Question

Associate to each sequence \(a=\left\\{\alpha_{n}\right\\}\), in which \(\alpha\), is 0 or 2 , the real number $$x(a)=\sum_{n=1}^{\infty} \frac{\alpha_{n}}{3^{n}}$$ Prove that the set of all \(x(a)\) is precisely the Cantor set described in Sec. \(2.44 .\)

Short answer

Answer: The sequence \(a\) consisting of only 0's and 2's can be used to represent real numbers in the Cantor set through the sum \(x(a) = \sum_{n=1}^{\infty} \frac{\alpha_n}{3^n}\), where \(x(a)\) is a real number in the Cantor set with ternary representation consisting only of 0's and 2's in its fractional part. This relationship shows that the set of all \(x(a)\) is precisely the Cantor set.

Step-by-step solution

Understanding the Cantor Set

The Cantor set is a subset of the interval \([0,1]\) which is obtained by iteratively removing middle thirds of intervals. First, we remove the middle third of the interval \([0,1]\) (i.e., the open interval \((1/3, 2/3)\)), obtaining two closed intervals: \([0, 1/3]\) and \([2/3, 1]\). Then, we remove the middle third of each of these intervals, obtaining four closed intervals: \([0, 1/9]\), \([2/9, 1/3]\), \([2/3, 7/9]\), and \([8/9, 1]\). Repeating this process infinitely many times, we obtain the Cantor set, which consists of those points in \([0,1]\) that do not get removed.

Relating the Cantor Set to the Sequence \(a\)

If a point belongs to the Cantor set, it does not get removed in any of the iterations. We can observe that to determine which interval a point belongs to, we need to consider its ternary (base-3) representation. In base-3, we write numbers as sums of multiples of powers of 3, for example: \(0.0213 = \frac{0}{3^1} + \frac{2}{3^2} + \frac{1}{3^3} + \frac{3}{3^4}\). Now, consider the correspondence between base-3 representation and the sequence \(a\). If a point in the interval \([0,1]\) has a ternary representation in which only 0's and 2's appear in the fractional part, then that point must belong to the Cantor set. This is because it would have escaped the removal process at each iteration. Conversely, any point in the Cantor set can be represented as a sum of multiples of powers of 3 with only 0's and 2's in the list of multiples.

Proving the Relationship Between \(x(a)\) and the Cantor Set

Now, let's prove that the set of all \(x(a)\) is precisely the Cantor set. Let \(x\) be a real number in \([0,1]\). If \(x\) belongs to the Cantor set, then its ternary representation will involve only the digits 0 and 2 in its fractional part. In other words, we can find a sequence \(a = \{\alpha_n\}\), where each \(\alpha_n\) is either 0 or 2, such that \(x = x(a) = \sum_{n=1}^{\infty} \frac{\alpha_n}{3^n}\). Conversely, if we have a sequence \(a = \{\alpha_n\}\), where each \(\alpha_n\) is either 0 or 2, then the number \(x(a) = \sum_{n=1}^{\infty} \frac{\alpha_n}{3^n}\) will have a ternary representation with only 0's and 2's in its fractional part. Therefore, \(x(a)\) belongs to the Cantor set. Hence, we have shown that the set of all \(x(a)\) is precisely the Cantor set described in Sec. 2.44.

Key concepts

Ternary Representation: Understanding the Base-3 System

Imagine you're representing numbers not by using our usual base-10 system, but rather by using base-3, known as the ternary system. In this system, numbers are expressed as sums of multiples of powers of 3. For example, the number 0.0213 in ternary can be expressed as \( \frac{0}{3^1} + \frac{2}{3^2} + \frac{1}{3^3} + \frac{3}{3^4} \). This means:

• The first position after the decimal, representing \( 3^1 \), holds a 0.
• The second position, representing \( 3^2 \), has a 2.
• The third position, denoting \( 3^3 \), contains a 1, and so on.

What's fascinating about this system is that it relates directly to the Cantor set. If a number has a ternary representation using only the digits 0 and 2, then it aligns perfectly with the Cantor set's criteria. This is because these numbers avoid the middle thirds of each interval as we repeatedly remove them to form the Cantor set.
Therefore, understanding ternary representation helps us see how points are kept within the Cantor set during its construction. Every number in the Cantor set has a ternary form limited to these two digits, demonstrating their preservation through each iterative removal step.

Sequence Convergence: Building the Cantor Set

When dealing with sequences that form real numbers, we often need to understand how these sequences converge. Convergence refers to the idea that as we keep adding more terms from a sequence, the sum gets closer and closer to a fixed number.

To relate this to the Cantor set, consider a sequence \( a = \{\alpha_n\} \) where \( \alpha_n \) is either 0 or 2. This sequence helps us create numbers through a sum like \( x(a) = \sum_{n=1}^{\infty} \frac{\alpha_n}{3^n} \). Each term takes a smaller and smaller number, but together they form a complete picture—a number within the range \([0,1]\).
The nature of convergence here ensures that as you keep adding terms, any sequence comprising of 0s and 2s will "fill in" gaps left by the removal process, representing one of those numbers we see remaining in the Cantor set. This convergence assures us that every time we construct such a sequence, it lands us squarely on a point that stays untouched by the process of repeatedly removing those pesky middle thirds.

Infinite Series and the Cantor Set

An infinite series is essentially a sum that goes on indefinitely. Think of it as adding an endless list of numbers. You might be curious how this is even possible, but mathematically, infinite series can converge to a specific value.
The Cantor set, interestingly, can be represented through an infinite series. Each number in the set equals a sum of terms like \( \frac{0}{3^1} + \frac{2}{3^2} + \frac{0}{3^3} + \cdots \), where each coefficient is either 0 or 2 and each term relates to powers of 3.
This setup is incredibly insightful. It tells us how numbers that make up the Cantor set are precisely those which can survive the rigorous process of chopping intervals into thirds. The infinite series allows us to view the Cantor set not just as a theoretical concept, but as a collection of real numbers finely crafted through this endless addition.
The infinite series proves a powerful tool—it frames how the Cantor set remains intact, rendering each number in that peculiar dance of endless removing yet ever-so-cautious preserving of middle third intervals.