \{ \omega \colon \omega =(a_1,a_2, \ldots ), a_i =0,1\} $$ has the same cardinality as the interval $[0,1)$. In a textbook I read the following explanation:

It is well known that every number $a \in [0,1)$ has a unique binary expansion (containing an infinite number of zeros) $$ a = \frac{a_1}{2} + \frac{a_2}{2^2} + \ldots \qquad (a_i=0,1). $$ Hence it is clear that there is a one-to-one correspondence between the points $\omega \in \Omega$ and the points $a$ of the set $[0,1)$, and therefore $\Omega$ has the cardinality of the continuum.

Now I have two questions:

• What is meant with "containing an infinite number of zeros"?
• The discussions in Set of points of $[0,1)$ that have a unique binary expansion and Cardinality: Set of all binary sequence equal c show that binary expansion is usually not unique. So I am a bit surprised, as I did not expect such a mistake in this text. Did I understand that incorrectly or is there a natural way to make the binary expansion unique for all numbers in $[0,1)$?

Answer 1

You are correct that there is a mistake in the text, though you are not quite correct about what the mistake is. By "containing an infinite number of zeros", they mean that when you have a number whose binary expansion is not unique, you should choose the one which has infinitely many $0$s (it will have one expansion ending in all $0$s, and one expansion ending in all $1$s; choose the first one). So this does give a well-defined function $f:[0,1)\to\Omega$.

However, because of the non-uniqueness of binary expansions, this function is not a bijection (contrary to what the text claims)! It is injective, but it is not surjective. After all, by construction, for every $a\in [0,1)$, the sequence $f(a)$ has infinitely many $0$s. So any sequence that has only finitely many $0$s will not be in the image of $f$.

Answer 2

An infinite number of zeros means that in the expansion $$a=\frac{a_1}{2}+\frac{a_2}{4}+...,$$ infinitely many of the numbers $a_i$ are zero. The reason we need this is because otherwise, as you say, the binary expansion is not unique. An example of this is $1/2=0+1/4+1/8+....$

This function is not a bijection, merely an injection. Its image is the set of all binary sequences with an infinite number of zeros. So perhaps this function is a step in the invocation of the Schroder Bernstein Theorem?

Answer 3

At first glance, it seems to me like the fact that we are limiting ourselves to binary sequences containing infinite zeroes means that we are guaranteed uniqueness.

In base 10, non uniqueness basically comes from the fact that $0.9999\ldots = 1$. Similarly in binary, non uniqueness comes from the fact that $0.111\ldots = 1$ (as an exercise, prove this using the $a_i$ expansion formula in your question). However, if we assert that there are infinitely many zeroes in our sequence, it means that the tail of the sequence cannot consist only of ones (why?). This in turn guarantees that our binary expansions are unique.