Discussion
Let the functions \(f: \mathbb{R}^{2} \to \mathbb{R}\) and \(g: \mathbb{R}^{2} \to \mathbb{R}\) be given by
\[ \begin{aligned} f(x_{1}, x_{2}) &= x_{1}^{2} + x_{2}^{2} - 2x_{1}x_{2}, \text{ and} \\ g(x_{1}, x_{2}) &= 2x_{1}^{2} + 2x_{2}^{2} - x_{1}x_{2}. \end{aligned} \]
Consider the following statements:
S1: For every compact subset K of \(\mathbb{R}\), \(f^{-1}(K)\) is compact.
S2: For every compact subset K of \(\mathbb{R}\), \(g^{-1}(K)\) is compact.
Then, which one of the following is correct?
Answer: B
S2 is TRUE and S1 is FALSE.
S1: \(f(x_{1}, x_{2}) = (x_{1} - x_{2})^{2}\). Then \(f^{-1}(\{0\})\) is the whole line \(x_{1} = x_{2}\), which is unbounded, so the preimage of the compact set {0} is not compact.
S2: g is a positive definite quadratic form, since its matrix \(\begin{bmatrix} 2 & -1/2 \\ -1/2 & 2 \end{bmatrix}\) has eigenvalues 1.5 and 2.5. Hence \(g(x) \ge 1.5\|x\|^{2}\), so the preimage of a bounded set is bounded; it is also closed because g is continuous. Closed and bounded in \(\mathbb{R}^{2}\) means compact, so S2 holds.
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