Tails, Subsequences, and Eventual Properties

Tails, Subsequences, and What Happens Eventually

Convergence is determined by the terms of a sequence far enough along, not by any fixed finite collection of early terms. Tails make this idea precise. Subsequences let us inspect selected terms, provided their original order is preserved. Together, these concepts explain two useful facts: deleting finitely many terms does not change convergence, and every subsequence of a convergent sequence converges to the same limit. The reverse inference from one convergent subsequence to the full sequence, however, can fail.

Let \(X\) be a topological space, and let \((x_n)\) be a sequence in \(X\). Recall that \(x_n\to x\) means that for every open neighborhood \(U\) of \(x\), there is an \(N\in\mathbb{N}\) such that \(x_n\in U\) whenever \(n\ge N\).

The indices of a subsequence must increase strictly: a subsequence retains the original order and cannot repeat a term by selecting the same index twice. Every tail is itself a subsequence, obtained from the indices \(N,N+1,\ldots\). More generally, a subsequence need not contain every term after some index; it may skip terms indefinitely.

“Eventually” does not mean “for most indices” or “after a large proportion of the sequence.” It means there is one index after which the property holds for every term. For example, \(x_n\in A\) eventually means that there is an \(N\) such that \(x_n\in A\) for all \(n\ge N\). The initial terms \(x_1,\ldots,x_{N-1}\) may or may not belong to \(A\).

Finite Changes and Convergence

Proof. Suppose \(x_n\to x\), and fix a tail starting at \(N_0\). Let \(U\) be any open neighborhood of \(x\). By convergence, there is an \(N\) such that \(x_n\in U\) whenever \(n\ge N\). For the tail, all its terms from the position \(\max(N,N_0)\) onward have original indices at least \(N\), so they belong to \(U\). Thus the tail converges to \(x\). Conversely, suppose the tail starting at \(N_0\) converges to \(x\). Given an open neighborhood \(U\) of \(x\), there is a position \(M\) in that tail such that its terms from position \(M\) onward lie in \(U\). Those are precisely original terms with indices at least \(N_0+M-1\), so the original sequence also converges to \(x\). A finite change leaves a tail unchanged, and the two directions just proved show that the original and changed sequences have the same convergence behavior. \(\square\)

The theorem concerns convergence to a specified point \(x\). It does not require \(X\) to be Hausdorff, so it remains valid even when a sequence can have more than one limit. The proof uses only the neighborhood definition of convergence.

The same reasoning applies to any property that is stated term by term. If a property holds eventually for a sequence, deleting some initial terms preserves that eventual property. Conversely, adding or changing finitely many initial terms cannot turn a property that fails at arbitrarily large indices into one that holds eventually.

Subsequences of Convergent Sequences

Proof. Let \(U\) be any open neighborhood of \(x\). Since \(x_n\to x\), there is an \(N\in\mathbb{N}\) such that \(x_n\in U\) whenever \(n\ge N\). A strictly increasing sequence of positive integers satisfies \(n_k\ge k\) for every \(k\): this follows by induction, since \(n_1\ge1\) and \(n_{k+1}\ge n_k+1\). Therefore, for every \(k\ge N\), we have \(n_k\ge k\ge N\), and hence \(x_{n_k}\in U\). Every open neighborhood of \(x\) eventually contains the subsequence terms, so \(x_{n_k}\to x\). \(\square\)

The increasing-index condition is what makes the proof work: once the original sequence has entered a neighborhood and stays there, a subsequence eventually selects only terms from that part of the sequence. The argument does not require the selected indices to be consecutive, or \(X\) to have a metric.

Eventual membership also passes from a sequence to each of its subsequences. If \(x_n\in A\) for all \(n\ge N\), then \(x_{n_k}\in A\) for all \(k\ge N\), because \(n_k\ge k\ge N\). The same argument applies to any condition on the terms, not just membership in a subset. For a convergent sequence, apply this to the condition that the terms lie in a given open neighborhood of the limit.

Why One Convergent Subsequence Is Not Enough

Theorem 2 says that convergence of the full sequence guarantees convergence of every subsequence to the same limit. It does not say that convergence of one subsequence guarantees convergence of the full sequence. A selected subsequence can skip terms that continue to behave differently.

A common mistake is to see \(x_{2k}=1\) and conclude that \(x_n\to1\). That calculation checks only the even indices. The odd indices give \(x_{2k-1}=-1\), so terms equal to \(-1\) occur arbitrarily far along the full sequence. For instance, the open neighborhood \((1/2,3/2)\) of \(1\) excludes every odd-indexed term. No matter how far along one starts, some later term remains outside that neighborhood. The convergence condition therefore fails.

The distinction is useful whenever a sequence is studied by selecting indices. A subsequence can confirm the behavior of its own selected terms, and every subsequence of a convergent sequence must inherit the limit. But a conclusion about the full sequence must account for all sufficiently late terms, not just one selection. No separation hypothesis is needed for the two preservation theorems; the examples showing failure of the converse use the usual Hausdorff topology on \(\mathbb{R}\).