Composition of Sequence with Mapping

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Theorem

Let $\sequence {a_j}_{j \mathop \in B}$ be a sequence.

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


Then $\sequence {a_j} \circ \sigma$ is a sequence whose value at each $k \in A$ is $a_{\map \sigma k}$.

Thus $\sequence {a_j} \circ \sigma$ is denoted $\sequence {a_{\map \sigma k} }_{k \mathop \in A}$.


Proof

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

Let the range of $\sequence {a_j}_{j \mathop \in B}$ be $S$.

Thus $\sequence {a_j}_{j \mathop \in B}$ can be expressed using the mapping $f: B \to S$ as:

$\forall j \in B: \map f j = a_j$


Let $k \in A$.

Then $\map \sigma k \in B$.


By definition of composition of mappings:

\(\ds f \circ \map \sigma k\) \(=\) \(\ds \map f {\map \sigma k}\)
\(\ds \leadsto \ \ \) \(\ds \sequence {a_j}_{j \mathop \in B}\) \(=\) \(\ds \sequence {a_{\map \sigma k} }_{\map \sigma k \mathop \in B}\)
\(\ds \) \(=\) \(\ds \sequence {a_{\map \sigma k} }_{k \mathop \in A}\) as $\map \sigma k \in B \implies k \in A$

$\blacksquare$