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:

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

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