Quiz: The Cantor Set — 16 questions

1. How is the Cantor set constructed from the interval $[0,1]$?

A. By repeatedly removing the closed middle third from each remaining open interval

B. By repeatedly removing the open middle third from each remaining closed interval

C. By removing the left third from each interval and retaining the middle portions

D. By dividing each interval into thirds and removing both outer thirds

Answer: (B)

Explanation

The Cantor set is formed by repeatedly deleting the open middle third while retaining the resulting closed intervals. Removing closed middle thirds or deleting outer portions describes different constructions.

2. What happens during the first step of the Cantor set construction?

A. The interval $[0,1]$ is divided into four equal closed intervals

B. The two outer thirds of $[0,1]$ are removed from the interval

C. The open interval $(1/3,2/3)$ is removed from $[0,1]$

D. The closed interval $[1/3,2/3]$ is removed from $[0,1]$

Answer: (C)

Explanation

The first step removes the open middle third, leaving the closed intervals $[0,1/3]$ and $[2/3,1]$. The endpoints remain because the removed interval is open.

3. Which expression defines the Cantor set in terms of the stage sets $I_n$?

A. $C=\bigcup_{n=1}^{\infty} I_n$

B. $C=\bigcap_{n=1}^{\infty} I_n$

C. $C=\lim_{n\to 0} I_n$

D. $C=I_1\cap I_2$

Answer: (B)

Explanation

The Cantor set consists of the points that remain in every stage, so it is the intersection of all the sets $I_n$. A union would include points removed at later stages.

4. Why is the Cantor set non-empty?

A. The first stage contains an interval that is never subdivided again

B. Every endpoint of every retained closed interval belongs to the Cantor set

C. Every interval removed during construction contributes its interior points

D. The construction eventually stops after all retained intervals are listed

Answer: (B)

Explanation

Endpoints of retained closed intervals survive every subsequent removal and therefore belong to the Cantor set. Points in removed intervals do not belong to the set, and the construction continues indefinitely.

5. Why is the Cantor set a closed subset of $[0,1]$?

A. It consists of a finite collection of isolated endpoints

B. It is the union of all open intervals removed during construction

C. It is the intersection of finite unions of closed intervals

D. It contains every point between any two of its endpoints

Answer: (C)

Explanation

Each stage set $I_n$ is a finite union of closed intervals, hence closed, and the intersection of these stage sets is closed. The removed intervals are open and do not describe the Cantor set.

6. What is the total length of the stage-$n$ set $I_n$?

A. $\ell(I_n)=\frac{1}{2^n3^n}$

B. $\ell(I_n)=\left(\frac{3}{2}\right)^n$

C. $\ell(I_n)=\left(\frac{2}{3}\right)^n$

D. $\ell(I_n)=\frac{n}{3^n}$

Answer: (C)

Explanation

There are $2^n$ intervals, each of length $1/3^n$, so their total length is $2^n\cdot(1/3^n)=(2/3)^n$. The reciprocal and alternative expressions do not result from multiplying the number and length of the intervals.

7. What does the limit $\ell(C)=\lim_{n\to\infty}(2/3)^n=0$ imply about the Cantor set?

A. Its total length remains equal to one at every stage

B. Its total length is zero, although the set is not empty

C. It becomes an open set after infinitely many removals

D. It contains no points because its total length is zero

Answer: (B)

Explanation

The stage lengths approach zero, so the Cantor set has length zero, while retained endpoints show that it is non-empty. Zero length therefore does not imply that a set contains no points.

8. How does a point in the Cantor set determine a sequence used in its coding?

A. It determines a unique sequence of interval lengths and endpoints

B. It determines a unique infinite sequence of left and right interval choices

C. It determines a sequence based on whether each point is rational or irrational

D. It determines a finite sequence of removed middle intervals

Answer: (B)

Explanation

At every construction stage, a Cantor-set point lies in one of the two retained subintervals, producing a unique infinite left/right sequence. Removed middle intervals do not generate this coding sequence because they contain no points of the Cantor set.

