Question

Let \( (\mathbb{Q}, d) \) be the metric space with \[ d(x, y) = |x - y|. \] Let \( E = \{ p \in \mathbb{Q} : 2<p^2<3 \}. \) Then, the set \( E \) is

• A. closed but not compact
• B. not closed but compact
• C. compact
• D. neither closed nor compact

Hint:

In metric spaces, a set may be closed in one space but not in another. In this case, \( E \) is closed in the rational numbers but not in the real numbers.

Answer 1

The Correct Option is A

We are given the set \( E = \{ p \in \mathbb{Q} : 2<p^2<3 \} \) in the metric space \( (\mathbb{Q}, d) \), where \( d(x, y) = |x - y| \). We need to determine whether the set \( E \) is closed, compact, or neither. Step 1: Understanding the set
The set \( E \) consists of rational numbers \( p \) such that \( 2<p^2<3 \). Solving for \( p \), we get: \[ \sqrt{2}<|p|<\sqrt{3}. \] Thus, the set \( E \) contains rational numbers between \( \sqrt{2} \) and \( \sqrt{3} \). Step 2: Checking if \( E \) is closed
A set is closed if it contains all of its limit points. Consider the real numbers between \( \sqrt{2} \) and \( \sqrt{3} \). The irrational numbers in this interval are limit points of the set \( E \), but \( E \) contains only rational numbers. Therefore, \( E \) does not contain all of its limit points and is not closed in the real numbers. However, in the context of the metric space \( (\mathbb{Q}, d) \), it is closed because there are no irrational numbers in \( E \). Step 3: Checking if \( E \) is compact
A set in a metric space is compact if it is closed and bounded. While \( E \) is bounded, it is not compact in \( \mathbb{Q} \) because it does not contain all its limit points. Thus, it is not compact in the rational numbers. Step 4: Conclusion
The set \( E \) is closed in \( \mathbb{Q} \), but not compact. Therefore, the correct answer is (A) closed but not compact.