Step 1: Concept:
This question evaluates point-set topology concepts on the real line $\mathbb{R}$, including limit points, open sets, compactness (Heine-Borel theorem), and connectedness.
Step 2: Key Formula or Approach:
1. Derived set $\mathbb{Q}'$: Every real number is a limit point of $\mathbb{Q}$ because $\mathbb{Q}$ is dense in $\mathbb{R}$.
2. Heine-Borel Theorem: A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded.
3. Connectedness in $\mathbb{R}$: A subset of $\mathbb{R}$ is connected if and only if it is an interval.
Step 3: Step-by-step Explanation:
• Statement A:
Since $\mathbb{Q}$ is dense in $\mathbb{R}$, every real number $x \in \mathbb{R}$ is a limit point of $\mathbb{Q}$, so $\mathbb{Q}' = \mathbb{R} \neq \emptyset$. Statement A is false.
• Statement B:
The open interval $(0, 1)$ is an open set in $\mathbb{R}$ under the standard topology. Statement B is correct.
• Statement C:
Let $S = \{\frac{1}{n} \mid n \in \mathbb{N}\} \cup \{0\}$.
The only limit point of $S$ is $0$, which belongs to $S$, so $S$ is closed.
Furthermore, $S \subseteq [0, 1]$, so $S$ is bounded.
By the Heine-Borel theorem, $S$ is closed and bounded, hence compact in $\mathbb{R}$. Statement C is correct.
• Statement D:
The set $\{x \in \mathbb{R} : |x| > 1\} = (-\infty, -1) \cup (1, \infty)$.
This is the union of two disjoint non-empty open sets, so it is disconnected. Statement D is false.
• Statement E:
The closed interval $[0, 1]$ contains all its limit points (closed) and is bounded. By the Heine-Borel theorem, it is compact. Statement E is correct.
Step 4: Final Answer:
Statements B, C, and E are correct. Therefore, option (C) is the correct answer.