Different spokes must route through the hub

Consider $\mathbb{R}^{2}$ with the Euclidean metric d. Let $O = (0, 0), P_{0} = (0, 1) \in \mathbb{R}^{2}$ , and for any integer $n \ge 1, P_{n} = (1/n, 1) \in \mathbb{R}^{2}$ . Let X = ⋃ $(n\ge0) L_{n}$ , where $L_{n}$ is the line segment joining $P_{n}$ and O. Define $d_X : X \times X \to \mathbb{R}$ as follows: d_X(a, b) = d(a, b) when $a, b \in L_{n}$ for some n, and d_X(a, b) = d(a, O) + d(b, O) otherwise. Let $\tau$ be the smallest topology such that the sets $B_(a,\varepsilon) =$ { $b \in X : d_X(a, b) < \varepsilon$ } are open in $\tau$ for all a ∈ X and $\varepsilon > 0$ . Which of the following statements are true?

A.d_X is a metric on X which induces the topology $\tau$ .✓ Correct
B.The topological space $(X, \tau)$ is connected.✓ Correct
C. $(P_{n})_{n}\ge_{1}$ converges to $P_{0}$ in the topological space $(X, \tau)$ .
D.The topological space $(X, \tau)$ is compact.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests standard counterexample.

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50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

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Last revised 12 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

Different spokes must route through the hub

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