Question

Which one of the following options is incorrect for the set \(S = \{x \in \mathbb{Q} : x^2 < 5\}\)?

• A. \(S\) is an open set in \(\mathbb{Q}\).
• B. The closure of \(S\) in \(\mathbb{Q}\) is itself.
• C. The limit points of \(S\) are \(-\sqrt{5}\) and \(\sqrt{5}\).
• D. \(S\) has no least upper bound in \(\mathbb{Q}\).

Hint:

Check whether S is clopen in Q; irrational numbers cannot be limit points of a subset of Q.

Answer 1

The Correct Option is C

Step 1: Identify the set. \(S = \{x \in \mathbb{Q} : x^2 < 5\}\) can be written as \(S = \mathbb{Q} \cap (-\sqrt5, \sqrt5)\), since \(x^2 < 5 \iff -\sqrt5 < x < \sqrt5\).

Step 2: Check openness in \(\mathbb{Q}\). The interval \((-\sqrt5,\sqrt5)\) is open in \(\mathbb{R}\), so its intersection with \(\mathbb{Q}\) is open in the subspace topology on \(\mathbb{Q}\). Hence option (A) is true.

Step 3: Check closedness in \(\mathbb{Q}\). The complement of \(S\) in \(\mathbb{Q}\) is \(\mathbb{Q} \cap \big((-\infty,-\sqrt5)\cup(\sqrt5,\infty)\big)\), which is also open in \(\mathbb{Q}\). So \(S\) is closed in \(\mathbb{Q}\) as well, meaning its closure in \(\mathbb{Q}\) is itself. Hence option (B) is true.

Step 4: Check the least upper bound. In \(\mathbb{R}\), \(\sup S = \sqrt5\), an irrational number. Since \(\mathbb{Q}\) is not complete, \(S\) has no supremum inside \(\mathbb{Q}\). Hence option (D) is true.

Step 5: Check the limit points. Since \(S\) is clopen in \(\mathbb{Q}\), every limit point of \(S\) computed within \(\mathbb{Q}\) already lies in \(S\). Moreover \(-\sqrt5\) and \(\sqrt5\) are irrational, so they are not even elements of \(\mathbb{Q}\) and cannot be limit points of a subset of \(\mathbb{Q}\). So option (C) is false, making it the incorrect statement.

\[\boxed{\text{Option (C) is incorrect}}\]