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 $$\sum_{j=1}^m \mu_j b_j = 0$$.

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

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.