Discussion
Let \(X = \{(x, y) \in \mathbf{R}^{2} : |x|\ |y| = 1\}\) and \(Y = \{(x, y) \in \mathbf{R}^{2} : |x| + |y| = 1\}\). Then :
Answer: C
The answer is Y is compact, but X is not compact.
By Heine-Borel, a subset of \(\mathbf{R}^{2}\) is compact exactly when it is closed and bounded. Both sets are closed, being level sets of continuous functions.
Y, defined by \(|x| + |y| = 1\), is the square with vertices \((\pm 1, 0)\) and \((0, \pm 1)\); it is bounded, so it is compact. X, defined by \(|x|\,|y| = 1\), is the four branches of the hyperbolas \(xy = \pm 1\); it contains \((n, 1/n)\) for every \(n\), so it is unbounded and not compact.
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