Subset of Linearly Independent Set is Linearly Independent

Theorem
Any subset of a linearly independent set is also linearly independent.

Proof
Let $G$ be an unitary $R$-module.

Then $\left \langle {a_n} \right \rangle$ is a linearly independent sequence iff $\left\{{a_1, a_2, \ldots, a_n}\right\}$ is a linearly independent set of $G$.

So suppose that $\left\{{a_1, a_2, \ldots, a_n}\right\}$ is a linearly independent set of $G$.

Then clearly $\left \langle {a_n} \right \rangle$ is a linearly independent sequence of $G$.

Conversely, let $\left \langle {a_n} \right \rangle$ be a linearly independent sequence of $G$.

Let $\left \langle {b_m} \right \rangle$ be a sequence of distinct terms of $\left\{{a_1, a_2, \ldots, a_n}\right\}$.

Let $\left \langle {\mu_m} \right \rangle$ be a sequence of scalars such that $\displaystyle \sum_{j=1}^m \mu_j b_j = 0$.

For each $k \in \left[{1 .. n}\right]$, let:
 * $\lambda_k = \begin{cases}

\mu_j & : j \mbox{ is the unique index such that } a_k = b_j \\ 0 & : a_k \notin \left\{{b_1, b_2, \ldots, b_m}\right\} \end{cases}$

Then:
 * $\displaystyle 0 = \sum_{j=1}^m \mu_j b_j = \sum_{k=1}^n \lambda_k a_k$

Thus:
 * $\forall k \in \left[{1 .. n}\right]: \lambda_k = 0$

As $\left\{{\mu_1, \ldots, \mu_m}\right\} \subseteq \left\{{\lambda_1, \ldots, \lambda_n}\right\}$, it follows that:
 * $\forall j \in \left[{1 .. m}\right]: \mu_j = 0$

and so $\left \langle {b_m} \right \rangle$ has been shown to be a linearly independent sequence.

Hence the result.