Question

Let \(S = \bigcup_{n=1}^{\infty} \left\{ (x, y) \in \mathbb{R}^2 : (x - n)^2 + y^2 = \dfrac{1}{n^2} \right\}\) with the usual topology. Which of the following statements is/are TRUE?

• A. \(S\) is not compact.
• B. \(S\) is compact.
• C. \(S\) is connected.
• D. \(S\) is not connected.

Hint:

Check boundedness first (the circles march off to \(x = \infty\), so \(S\) cannot be compact), then check which circles actually touch: only \(C_1\) and \(C_2\) intersect, every other circle stands alone, so \(S\) is not connected.

Answer 1

The Correct Option is A, D

Step 1: Understanding the Question.
\(S\) is the union of infinitely many circles \(C_n\), one for each \(n = 1, 2, 3, \dots\), where \(C_n\) is centred at \((n, 0)\) with radius \(\dfrac{1}{n}\). We need to check whether \(S\) is compact and whether it is connected, as a subspace of \(\mathbb{R}^2\) with the usual topology.

Step 2: Check boundedness (needed for compactness).
By the Heine-Borel theorem, a subset of \(\mathbb{R}^2\) is compact if and only if it is closed and bounded. The centre of \(C_n\) is at \(x=n\), so as \(n\) grows without bound, points of \(S\) appear arbitrarily far to the right along the \(x\)-axis (for instance, the point \((n + \tfrac{1}{n}, 0)\) lies on \(C_n\) for every \(n\)). So \(S\) contains points with arbitrarily large \(x\)-coordinate, which means \(S\) is unbounded.

Step 3: Conclude on compactness.
Since \(S\) is unbounded, it cannot be compact, regardless of whether it is closed. So option (A), "\(S\) is not compact," is TRUE, and option (B), "\(S\) is compact," is FALSE.

Step 4: Check whether adjacent circles meet.
Two circles \(C_m\) and \(C_n\) (say \(m < n\)) have centres a distance \(n - m\) apart and radii \(\tfrac1m\) and \(\tfrac1n\). Two circles intersect exactly when the distance between centres is between the difference and the sum of their radii, that is
\[ \left| \tfrac1m - \tfrac1n \right| \le n - m \le \tfrac1m + \tfrac1n \]
For \(m=1, n=2\): distance \(=1\), and \(\tfrac11+\tfrac12 = 1.5 \ge 1\), so \(C_1\) and \(C_2\) do intersect (in fact at two points, near \(x \approx 1.875\)). But for any \(m \ge 2\), \(\tfrac1m + \tfrac1{m+1} \le \tfrac12+\tfrac13 = \tfrac56 < 1 = \) the distance between adjacent centres, so \(C_m\) and \(C_{m+1}\) do NOT meet once \(m \ge 2\). Circles that are further apart (\(n - m \ge 2\)) are even less likely to meet, since the distance only grows while the radii only shrink, so no two non-adjacent circles ever intersect either.

Step 5: Count the connected pieces.
So \(C_1\) and \(C_2\) touch and fuse into one connected piece, while \(C_3, C_4, C_5, \dots\) each stand completely on their own, disjoint from every other circle in \(S\). This gives infinitely many separate, non-touching pieces, so \(S\) splits into infinitely many disjoint nonempty pieces, and a space with more than one such piece is by definition not connected.

Step 6: Conclude on connectedness.
So option (D), "\(S\) is not connected," is TRUE, and option (C), "\(S\) is connected," is FALSE.

Final Answer:
\(S\) is neither compact nor connected.
\[ \boxed{\text{(A) and (D)}} \]