Sequence Spaces as Vector Spaces

Sequences as Vectors

A vector need not have only finitely many entries. A real sequence has one real entry for each positive integer, and its entries can be added or scaled one at a time. This gives a large example of a vector space: it includes familiar sequences such as \((1,1,1,\dots)\), \((1,\tfrac12,\tfrac13,\dots)\), and \((1,2,3,\dots)\), as well as sequences with no simple pattern. We will verify the vector-space axioms directly, then examine two important collections inside this space: the bounded sequences and the sequences with only finitely many nonzero terms.

We index sequences by \(\mathbb{N}=\{1,2,3,\dots\}\). Write \(\mathbb{R}^{\mathbb{N}}\) for the set of all functions from \(\mathbb{N}\) to \(\mathbb{R}\). We write a sequence as \(\mathbf{x}=(x_n)_{n\in\mathbb{N}}\), where \(x_n\) is its \(n\)th term. Sequences are equal exactly when their corresponding terms are equal: \(\mathbf{x}=\mathbf{y}\) means \(x_n=y_n\) for every \(n\in\mathbb{N}\).

These operations are well-defined: each term \(x_n+y_n\) and \(ax_n\) is real, so the results are again in \(\mathbb{R}^{\mathbb{N}}\). Because the operations are defined term by term, the vector-space identities can be checked at each index. The next result makes that verification explicit.

Proof. Let \(\mathbf{x}=(x_n)\), \(\mathbf{y}=(y_n)\), and \(\mathbf{z}=(z_n)\) be arbitrary real sequences, let \(a,b\in\mathbb{R}\), and fix any \(n\in\mathbb{N}\). The \(n\)th terms of \(\mathbf{x}+\mathbf{y}\) and \(\mathbf{y}+\mathbf{x}\) are \(x_n+y_n\) and \(y_n+x_n\), which are equal by commutativity of real addition. The \(n\)th terms of \((\mathbf{x}+\mathbf{y})+\mathbf{z}\) and \(\mathbf{x}+(\mathbf{y}+\mathbf{z})\) are \((x_n+y_n)+z_n\) and \(x_n+(y_n+z_n)\), which are equal by associativity of real addition. The \(n\)th term of \(\mathbf{x}+\mathbf{0}\) is \(x_n+0=x_n\), and the \(n\)th term of \(\mathbf{x}+(-\mathbf{x})\) is \(x_n+(-x_n)=0\).

The scalar identities follow from the real-number identities at the same index: the \(n\)th terms of \(a(\mathbf{x}+\mathbf{y})\) and \(a\mathbf{x}+a\mathbf{y}\) are \(a(x_n+y_n)\) and \(ax_n+ay_n\); those of \((a+b)\mathbf{x}\) and \(a\mathbf{x}+b\mathbf{x}\) are \((a+b)x_n\) and \(ax_n+bx_n\); those of \(a(b\mathbf{x})\) and \((ab)\mathbf{x}\) are \(a(bx_n)\) and \((ab)x_n\); and the \(n\)th term of \(1\mathbf{x}\) is \(1x_n=x_n\). Each pair is equal by a corresponding distributive, associative, or identity law in \(\mathbb{R}\). Since these equalities hold for every \(n\), they are equalities of sequences. Thus all the vector-space axioms hold. \(\square\)

Computing with Sequences

Two collections of sequences will be considered as subsets of \(\mathbb{R}^{\mathbb{N}}\). A sequence is bounded if all its terms have a common finite bound in absolute value. A sequence has finite support if only finitely many of its terms are nonzero; its support is the set of indices where those terms are nonzero.

Bounded Sequences Form a Subspace

A subset of a vector space is a subspace when it is itself a vector space under the operations inherited from the ambient space. For the bounded sequences, it is enough to show that the zero sequence belongs to the set and that every linear combination of two members remains in the set. The vector-space identities then hold because they already hold in \(\mathbb{R}^{\mathbb{N}}\).

