Discussion
Let X, Y be metric spaces. Suppose \(f, g : X \to Y\) are continuous functions and \(D = \{x \mid f(x) = g(x)\}\). Then \(f \equiv g\) if D is :
Answer: A
The functions agree everywhere if D is dense in X.
The set \(D = \{x : f(x) = g(x)\}\) is always closed: if \(x_{n} \in D\) and \(x_{n} \to x\), continuity gives \(f(x) = \lim f(x_{n}) = \lim g(x_{n}) = g(x)\). A closed set that is dense is the whole space, so \(D = X\) and \(f \equiv g\).
The other properties are not enough. On \(X = \mathbf{R}\), \(f(x) = x\) and \(g(x) = 0\) agree only on \(D = \{0\}\), which is compact and closed; and \(f(x) = \sin x\), \(g(x) = 0\) agree on the multiples of \(\pi\), an unbounded set, while still being different functions.
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