Calculation
Analyzing the Set \(E\)
The set \(E\) consists of all rational numbers \(q\) such that \(-\sqrt{2} < q < \sqrt{2}\). Note that \(\pm \sqrt{2}\) are irrational numbers, so they are not in \(\mathbb{Q}\).
Option 4: Is \(E\) Open in \(\mathbb{Q}\)?
By the definition of relative topology, a set \(E\) is open in \(\mathbb{Q}\) if \(E = V \cap \mathbb{Q}\) for some open set \(V\) in \(\mathbb{R}\).
Since \((-\sqrt{2}, \sqrt{2})\) is an open interval in \(\mathbb{R}\), \(E\) is open in \(\mathbb{Q}\).
Option 3: Is \(E\) Closed in \(\mathbb{Q}\)?
A set is closed in \(\mathbb{Q}\) if its limit points that are rational are contained in the set.
The limit points of \(E\) in \(\mathbb{R}\) are the interval \([-\sqrt{2}, \sqrt{2}]\).
The limit points of \(E\) in \(\mathbb{Q}\) are \([-\sqrt{2}, \sqrt{2}] \cap \mathbb{Q}\).
Since \(\pm \sqrt{2} \notin \mathbb{Q}\), the set \((-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q}\) is exactly the same as \([-\sqrt{2}, \sqrt{2}] \cap \mathbb{Q}\).
Therefore, \(E\) contains all its limit points in \(\mathbb{Q}\), so \(E\) is closed in \(\mathbb{Q}\).
Option 2: Is \(E\) Compact in \(\mathbb{R}\)?
In \(\mathbb{R}\), by the Heine-Borel theorem, a set is compact if and only if it is closed and bounded.
\(E\) is bounded, but it is not closed in \(\mathbb{R}\) because it does not contain its limit points (like \(0.5, 1, \dots\) or even irrational ones in the full real context). More specifically, any sequence of rationals converging to \(\sqrt{2}\) has its limit outside \(E\).
Thus, \(E\) is not compact in \(\mathbb{R}\). This statement is correct.
Option 1: Is \(E\) Compact in \(\mathbb{Q}\)?
A set is compact in a metric space if every open cover has a finite subcover.
Consider an increasing sequence of rational numbers \(r_n\) converging to \(\sqrt{2}\) from below. The open cover \(U_n = (-\sqrt{2}, r_n) \cap \mathbb{Q}\) covers \(E\) but has no finite subcover because any finite union would only go up to the largest \(r_k < \sqrt{2}\), leaving out rationals between \(r_k\) and \(\sqrt{2}\).
Therefore, \(E\) is NOT compact in \(\mathbb{Q}\).
**Correct Option (The INCORRECT statement):
(1) E is compact in \(\mathbb{Q}\).**