# 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 $4$