Discussion
Consider \(\mathbb{R}^{2}\) with the usual metric. Let \(A = \{(x,y) \in \mathbb{R}^{2} : x^{2} + y^{2} \leq 1\}\) and \(B = \{(x,y) \in \mathbb{R}^{2} : (x-2)^{2} + y^{2} \leq 1\}\). Let \(M = A \cup B\) and \(N = \text{interior}(A) \cup \text{interior}(B)\). Then, which of the following statements is TRUE?
Answer: C
The answer is M is connected and N is not connected.
A is the closed unit disc at the origin and B the closed unit disc centred at (2, 0). The centres are 2 apart, equal to the sum of the radii, so the discs touch at exactly one point, (1, 0).
M = A ∪ B is a union of two connected sets sharing the point (1, 0), so it is connected. N is the union of the two open discs, which do not contain (1, 0) and are disjoint open sets; they separate N, so N is not connected.
That rules out 'M and N are connected' and 'N is connected', and 'neither is connected' fails because M is connected.
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