Linear Independence of Infinite Sets

Linear Independence of Infinite Sets

The jump from finite independence to inifinite independence requires just a little bit imagination: $$ \begin{array}{cccc} 1&0&0&\cdots \\ 0&1&0&\cdots \\ 0&0&1&\cdots \\ \vdots &\vdots &\vdots &\ddots \end{array} $$

The above theorem is meant for any S, both finite and inifinite. I am a little suspicious whether it's really true. For example: $$ S = \begin{array}{cccc} 1&0&0&\cdots \\ 0&1&0&\cdots \\ 0&0&1&\cdots \\ \vdots &\vdots &\vdots &\ddots \end{array} $$ \begin{align*} v &= (1, 1, 1, 1, \cdots) \\ \text{ or } \\ v &= (1, 0, 1, 0, 1, 0, \cdots) \end{align*} Can you express $v$ as a finite linear combination of the elements of S? On the other hand, if $v$ can be written as a finite linear combination of the elements of S, then this combination must be unique. In the simplest case, suppose it's not unique, and two forms exist: \begin{align*} v &= as_1 + bs_2 \\ v &= cs_3 + ds_4 \\ \end{align*} Let $W = [s_1, s_2, s_3, s_4]$, then $W$ is an independent set. But if we let \begin{align*} u_1 &= \begin{pmatrix} a \\ b \\ 0 \\ 0 \end{pmatrix} \\ u_2 &= \begin{pmatrix} 0 \\ 0 \\ c \\ d \end{pmatrix} \end{align*} then we can see \begin{align*} v &= Wu_1 \\ \text{and} \\ v &= Wu_2 \end{align*} which contradicts the fact that W is an independent set.