Sequences in R^n

Sequences as Lists of Vectors

A sequence in \(\mathbb{R}^n\) is an ordered list of vectors. Its order and its indexing matter: the same vector may occur at several different indices, and a subsequence is selected by choosing indices in increasing order. This viewpoint lets us transfer familiar questions about real sequences to vectors by examining one coordinate at a time. In particular, boundedness of a vector sequence can be checked through its finitely many coordinate sequences.

Throughout, \(n\) is a positive integer, indices \(k\) are positive integers, and unless stated otherwise we measure distances and norms using the Euclidean norm \(\|\cdot\|_2\). Write the \(k\)-th vector as \(x^{(k)}\), with coordinates \(x^{(k)}=(x_1^{(k)},\ldots,x_n^{(k)})\). The superscript labels the term; the subscript labels its coordinate.

The coordinate sequences are obtained by applying the coordinate projections to the terms. These are instances of the projections from product spaces introduced earlier in this course. A sequence is therefore not merely a subset of \(\mathbb{R}^n\): its indexing retains information that its range may discard.

Boundedness Coordinate by Coordinate

A sequence is bounded when all of its terms remain within some fixed distance of the origin. The bound must work for every index; a bound that changes with \(k\) does not establish boundedness. Since there are only finitely many coordinates, a common bound on each coordinate gives a bound on the vectors, and a bound on the vectors gives a bound on each coordinate.

Proof. Suppose first that there is an \(M\geq0\) with \(\|x^{(k)}\|_2\leq M\) for every \(k\). For each coordinate \(i\), the definition of the Euclidean norm gives \(|x_i^{(k)}|\leq\|x^{(k)}\|_2\leq M\). Thus every coordinate sequence is bounded.

Conversely, suppose that each coordinate sequence is bounded. For each \(i\in\{1,\ldots,n\}\), choose \(M_i\geq0\) with \(|x_i^{(k)}|\leq M_i\) for every \(k\). Since there are finitely many coordinates, \(M=\max\{M_1,\ldots,M_n\}\) exists and is finite. Then, for every \(k\), \(\|x^{(k)}\|_\infty\leq M\), and \(\|x^{(k)}\|_2\leq\sqrt{n}\,\|x^{(k)}\|_\infty\leq\sqrt{n}M\). Also, \(\|x^{(k)}\|_1\leq nM\), because each of the \(n\) coordinate magnitudes is at most \(M\). The sequence is therefore bounded in all three standard norms. These comparisons agree with the norm comparison theorem from “Norms on \(\mathbb{R}^n\).” \(\square\)

Subsequences and Index Selection

A subsequence keeps selected terms in their original order. It may skip terms, but it cannot reorder them or use the same index twice. Distinct indices may nevertheless produce the same vector, so a subsequence can contain repeated vector values.

The strict increase ensures that the selected terms occur in the same order as in the original sequence. It also ensures that every selected index is at least \(k_1\), a fact useful when reasoning about all sufficiently late terms. Merely listing some vectors that occur in the original range is not enough to specify a subsequence: their original indices and order matter.

Proof. Suppose \((x^{(k)})\) is bounded, so there is an \(M\geq0\) such that \(\|x^{(k)}\|_2\leq M\) for every positive integer \(k\). Let \((x^{(k_j)})\) be any subsequence. Each \(k_j\) is a positive integer, so the original bound applies at that index: \(\|x^{(k_j)}\|_2\leq M\) for every \(j\). Thus the same \(M\) bounds the subsequence. \(\square\)

This theorem is one-way: selecting fewer terms cannot break a bound that already holds. It does not say that a bounded sequence must have a constant subsequence, or that a particular selection of indices has any additional property. Those are separate questions, and boundedness alone should not be confused with them.

Unbounded Sequences Have Terms Far from the Origin

If a sequence is unbounded, no fixed ball centered at the origin contains all its terms. A stronger and useful fact is that one can choose terms at strictly increasing indices whose norms become arbitrarily large. The construction must move to later indices at each step; otherwise it might simply select the same early large term repeatedly.

Proof. We construct indices recursively. Since the original sequence is unbounded, choose \(k_1\) such that \(\|x^{(k_1)}\|_2>1\). Suppose \(k_j\) has been chosen. The terms after index \(k_j\) must still have unbounded norms. Indeed, if the tail \(\{x^{(k)}:k>k_j\}\) had norms bounded by some \(C\), then the finitely many earlier terms \(x^{(1)},\ldots,x^{(k_j)}\) would have a finite maximum norm \(D\). The whole sequence would then be bounded by \(\max\{C,D\}\), contrary to the hypothesis.

Therefore the tail after \(k_j\) contains an index \(k_{j+1}>k_j\) with \(\|x^{(k_{j+1})}\|_2>j+1\). Repeating this construction gives strictly increasing indices and \(\|x^{(k_j)}\|_2>j\) for every \(j\). For any fixed \(R\geq0\), choose an integer \(J>R\). Whenever \(j\geq J\), we have \(\|x^{(k_j)}\|_2>j\geq J>R\). This proves that the selected norms eventually exceed \(R\), as required. \(\square\)

What Boundedness Does—and Does Not—Tell Us

Boundedness describes the distance of every term from the origin; it does not describe how the terms relate to one another. For example, the sequence \(((-1)^k,0)\) in \(\mathbb{R}^2\) is bounded, since every term has norm \(1\), even though it takes two distinct values repeatedly. A bound supplies a region containing the entire sequence, not a claim that the sequence settles near one point.

Likewise, the range and the indexed sequence answer different questions. The range of the alternating example is just \(\{(-1,0),(1,0)\}\), while the sequence records the order in which these points occur. Subsequence arguments depend on that order, so it is important to keep the index notation visible. The coordinate criterion is especially effective when a vector formula is complicated: bound each coordinate uniformly, then combine the finitely many bounds using a norm inequality.

The results here prepare for the study of convergence in \(\mathbb{R}^n\). Boundedness and subsequence selection are useful tools in that study, but neither definition by itself asserts that a sequence approaches a vector. The next step is to formulate precisely what it means for the terms of a vector sequence to approach a limit.