Supremum, Accumulation Points, and a Misused Existential Quantifier
In a real-analysis review session, a student presents the following argument. Let A={1}∪{1/n:n≥2}. Since sup A=1, for each n∈N choose a_n∈A with 1−1/n<a_n≤1. Then a_n→1. The student concludes that 1 is an accumulation point of A. The instructor objects that the chosen sequence might be constant. The student replies that A has infinitely many points below 1, so the choice can always be made with a_n≠1. A second student claims that any infinite bounded-above set must have its supremum as an accumulation point, citing the same sequence construction. You are asked to judge both claims, identify exactly where the quantifier 'for each ε>0 there exists x∈A' is being misused, and state a correct sufficient condition under which sup A is an accumulation point.
1. Is 1 an accumulation point of A? Diagnose the quantifier error in the sequence argument.
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No. For every n≥2, the interval (1−1/n,1] contains only the point 1 from A: the next largest element below 1 is 1/2, and 1−1/n≥1/2. Thus any choice a_n∈A with 1−1/n<a_n≤1 must have a_n=1 for all n≥2. The student's sequence converges to 1, but it is eventually constant, so it gives no evidence of accumulation.
The logical error is treating 'for each n there exists a_n' as if the chosen a_n must be distinct from 1. The existence statement only guarantees a_n∈A; it may return the limit point itself. Convergence a_n→1 with a_n∈A proves 1∈closure(A), not that 1 is an accumulation point. In R, closure(A)=A∪A′, so an isolated point of A lies in the closure without being in A′. The correct definition requires every neighborhood of 1 to contain a point of A different from 1, equivalently a sequence in A∖{1} converging to 1. For A, the neighborhood (0.9,1.1) meets A only at 1, so 1 is isolated, not an accumulation point.
2. Does every infinite bounded-above set have its supremum as an accumulation point? Give a counterexample and a correct sufficient condition.
Read the worked answer
No. The same set A={1}∪{1/n:n≥2} is infinite and bounded above, and sup A=1, but 1 is isolated. Infinity alone does not force points of A to approach the supremum from below; A can have a gap below its largest element.
A correct sufficient condition is sup A∉A. If a=sup A and a∉A, then for each n choose a_n∈A with a−1/n<a_n≤a. Since a_n≠a, we get a−1/n<a_n<a, so a_n→a through points of A distinct from a. Hence a is an accumulation point. Equivalently, A has no maximum: if sup A belonged to A, it would be the maximum, so absence of a maximum forces sup A∉A. Another direct sufficient condition is that A∩(a−ε,a) is nonempty for every ε>0, i.e. A approaches a strictly from below. The counterexample shows why the hypothesis must be about the local structure near the supremum, not merely about cardinality.