$\begingroup$Define $\mathbb{R}^{\infty}:=\lbrace (a_1,a_2,a_3,\ldots) \mid a_i \in \mathbb{R} \rbrace$ and let $\phi_i((a_1,a_2,a_3,\ldots))=a_i$. Show that $S=\lbrace \phi_i \mid i \geq 1 \rbrace$ does not span the dual space $L(\mathbb{R}^{\infty},\mathbb{R})$, the space of linear functionals on $\mathbb{R}^{\infty}$.
So I assumed that if $S$ spans $L(\mathbb{R}^{\infty},\mathbb{R})$, then every linear functional $f$ on $\mathbb{R}^{\infty}$ can be expressed as a finite linear combinations of $\phi_i$. But consider:
$f(e_i)= \begin{cases} 1 & \quad \text{if } i \text{ is even}\\ 0 & \quad \text{if } i \text{ is odd}\\ \end{cases}$
Then $f=\sum_{k=1}^{\infty}\phi_{2k}$, which clearly is an infinite linear combination. Is my proof correct? Do I need to assume finite linear combination? Thanks in advance!
$\endgroup$3 Answers
$\begingroup$"Linear combination"---unless stated otherwise---means "finite linear combination."
Additionally, you simply wrote $f$ as an infinite sum of the $\phi_i$'s; you need to prove that there does not exist a linear combination of the $\phi_i$'s that sums up to $f$. You can show this by contradiction; seeking a contradiction, suppose there exist $\{a_k\}_{k = 1}^n, \{i_k\}_{k = 1}^n$ so that $$ f = \sum\limits_{k = 1}^n a_k \phi_{i_k}.$$
Relabel the $i_k$'s so that $i_1 < i_2 < \ldots < i_n$. Then we note that $f(e_{2i_n}) = 1$ but $\phi_{i_j}(e_{2i_n}) = 0$ for all $j$ implying $$1 = f(e_{2i_n}) = \sum\limits_{k = 1}^n a_k \phi_{i_k}(e_{2i_n}) = 0,$$
a contradiction.
$\endgroup$$\begingroup$Infinite linear combinations aren't defined. You need to prove that, for any finite subset $F$ of $S$, your $f$ does not belong to the span of $F$. Note that it's not restrictive to consider $F_n=\{\phi_1,\phi_2,\dots,\phi_n\}$, because any finite subset of $S$ is contained in $F_n$, for some $n$.
Suppose $f$ belongs to the span of $F_n$; then $$ f=\sum_{k=1}^n c_k\phi_k, $$ and so, for $m>n$, $$ f(e_m)=0, $$ which is absurd, by taking $m=2n+1$.
$\endgroup$$\begingroup$In the algebraic theory of modules over a ring $A$ (or over a field $K$ a.k.a vector spaces) linear combinations of finite support are essential to consider. A finite linear combination is of finite support but the converse is not necessarly true.
The support of a family $(\lambda_i)_{i\in I}$ of elements in a ring $A$ (or field $K$) or more generally in a set $E$ is the set of all indices such that $\lambda_i\neq 0$ where $0$ is the unit element for some law of composition to be considered on ring $A$ (or field $K$) more generally on set $E$. Hence a family can be non-finite but finitely supported.
One also say that $x$ enters as a linear combination of a family $(x_i)_{i\in I}$ of elements in module or vector space $E$ with coefficients to be found in the underlying ring $A$ (or field $K$), if there exists a family of finite support of such coefficients $(\lambda_i)_{i\in I}\quad \lambda_i\in A$ (or $K$) such that, $x=\sum_i \lambda_i x_i$. The unit element to be considered here is $0$ the unit of the addition law on the set of scalars i.e. on the ring $A$ (or the field $K$).
Given any family $(x_i)_{i\in I}$ of elements (finite or not) of a module or a vector space $E$, the subspace generated by this family is the space of all the linear combinations (i.e. made with a family of coefficients of finite support) of this family.
Hope this helps.
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