Discussion
Consider \(\mathbb{R}^{2}\) with the usual Euclidean metric. Let
\[ \begin{aligned} X = &\{(x, x\sin\tfrac{1}{x}) \in \mathbb{R}^{2} : x \in (0,1]\} \\ &\cup \{(0,y) \in \mathbb{R}^{2} : -\infty \lt y \lt \infty\} \end{aligned} \]
and
\[ \begin{aligned} Y = &\{(x, \sin\tfrac{1}{x}) \in \mathbb{R}^{2} : x \in (0,1]\} \\ &\cup \{(0,y) \in \mathbb{R}^{2} : -\infty \lt y \lt \infty\}. \end{aligned} \]
Consider the following statements:
P: X is a connected subset of \(\mathbb{R}^{2}\).
Q: Y is a connected subset of \(\mathbb{R}^{2}\).
Then
Answer: A
Both P and Q are TRUE.
Use the fact that if A is connected and \(A \subseteq B \subseteq \overline{A}\), then B is connected, and that two connected sets with a common point have a connected union.
X: the graph of \(x\sin(1/x)\) on (0, 1] is connected (a continuous image of an interval) and tends to (0, 0) as \(x \to 0\). Adding the point (0, 0) keeps it connected, and the y-axis, which is connected and contains (0, 0), joins it. So X is connected.
Y: the graph of \(\sin(1/x)\) has closure containing the segment \(\{0\} \times [-1, 1]\), so the graph plus that segment is connected (the topologist's sine curve). The whole y-axis is connected and contains the segment, so Y is connected too. Y is not path-connected, but connectedness is all that is asked.
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