Discussion
Let
\(A = \{(x, y) \in \mathbf{R}^{2}/x = 0, -1 \le y \le 1\}\)
\[ \begin{aligned} B = \{&(x, y) \in \mathbf{R}^{2}/ \\ &0 \lt x \le \tfrac{1}{\pi}, y = \sin\tfrac{1}{x}\} \end{aligned} \]
and \(X = A \cup B\). Then :
Answer: B
X is connected but not path connected.
X is the topologist's sine curve: the graph B of \(\sin(1/x)\) on \((0, 1/\pi]\) together with the segment A on the \(y\)-axis from \(-1\) to 1. B is connected (a continuous image of an interval), and every point of A is a limit of points of B, because \(\sin(1/x)\) takes every value in \([-1, 1]\) on every interval \((0, \epsilon)\). So \(X = A \cup B\) lies between B and its closure and is connected.
No path joins a point of A to a point of B: a path would have to reach \(x = 0\) while its \(y\)-coordinate oscillates between \(-1\) and 1 infinitely often, which contradicts continuity. So X is not path connected, ruling out A, C and D.
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