Question

Which of the following set is open in the real line $\mathbb{R}$?

• A. $\{x \mid |x| \geq 2\}$
• B. $\{x \mid |x| > 2\}$
• C. $\{x \mid 0 \leq x < 2\}$
• D. $\{x \mid -\infty < x \leq 0\}$

Hint:

Strict inequalities ($>$ or $<$) generate open sets/intervals like $(a, b)$. Non-strict inequalities ($\geq$ or $\leq$) include boundary points and generate closed sets/intervals like $[a, b]$.

Answer 1

The Correct Option is B

Step 1: Concept
In the real topology on $\mathbb{R}$: - A subset $S \subseteq \mathbb{R}$ is called an open set if for every $x \in S$, there exists an $\epsilon > 0$ such that the open interval $(x - \epsilon, x + \epsilon) \subseteq S$.
- Equivalently, any open set in $\mathbb{R}$ can be expressed as a union of open intervals $(a, b)$.
- A set is closed if its complement is open, or if it contains all its limit points. Closed intervals $[a, b]$, $[a, \infty)$, $(-\infty, b]$ are closed sets.

Step 2: Key Formulas and Approach
Write each given set in interval notation and check whether it consists entirely of open intervals.

Step 3: Step-by-step Explanation

• Option (A): $S_1 = \{x \mid |x| \geq 2\ = (-\infty, -2] \cup [2, \infty)$}
This set contains its boundary points $-2$ and $2$. For $x = 2$, any neighborhood $(2-\epsilon, 2+\epsilon)$ contains points like $2-\frac{\epsilon}{2} \notin S_1$. Thus $S_1$ is closed, not open.

• Option (B): $S_2 = \{x \mid |x| > 2\ = (-\infty, -2) \cup (2, \infty)$}
$(-\infty, -2)$ and $(2, \infty)$ are both open intervals. Since the union of any family of open sets is open, $S_2$ is an open set in $\mathbb{R}$.

• Option (C): $S_3 = \{x \mid 0 \leq x < 2\ = [0, 2)$}
The point $0 \in S_3$, but no open interval around $0$ is contained in $[0, 2)$ (since points $0 - \epsilon < 0 \notin S_3$). So $S_3$ is neither open nor closed.

• Option (D): $S_4 = \{x \mid -\infty < x \leq 0\ = (-\infty, 0]$}
This is a closed ray containing $0$, so it is a closed set, not open.

Step 4: Final Answer
The set $\{x \mid |x| > 2\} = (-\infty, -2) \cup (2, \infty)$ is open in $\mathbb{R}$. Thus, Option (B) is correct.