Discussion
Let \(\{f : \mathbf{R} \to \mathbf{R} : f\) is continuous such that \(f(t) = 0\) for all \(t\) in a subfield of \(\mathbf{R}\} = X\). Then X contains :
Answer: D
X contains only one element.
Every subfield of \(\mathbf{R}\) contains 1, hence all integers and all quotients of integers: it contains \(\mathbf{Q}\). So any \(f\) in X vanishes on \(\mathbf{Q}\). Since \(\mathbf{Q}\) is dense in \(\mathbf{R}\) and \(f\) is continuous, \(f(x) = \lim f(q_{n}) = 0\) for every real \(x\), taking rationals \(q_{n} \to x\).
So the only such function is the zero function, and X has exactly one element. The options with more than one element would need a non-zero continuous function vanishing on a dense set, which is impossible.
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