# Index of Subsequence not Less than its Index

## 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

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 5$: Subsequences: $\S 5.1$: Subsequences