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?
Show 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.
Step 1: Understanding the Concept.
$S$ is built from infinitely many circles $C_n$ centred at $(n,0)$ with radius $1/n$, for $n = 1, 2, 3, \dots$. To test compactness we can use sequences, and to test connectedness we can look for a clopen (both open and closed) subset of $S$ other than $S$ itself and the empty set.
Step 2: A sequence with no convergent subsequence in S.
Pick $p_n = (n+\tfrac1n, 0)$, the rightmost point of $C_n$, which is a genuine point of $S$ for every $n$. As $n \to \infty$, $p_n \to \infty$ along the $x$-axis, so $(p_n)$ has no subsequence converging to any point of $\mathbb{R}^2$, let alone a point of $S$. A compact metric space is sequentially compact, meaning every sequence must have a subsequence converging to a point of the space. Since $(p_n)$ fails this, $S$ is not compact. This makes option (A) true and option (B) false.
Step 3: Look for a natural split of S.
Consider $U = C_1 \cup C_2$ (the two circles that actually touch each other) and $V = C_3 \cup C_4 \cup C_5 \cup \dots$ (all the rest). Since $C_1$ and $C_2$ meet only each other, and every $C_n$ for $n \ge 3$ stays a fixed positive distance away from $U$ and from every other $C_m$ (because non-adjacent, and even adjacent for $m\ge 2$, circles never touch, as shown by comparing the centre-distance $n-m$ to the shrinking sum of radii $1/m+1/n$), $U$ and $V$ are disjoint, their union is all of $S$, and each is a union of circles that is both closed and open in $S$ (since each stays a positive distance away from the rest of $S$).
Step 4: Conclude on connectedness.
Since $U$ and $V$ are both nonempty, disjoint, and clopen in $S$ with $U \cup V = S$, this is a genuine separation of $S$ into two pieces. A space that admits such a separation is, by definition, not connected. So option (D) is true and option (C) is false. In fact the same style of separation can be repeated inside $V$, splitting off $C_3$, then $C_4$, and so on, showing $S$ actually has infinitely many connected pieces, not just two.
Step 5: Final Answer.
$S$ is unbounded, so it is not compact, and it splits into infinitely many separate circles (with only $C_1, C_2$ fused together), so it is not connected.
\[ \boxed{\text{(A) and (D)}} \]
