Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : The set of rationals $\mathbb{Q}$ is closed in the real line $\mathbb{R}$. Reason R : The closure of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$. In the light of the above statements, choose the correct answer from the options given below

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is NOT the correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Hint:

$\mathbb{Q}$ is neither open nor closed in $\mathbb{R}$! Its interior is empty ($\text{Int}(\mathbb{Q}) = \emptyset$) and its closure is all of $\mathbb{R}$ ($\bar{\mathbb{Q}} = \mathbb{R}$).

Answer 1

The Correct Option is D

Step 1: Concept
In topology, a subset $S \subseteq \mathbb{R}$ is closed if and only if $S$ contains all its limit points, which is equivalent to saying that $S$ equals its closure, i.e., $S = \bar{S}$.

Step 2: Key Formulas and Approach
The density property of $\mathbb{Q}$ in $\mathbb{R}$ states that between any two real numbers, there exists a rational number. Consequently, every real number is a limit point of $\mathbb{Q}$.

Step 3: Step-by-step Explanation

• Evaluating Reason R: Since $\mathbb{Q}$ is dense in $\mathbb{R}$, every real number $x \in \mathbb{R}$ is a limit point of $\mathbb{Q}$.
Thus, the closure of $\mathbb{Q}$ in $\mathbb{R}$ is:
\[ \bar{\mathbb{Q}} = \mathbb{R} \] Hence, Reason R is true.

• Evaluating Assertion A:
A set $S$ is closed if $S = \bar{S}$.
Here $\bar{\mathbb{Q}} = \mathbb{R} \neq \mathbb{Q}$ (since irrational numbers like $\sqrt{2} \in \mathbb{R}$ are limit points of $\mathbb{Q}$ but $\sqrt{2} \notin \mathbb{Q}$).
Since $\mathbb{Q}$ does not contain all its limit points, $\mathbb{Q}$ is not closed in $\mathbb{R}$.
Hence, Assertion A is false.

Step 4: Final Answer
Assertion A is false, but Reason R is true. Thus, Option (D) is correct.