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 $\map \sigma k \in B \implies k \in A$ |
$\blacksquare$