Discussion
Let \(f : [0, 1] \to S^{1}\) be defined as \(f(x) = e^{2\pi i x}\).
If Q denotes the set of rational numbers and \(S^{1}\) is unit circle, then :
Answer: C
\(f([0, 1] \cap Q)\) is dense in S1.
The map \(f(x) = e^{2\pi i x}\) is continuous from [0, 1] onto the circle. The rationals in [0, 1] are dense in [0, 1], and a continuous surjection carries a dense set onto a dense subset of the image, so the points \(e^{2\pi i q}\), the roots of unity, are dense in \(S^{1}\).
The image is countable, so it cannot be all of \(S^{1}\) or any closed set containing an arc; being dense but not everything, it is not closed. It is not open, because every arc of the circle contains points \(e^{2\pi i x}\) with \(x\) irrational, which are not in the image. And a countable subset of the circle with more than one point is not connected: removing two points with irrational angles, one on each arc between two of its points, separates it into two relatively open pieces.
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