264 views
1 1 vote

WHICH OF THE FOLLOWING SETS IS/ARE UNCOUNTABLE?

  • I. THE SET OF ALL POLYNOMIALS WITH RATIONAL COEFFICIENTS.
     
  • II. THE SET OF ALL SUBSETS OF THE NATURAL NUMBERS $\mathbb{N}$.
     
  • III. THE SET OF ALL FUNCTIONS FROM $\{0,1\}$ TO THE SET OF NATURAL NUMBERS $\mathbb{N}$.
     
  • IV. THE SET OF ALL INFINITE-LENGTH BINARY STRINGS.
     
  1. I AND III
     
  2. II AND IV
     
  3. I, II, AND III
     
  4. IV ONLY

1 Answer

0 0 votes

I. THE SET OF ALL POLYNOMIALS WITH RATIONAL COEFFICIENTS.

  • COUNTABLE. A polynomial is defined by a finite sequence of coefficients (e.g., $a_n x^n+ \left.\cdots+a_1 x+a_0\right)$. Each coefficient is a rational number.
     
  • The set of rational numbers $\mathbb{Q}$ is countable.
     
  • The set of all finite-length tuples of rational numbers $\left(\mathbb{Q}^k\right.$ for any finite $\left.k\right)$ is countable.
     
  • The set of all polynomials is a countable union of these countable sets (one set for each degree $k$ ). A countable union of countable sets is countable.

II. THE SET OF ALL SUBSETS OF THE NATURAL NUMBERS $\mathbb{N}$.
 
  • UNCOUNTABLE. This is the definition of the power set of $\mathbb{N}$, denoted $P(\mathbb{N})$.
     
  • The set $\mathbb{N}$ is countably infinite (its cardinality is $\aleph_0$ ).
     
  • Cantor's Theorem states that the power set of any set $A$ has a strictly greater cardinality than $A$.
     
  • Therefore, the cardinality of $P(\mathbb{N})$ is $2^{K_0}$, which is uncountable.

III. THE SET OF ALL FUNCTIONS FROM $\{0,1\}$ TO $\mathbb{N}$.
 
  • COUNTABLE. A function $f:\{0,1\} \rightarrow \mathbb{N}$ is just a mapping of two values: $f(0)$ and $f(1)$.
     
  • $f(0)$ must be an element of $\mathbb{N}$.
     
  • $f(1)$ must be an element of $\mathbb{N}$.
     
  • The set of all such functions is therefore in a one-to-one correspondence with the set of all ordered pairs $\left(n_1, n_2\right)$ where $n_1, n_2 \in \mathbb{N}$. This is the set $\mathbb{N} \times \mathbb{N}$.
     
  • The Cartesian product of two countable sets is countable.
     
IV. THE SET OF ALL INFINITE-LENGTH BINARY STRINGS.
 
  • UNCOUNTABLE. This set is famously proven to be uncountable using Cantor's diagonalization argument.
     
  • This set can also be shown to have the same cardinality as the set of real numbers between 0 and 1 (by treating each string as a binary expansion).
Answer:
Position:
Show:

Related questions

1 1 vote
1 1 answer
342
342 views
GO Classes asked Nov 11, 2025
342 views
WHICH OF THE FOLLOWING SETS IS/ARE UNCOUNTABLE?I. THE SET OF ALL FINITE SUBSETS OF $\mathbb{Q}$ (THE SET OF RATIONAL NUMBERS). II. THE SET OF ALL SUBSETS OF $\mathbb{R}$ ...
2 2 votes
1 1 answer
289
289 views
GO Classes asked Nov 11, 2025
289 views
WHICH OF THE FOLLOWING SETS IS/ARE COUNTABLE?I. THE SET OF ALL REGULAR LANGUAGES OVER THE ALPHABET $\{0,1\}$. II. THE SET OF ALL TURING MACHINES THAT HALT ON THE EMPTY ST...
3 3 votes
2 2 answers
346
346 views
GO Classes asked Nov 11, 2025
346 views
CONSIDER THE FOLLOWING CONTEXT-FREE GRAMMAR $G$, WHERE $S, X, Y$ ARE THE VARIABLES, $a, b$ ARE THE TERMINAL SYMBOLS, AND $S$ IS THE START VARIABLE:$S \rightarrow a X \mid...
3 3 votes
2 2 answers
298
298 views
GO Classes asked Nov 11, 2025
298 views
Consider the following context-free grammar $G^{\prime}$, where $S, X$, and $Y$ are the variables (nonterminals), $0$ and $1$ are the terminal symbols, $S$ is the start v...