Question

Let \( S \) be an infinite subset of \( \mathbb{R} \) such that \( S \setminus \{a\} \) is compact for some \( a \in S \). Then which one of the following is TRUE?

• A. Every sequence in \( S \) has a subsequence converging to an element in \( S \)
• B. \( S \) contains no limit points
• C. \( S \) is a union of open intervals
• D. \( S \) is a connected set

Hint:

A compact subset of \( \mathbb{R} \) is always closed and bounded, and removing a single point from a compact set does not disconnect it.

Answer 1

The Correct Option is A

Step 1: Understanding the condition.
We are given that \( S \) is an infinite subset of \( \mathbb{R} \) such that \( S \setminus \{a\} \) is compact for some point \( a \in S \). This means \( S \setminus \{a\} \) is closed and bounded in \( \mathbb{R} \). Compact sets in \( \mathbb{R} \) have the property that every sequence within them has a subsequence converging to a limit within the set.
Step 2: Analyzing the options.
(A) \( S \) is a connected set: This is not necessarily true. While compact sets are connected, the removal of a point from \( S \) could potentially disconnect it. Thus, \( S \) may not be connected.
(B) \( S \) contains no limit points: This is incorrect. Since \( S \setminus \{a\} \) is compact, it must contain limit points.
(C) \( S \) is a union of open intervals: This is incorrect. Compact sets in \( \mathbb{R} \) cannot be expressed as a union of open intervals.
(D) Every sequence in \( S \) has a subsequence converging to an element in \( S \): This is true. Since \( S \setminus \{a\} \) is compact, any sequence in \( S \) will have a subsequence converging to a limit point in \( S \).

Step 3: Conclusion.
Thus, the correct answer is (A).