Discussion
Let X and Y be metric spaces and \(f, g : X \to Y\) be continuous. Then \(\{x \mid f(x) = g(x)\}\) is :
Answer: B
The set where two continuous maps agree is closed.
Consider the map \(h : X \to Y \times Y\), \(h(x) = (f(x), g(x))\), which is continuous because both components are. The set \(\{x \mid f(x) = g(x)\}\) is \(h^{-1}(\Delta)\), where \(\Delta = \{(y, y)\}\) is the diagonal of \(Y \times Y\). In a metric space, indeed in any Hausdorff space, the diagonal is closed, so its preimage under a continuous map is closed.
A direct argument works too: if \(x_{n} \to x\) with \(f(x_{n}) = g(x_{n})\), continuity gives \(f(x) = \lim f(x_{n}) = \lim g(x_{n}) = g(x)\).
The set need not be open (take \(f(x) = x\), \(g(x) = 0\) on R: the set is {0}), connected (\(f(x) = x^{2}\), \(g(x) = 1\) gives {-1, 1}) or compact (\(f = g\) gives all of R).
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