9. What ternary digits result when left choices are encoded by 0 and right choices by 2?

A. A ternary expansion containing 0s, 1s, and 2s in arbitrary positions

B. A binary expansion containing only 0s and 1s

C. A ternary expansion containing only 0s and 2s

D. A ternary expansion containing only 1s and 2s

Answer: (C)

Explanation

Replacing each left choice with 0 and each right choice with 2 gives a ternary expansion whose digits are restricted to 0 and 2. The digit 1 corresponds to the removed middle thirds rather than to retained interval choices.

10. In the diagonal construction, how is the ith digit of the new ternary sequence chosen from the listed expansions?

A. It is chosen by copying the first digit that differs among the listed expansions

B. It is chosen as 1 whenever the ith listed digit is 0 or 2

C. It is chosen to equal the ith digit of the ith listed expansion

D. It is chosen as 0 when the ith listed digit is 2, and as 2 when it is 0

Answer: (D)

Explanation

The construction sets the new digit to 0 when the diagonal digit is 2 and to 2 when the diagonal digit is 0. This guarantees a disagreement with the nth listed expansion at position n, whereas copying diagonal digits would not create that disagreement.

11. Why does the diagonal construction establish that the Cantor set is uncountable?

A. It creates a Cantor-set element differing from the nth listed element at its nth digit

B. It creates an element outside the Cantor set by inserting a digit 1 at each diagonal position

C. It shows that every listed element has two different ternary expansions

D. It proves that the Cantor set contains an interval between each pair of listed elements

Answer: (A)

Explanation

The constructed sequence uses only 0s and 2s and differs from the nth listed sequence at its nth digit, so it cannot appear anywhere on the proposed list. The other claims do not establish an element missing from every enumeration.

12. Which statement correctly describes the intervals contained in the Cantor set?

A. It contains no points because all intervals are removed

B. It contains one nontrivial interval at each construction stage

C. It contains every interval whose endpoints lie in the Cantor set

D. It contains points but no nontrivial interval

Answer: (D)

Explanation

The Cantor set is nonempty and contains many points, but it contains no nontrivial interval. The removal process eliminates intervals while leaving a highly structured collection of individual points.

13. What does it mean for the Cantor set to be nowhere dense in [0, 1]?

A. Its total length is zero as a measure-theoretic property

B. It has no connected subset containing two distinct points

C. Its closure has empty interior in the usual topology

D. It is closed and bounded as a subset of the real line

Answer: (C)

Explanation

Nowhere dense means that the closure has empty interior; for the Cantor set, this expresses its topological thinness. Zero length concerns measure, while total disconnectedness and compactness are different properties.

14. What does the equality $(0.1)_3=(0.022\ldots)_3=1/3$ demonstrate about ternary expansions?

A. A ternary expansion is unique whenever its value lies in [0, 1]

B. A real number can have more than one ternary expansion

C. Expansions using only 0s and 2s are never valid

D. Every ternary expansion must contain a digit 1

Answer: (B)

Explanation

The equality gives two distinct ternary representations of the same number, showing that ternary expansions are not always unique. This does not imply that every expansion contains 1 or that expansions in the unit interval are unique.

15. How can membership in the Cantor set be characterized using ternary expansions?

A. A number in [0, 1] belongs to it exactly when it has an expansion using only 0 and 2

B. A number in [0, 1] belongs to it when its ternary expansion contains a finite number of digits

C. A number in [0, 1] belongs to it exactly when its expansion uses equal amounts of 0, 1, and 2

D. A number in [0, 1] belongs to it exactly when every expansion contains the digit 1

Answer: (A)

Explanation

The Cantor set consists precisely of numbers in [0, 1] that have a ternary expansion containing only 0s and 2s. The presence of the digit 1 in a representation corresponds to entering a removed middle third, so the other criteria do not characterize membership.

16. Why does a ternary expansion containing only 0s and 2s imply membership in the Cantor set?

A. The number belongs to one retained interval, which is enough for membership

B. The digit restriction places the number outside every removed interval at one stage

C. The number has a finite ternary expansion and therefore lies in the initial interval

D. Induction shows that the number belongs to every finite stage of the construction

Answer: (D)

Explanation

An inductive argument shows that the number remains in every stage interval set, so it belongs to their intersection, the Cantor set. Belonging to a single stage does not suffice, and the expansion need not be finite.