Subsequence of Subsequence
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Theorem
Let $s$ be a set.
Let $\sequence {s_n}$ be a sequence in $S$.
Let $\sequence {s_m}$ be a subsequence of $\sequence {s_n}$.
Let $\sequence {s_k}$ be a subsequence of $\sequence {s_m}$.
Then $\sequence {s_k}$ is a subsequence of $\sequence {s_n}$.
Proof
By definition, there exists a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that:
- $\forall m \in \N: s_m = s_{n_r}$
Similarly, there exists a strictly increasing sequence $\sequence {m_s}$ in $\N$ such that:
- $\forall k \in \N: s_k = s_{m_s}$
We have that:
- $\forall k \in \N: s_k \in \sequence {s_m}$
and that:
- $\forall m \in \N: s_m \in \sequence {s_n}$
hence:
- $\forall k \in \N: s_k \in \sequence {s_n}$
Because $\sequence {s_k}$ is a subsequence of $\sequence {s_m}$, we have that:
- $k = m_s \implies k + 1 = m_s + p$ for some $p \in \Z_{>0}$
and:
- $m_s = n_r \implies m_s + p = n_r + q$ for some $q \in \Z_{>0}$
Hence:
- $k = n_r \implies k + 1 = n_r + q$
and so $\sequence {s_k}$ is a subsequence of $\sequence {s_n}$.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $4$