Composition of Sequence with Mapping

Theorem
Let $\left \langle {a_j} \right \rangle_{j \in B}$ be a sequence.

Let $\sigma: A \to B$ be a mapping, where $A \subseteq \N$.

Then $\left \langle {a_j} \right \rangle \circ \sigma$ is a sequence whose value at each $k \in A$ is $a_{\sigma \left({k}\right)}$.

Thus $\left \langle {a_j} \right \rangle \circ \sigma$ is denoted $\left \langle {a_{\sigma \left({k}\right)}} \right \rangle_{k \in A}$.

Proof
By definition, a sequence is a mapping whose domain is a subset of $\N$.

Let the range of $\left \langle {a_j} \right \rangle_{j \in B}$ be $S$.

Thus $\left \langle {a_j} \right \rangle_{j \in B}$ can be expressed using the mapping $f: B \to S$ as:
 * $\forall j \in B: f \left({j}\right) = a_j$

Let $k \in A$.

Then $\sigma \left({k}\right) \in B$.

By definition of composition of mappings: