Proof. Case 1: the sequence takes only finitely many distinct values. Then at least one value v is taken for infinitely many indices, since finitely many values covering infinitely many indices leaves some value used infinitely often. Listing those indices in increasing order gives a subsequence all of whose terms equal v; a constant sequence converges to v.
Case 2: the sequence takes infinitely many distinct values, so the set S₁ is infinite; it is also bounded. Theorem 10.6 therefore provides a cluster point x* of S₁. Observe first that discarding finitely many points cannot destroy this: if F is finite and V is a neighborhood of x*, then V already holds infinitely many points of S₁, so it still holds one lying outside F and differing from x*. Hence x* remains a cluster point of every set S_r = {xₘ : m > n_{r−1}}.
Now construct the indices. Pick n₁ with ‖x_{n₁} − x*‖ < 1. Having chosen n_{r−1}, apply the observation to S_r = {xₘ : m > n_{r−1}} and to the ball V_r = {y : ‖y − x*‖ < 1/r}: some member of S_r lies in V_r, that is, ‖x_{n_r} − x*‖ < 1/r for some n_r > n_{r−1}. The indices increase strictly, so X′ = (x_{n_r}) is a subsequence, and Theorem 14.9 applied with a_r = 1/r and C = 1 gives lim X′ = x*.
收網的每一輪同時辦兩件事：把半徑減半再減三分之一地縮小（第 r 輪用半徑 1/r），並且只從編號比上一輪更大的項裡挑。第二件事保證了編號嚴格遞增，挑出來的東西才配稱子數列；第一件事保證誤差被 1/r 壓住。最後一步不必再做 ε-K——§14-4 的 14.9（誤差被一條已知趨零的數列壓住即收斂）直接收尾，那條趨零的數列就是 (1/r)，常數 C 取 1。具體：第 100 輪挑到的那一項，離 x* 不到 0.01。
兩種情形都交出了收斂的子數列。∎
這張圖在說 16.4 為什麼非分兩案不可：10.6 只對無限集有話講，而數列可以重複取值——((−1)ⁿ) 有無限多項卻只有兩個值，值集合連 cluster point 都沒有。左路用鴿籠原理直接造常數子數列，右路才輪得到 10.6。
Proof. The construction in Case 2 above used only that x* is a cluster point of the value set: from that fact alone it produced indices n₁ < n₂ < ⋯ with ‖x_{n_r} − x*‖ < 1/r, whence (x_{n_r}) converges to x*. Since a cluster point of a set requires every neighborhood to meet it in infinitely many points, that hypothesis is available here by assumption rather than via Theorem 10.6.