To determine which subset of $\mathbb{R}$ has a non-empty interior, we need to identify the set that contains an open interval, meaning there are points around each element in the set that are also part of the set.
Let's examine each subset:
(A) The set of all irrational numbers in $\mathbb{R}$:
This set does not have a non-empty interior because between any two irrational numbers, there are infinitely many irrational numbers, but there are also infinitely many rational numbers. So, there are no open intervals within this set.
(B) The set ${a \in \mathbb{R} : \sin(a) = 1}$:
The equation $\sin(a) = 1$ has solutions for $a = \frac{\pi}{2} + 2\pi k$, where $k$ is an integer. This set consists of all these solutions, and it forms an infinite set of isolated points. Since there are no open intervals around each point, the interior of this set is empty.
(C) The set $\{b \in \mathbb{R} : x^2 + bx + 1 = 0$ has distinct roots$\}$:
For the quadratic equation $x^2 + bx + 1 = 0$ to have distinct roots, its discriminant $(b^2 - 4ac)$ must be greater than zero. Here, $a=1$, $b$, and $c=1$. So the discriminant is $b^2 - 4(1)(1) = b^2 - 4$.
To have distinct roots:
$b^2 - 4 > 0$
$b^2 > 4$
$|b| > 2$ or $b < -2$ or $b > 2$
The set $\{b \in \mathbb{R} : x^2 + bx + 1 = 0$ has distinct roots$\}$ includes all real numbers such that $b < -2$ or $b > 2$. This set forms an open interval between $-\infty$ and $-2$ and another open interval between $2$ and $+\infty$. Therefore, this set has a non-empty interior.
(D) The set of all rational numbers in $\mathbb{R}$:
This set does not have a non-empty interior because between any two rational numbers, there are infinitely many rational numbers, but there are also infinitely many irrational numbers. So, there are no open intervals within this set.
In conclusion, the subset (C) $\{b \in \mathbb{R} : x^2 + bx + 1 = 0$ has distinct roots$\}$ has a non-empty interior.
This answer is AI-generated and verified by expert.