I am solving the following exercise:

for $n \in \mathbb{N}$ and $a_1,a_2, \ldots ,a_n \in \{0,1\}$ we define: $$ I(a_n, \ldots , a_n) := \left \lbrack \sum_{i=1}^n \frac{a_i}{2^i}, \sum_{i=1}^n \frac{a_i}{2^i} + \frac{1}{2^n} \right \rbrack$$

(a) show that there is a transformation $g: \{0,1\}^\mathbb{N} \rightarrow \lbrack 0,1 \rbrack$ such that: $$ g\left ( \left(a_n\right)_{n\in \mathbb{N}}\right) \in \bigcap_{n \in \mathbb{N}} I(a_1, \ldots, a_n)$$

(b) show that $g$ is surjective (and therefore defines a binary representation of the real numbers within the unit interval).

(c) is $g$ injective (meaning is the binary representation unique)?

My Problem: This was an exercise asked by our analysis professor. (a) I do not understand what this transformation $g$ does. I understand that we should take an element $a_i$ and do the transformation $g$ and then we should get an result which lies in the interval $\lbrack 0,1 \rbrack$. If I assume any value for $n$ and then just randomly decide which $a_i$ are $0$ or $1$, I always get a result which lies in that interval. Nevertheless I do not see what I am supposed to do. Further I am not sure what do they mean with $\bigcap_{n \in \mathbb{N}} I(a_1, \ldots, a_n)$? I see that this should be the intersection of all those values I can calculate with the sum formula? Even if so that somehow does not help me to clarify the exercise.

I would be very thankful if someone could explain to me step-by-step what I am supposed to do in this exercise. I appreciate every little hint or help, because this exercise confuses me a lot.


1 Answer 1


Note that $I(a_1,\ldots,a_n)$ is an interval, not a number. It is the interval of those numbers whose binary representation is $0.a_1a_2\ldots a_n\ldots$.

A graphical approach to understand why, is:

  • Take a random sequence of four or five $0$'s and $1$'s.
  • Divide the interval $[0,1]$ in two equal parts. Note that the left part is the set of the numbers whose binary representation begins with $0.0\ldots$ and the other, the same for $0.1\ldots$. So choose the left if $a_1=4$ and the right if $a_1=1$.
  • Divide the chosen part and select left or right depending on the value of $a_2$ (same criterion: $0=$left, $1=$right).
  • When you finish with $a_n$, the selected interval is $I(a_1,\ldots,a_n)$.

The intersection of infinitely many intervals like this is a real number, because the length of the intervals tends to $0$.

Note that this real number is the limit of a sequence of sums, each of them has a term more than its predecesor. In other words, a convergent series.

I hope these tips help.

  • $\begingroup$ I understood that I is not an number, I just meamt that with calculating the defined sum, we always get a number in the interval. Still I do not see what $g$ does? $\endgroup$
    – Mainviel
    Oct 17, 2014 at 14:20

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