Index of Subsequence not Less than its Index

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Theorem

Let $\sequence {x_n}_{n \mathop \ge 1}$ be a sequence in a set $S$.

Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.


Then:

$\forall n \in \N_{>0}: n_r \ge r$


Proof

The proof proceeds by induction.

For all $r \in \Z_{\ge 1}$, let $\map P r$ be the proposition:

$n_r \ge r$


Basis for the Induction

The first term of $\sequence {x_{n_r} }$ by definition cannot be less than $1$.

That is:

$n_1 \ge 1$

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$n_k \ge k$


from which it is to be shown that:

$n_{k + 1} \ge {k + 1}$


Induction Step

This is the induction step:

\(\ds n_{k + 1}\) \(>\) \(\ds n_k\) Definition of Strictly Increasing Sequence
\(\ds \) \(>\) \(\ds k\) Induction Hypothesis
\(\ds \) \(\ge\) \(\ds k + 1\) as $k$ is an integer


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall r \in \Z_{\ge 1}: n_r \ge r$

$\blacksquare$


Sources

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