Discussion
If
\(A = \{(x, y) / -1 \le x \lt 1, -1 \le |y| \le 1\}\)
and
\(B = \{(x, y) / (x - 1)^{2} + y^{2} \le 1/2\}\)
then :
Answer: B
The compact set is \(B \setminus A\).
A is the strip \(-1 \le x \lt 1\), \(|y| \le 1\) (the condition \(-1 \le |y|\) is automatic), which is not closed because its right edge \(x = 1\) is missing. B is the closed disk of radius \(1/\sqrt{2} \approx 0.707\) centred at (1, 0).
Every point of B has \(|y| \le 0.707\) and \(x \ge 0.29 \gt -1\), so a point of B fails to lie in A only when \(x \ge 1\). Hence \(B \setminus A = B \cap \{x \ge 1\}\), a closed half-disk: closed and bounded, so compact.
The others are not closed. The points \((1 - 1/n, 0)\) lie in \(A \cap B\) but their limit (1, 0) is not in A, so \(A \cap B\) is not closed. The points \((1 - 1/n, 0.9)\) lie in A but outside B (distance about 0.9 from the centre), and their limit (1, 0.9) lies in neither set, so \(A \setminus B\) and \(A \cup B\) are not closed either.
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