Math Problem Statement

prove that the set [0,1] is not countable

Solution

To prove that the set [0,1][0,1] is not countable, we use Cantor's diagonalization argument. Here's the proof:


Definitions:

  1. Countable Set: A set is countable if its elements can be listed in a sequence {a1,a2,a3,}\{a_1, a_2, a_3, \dots\} (i.e., there exists a bijection between the set and the natural numbers N\mathbb{N}).
  2. Uncountable Set: A set is uncountable if it is not countable.

Proof:

Assume, for the sake of contradiction, that the set [0,1][0,1] is countable. Then, all the numbers in [0,1][0,1] can be listed as a sequence: x1,x2,x3,x_1, x_2, x_3, \dots where each xix_i is a real number in the interval [0,1][0,1].

Step 1: Represent numbers in decimal expansion

Each real number xix_i in the interval [0,1][0,1] has a decimal expansion. For example: x1=0.a11a12a13x_1 = 0.a_{11}a_{12}a_{13}\dots x2=0.a21a22a23x_2 = 0.a_{21}a_{22}a_{23}\dots x3=0.a31a32a33x_3 = 0.a_{31}a_{32}a_{33}\dots where aija_{ij} is the jj-th decimal digit of xix_i.

Step 2: Construct a new number yy (diagonalization)

We now construct a new real number y[0,1]y \in [0,1] such that yy differs from every xix_i in at least one decimal place. Specifically, define the decimal digits of y=0.b1b2b3y = 0.b_1b_2b_3\ldots as follows:

\begin{cases} 1 & \text{if } a_{jj} \neq 1, \\ 2 & \text{if } a_{jj} = 1. \end{cases}$$ In other words, $$b_j$$ is chosen to differ from $$a_{jj}$$ (the $$j$$-th digit of $$x_j$$). #### Step 3: Verify $$y \notin \{x_1, x_2, x_3, \dots\}$$ By construction, $$y$$ differs from each $$x_i$$ in at least one decimal place: - $$y$$ differs from $$x_1$$ in the first decimal place ($$b_1 \neq a_{11}$$). - $$y$$ differs from $$x_2$$ in the second decimal place ($$b_2 \neq a_{22}$$). - $$y$$ differs from $$x_3$$ in the third decimal place ($$b_3 \neq a_{33}$$). - And so on. Thus, $$y$$ is not equal to any $$x_i$$ in the list. #### Step 4: Contradiction Since $$y \in [0,1]$$ but $$y$$ is not in the list $$\{x_1, x_2, x_3, \dots\}$$, the assumption that $$[0,1]$$ is countable is false. --- ### Conclusion: The interval $$[0,1]$$ is **uncountable**. --- Would you like further clarification or details? 😊 --- ### Related Questions: 1. What does "uncountable" mean, and how does it differ from "infinite"? 2. Can you give another example of an uncountable set? 3. How does Cantor's diagonalization apply to functions or other sets? 4. Can you explain why the decimal expansion construction avoids duplicates like $$0.999\ldots$$? 5. How does this proof relate to the cardinality of the real numbers? --- ### Tip: When dealing with infinite sets, remember that "infinite" doesn't always mean the same size—some infinities (like $$\mathbb{R}$$) are larger than others (like $$\mathbb{N}$$).

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Math Problem Analysis

Mathematical Concepts

Set Theory
Uncountability
Cantor's Diagonalization Argument
Real Numbers

Formulas

Cantor's diagonalization method

Theorems

Cantor's diagonalization theorem

Suitable Grade Level

Undergraduate