Proof. The zero sequence is bounded, since \(|0|\leq0\) at every index. Let \(\mathbf{x},\mathbf{y}\in B\). There are bounds \(M_x,M_y\geq0\) such that \(|x_n|\leq M_x\) and \(|y_n|\leq M_y\) for every \(n\). For arbitrary \(a,b\in\mathbb{R}\), the triangle inequality gives \(|ax_n+by_n|\leq |a||x_n|+|b||y_n|\leq |a|M_x+|b|M_y\) for every \(n\). The right-hand side is a fixed finite number, so \(a\mathbf{x}+b\mathbf{y}\) is bounded and belongs to \(B\). Thus \(B\) contains zero and is closed under linear combinations. Its inherited operations satisfy the vector-space axioms, so \(B\) is a subspace. \(\square\)

For instance, \(\mathbf{x}=((-1)^n)\) is bounded by \(1\), and \(\mathbf{y}=(1/n)\) is bounded by \(1\). The theorem guarantees that any real linear combination of these sequences is bounded; the previous example gave the explicit bound \(5\) for one such combination. By contrast, \((1,2,3,\dots)\) is not bounded: for any proposed bound \(M\), there is a positive integer \(n>M\), and then its \(n\)th term exceeds \(M\).

Finite-Support Sequences Form a Subspace

Finite support is preserved for a different reason. Outside the union of the indices where two sequences are nonzero, both sequences vanish, so every linear combination vanishes there too. The union of two finite sets is finite, which is the key point in the proof.

Proof. The zero sequence has empty support, which is finite. Let \(\mathbf{x},\mathbf{y}\in F\), and write \(S_x=\{n:x_n\ne0\}\) and \(S_y=\{n:y_n\ne0\}\). Both sets are finite. If \(n\notin S_x\cup S_y\), then \(x_n=y_n=0\), so \(ax_n+by_n=0\) for any \(a,b\in\mathbb{R}\). Consequently, the support of \(a\mathbf{x}+b\mathbf{y}\) is contained in \(S_x\cup S_y\), which is finite. Thus \(F\) contains zero and is closed under linear combinations. The inherited operations satisfy the vector-space axioms, so \(F\) is a subspace. \(\square\)

How the Two Subspaces Compare

Every finite-support sequence is bounded, but not every bounded sequence has finite support. To see the first claim, if the support is empty then every term is zero and \(0\) is a bound. If the support is a nonempty finite set, the absolute values of the terms on that set have a largest value; outside the set all terms are zero. That largest value therefore bounds the whole sequence. For the second claim, the sequence \((1,\tfrac12,\tfrac13,\dots)\) is bounded by \(1\) but has nonzero terms at every positive integer, so its support is infinite.

Common Mistake: Bounded Does Not Mean Finite Support

A common error is to assume that a sequence tending toward zero must eventually become zero. For example, \((1,\tfrac12,\tfrac13,\dots)\) is bounded, and its terms approach zero, but its \(n\)th term is \(1/n\ne0\) for every \(n\). Its support is therefore all of \(\mathbb{N}\), not a finite set. Conversely, \((1,2,3,\dots)\) has finite terms at every index, but that fact alone does not make it bounded: its terms grow without a common bound.

Another tempting but incorrect bound illustrates why absolute values and the triangle inequality matter. Since \((-1)^n\) and \(1/n\) are each bounded by \(1\), one might claim their sum is also bounded by \(1\). But at \(n=1\), \((-1)^1+1=-1+1=0\), while at \(n=2\), \((-1)^2+\tfrac12=1+\tfrac12=\tfrac32\), which already exceeds \(1\). The valid general estimate is \(|(-1)^n+1/n|\leq1+1=2\), not \(1\).

The distinction between the two subspaces is useful when selecting a space that captures a problem's sequences. Boundedness restricts the size of the terms but allows infinitely many nonzero terms. Finite support restricts how many terms can be nonzero, and therefore automatically gives boundedness. Both restrictions are compatible with termwise addition and scalar multiplication, which is why each defines a vector space in its own right.