Compactness of a Convergent Sequence and Its Terms
In R with the usual metric, let S={0}∪{1/n:n∈N} and T={1/n:n∈N}. A student writes: 'Both sets are bounded. T is closed in itself because its complement in T is empty, and S is closed because its only limit point is 0, which belongs to S. By the Heine-Borel theorem, closed and bounded subsets of R are compact, so both S and T are compact.' The instructor says the conclusion is half right and that the phrase 'closed in itself' is doing illegitimate work. A second student says that if T is regarded as a metric space with the subspace metric, then it is automatically closed and bounded, so Heine-Borel applies. You are asked to adjudicate the compactness of S and T, explain the relative-topology error, and give a concrete open cover showing failure of compactness for the noncompact set.
1. Determine which of S and T is compact and justify both conclusions.
Read the worked answer
S is compact. It is bounded, and it is closed in R: its complement is (-∞,0)∪⋃_{n≥1}(1/(n+1),1/n)∪(1,∞), a union of open intervals. By Heine-Borel, closed and bounded in R implies compact. One can also prove this directly: in any open cover, the set containing 0 contains all but finitely many of the points 1/n, and the remaining finitely many points can be covered by finitely many additional cover elements.
T is not compact. It is not closed in R, because 0 is a limit point of T but 0∉T. Since compact subsets of R are closed, T cannot be compact. A direct open-cover proof is also available: let U_n=(1/(n+1),2) for n≥1. The family {U_n} covers T, since 1/k∈U_k. But any finite subfamily has a largest index M, and its union is U_M, which misses 1/(M+2)∈T. Hence no finite subcover exists.
2. Explain the relative-topology error and why the second student's subspace-metric argument fails.
Read the worked answer
The statement 'T is closed in itself' is true but vacuous: every metric space is closed in itself, because its complement in itself is empty. Heine-Borel is not a statement about a set being closed in its own subspace topology. It says that a subset of R^n is compact if and only if it is closed in the ambient R^n and bounded. The relevant closedness is closedness in R, not in T.
The second student makes the same mistake. Giving T the subspace metric does not make Heine-Borel apply with 'closed in itself' as the hypothesis. Compactness is intrinsic, but the Heine-Borel criterion uses the ambient Euclidean space. If T were compact as a subspace, its inclusion image in R would be compact and hence closed in R, contradicting the fact that 0 is a limit point of T not in T. Thus T is not compact even though it is bounded and closed as a subspace of itself.