## Linear Independence of Infinite Sets

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.

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