GO Classes DPP | GATE CS | Discrete Mathematics| Functions

Answer 1

We are given that $f: [0,1] \to [0,1]$ is a one-to-one (injective) function.

1. Cardinality of the domain: The domain of the function is the interval $[0,1]$, which contains all real numbers between $0$ and $1$. This set is uncountably infinite and has the cardinality of the continuum, denoted as $c$.

2. Cardinality of the range: Because $f$ is a one-to-one function, each element in the domain maps to a unique element in the range. Therefore, the range of $f$, denoted as $f([0,1])$, must have the exact same cardinality as the domain. This means the range is also an uncountably infinite set.

Now let's evaluate each option:

• A. $f$ must be onto $[0,1]$: This is false. We can easily find a one-to-one function that does not cover the entire interval $[0,1]$. For example, $f(x) = \frac{x}{2}$ maps $[0,1]$ to $[0, 0.5]$, which is not onto $[0,1]$.

• B. The range of $f$ must contain a rational number: This is false. The set of all irrational numbers in $[0,1]$ is uncountably infinite (cardinality $c$). Because both the domain $[0,1]$ and the set of irrationals in $[0,1]$ share the same cardinality, it is mathematically possible to construct a one-to-one mapping (a bijection) from $[0,1]$ entirely into the irrational numbers of $[0,1]$. In such a case, the range would contain zero rational numbers.

• C. The range of $f$ must contain infinitely many irrational numbers: This is true. The set of all rational numbers is countably infinite. If the range of $f$ contained only finitely many (or even countably many) irrational numbers, the entire range would be restricted to the sum of a countable set (rationals) and a finite/countable set (irrationals). This would make the entire range countable. However, we established earlier that the range must be uncountably infinite. Therefore, to make up this uncountable size, the range must contain an uncountable number of irrational numbers, which implies it contains infinitely many irrational numbers.

• D. The range of $f$ must be an interval: This is false. We can construct a one-to-one function that maps $[0,1]$ to a disconnected set. For example, a bijection could exist between $[0,1]$ and the union of two separate intervals like $[0, 0.2] \cup [0.8, 1]$.