Inertia Principle

Theorem

Let $\sequence {a_n}$ be a sequence in $\R$.

Let $a_n \to l$ as $n \to \infty$.

Let $c \in \R$: such that $c < l$.

Then $\exists N \in \N$ such that:

$\forall n \in \N: n \ge N \implies c < a_n$

Proof

Pick $\epsilon = l - c > 0$ (as $c < l$).

As $a_n \to l$ as $n \to \infty$, then $\exists N \in \N$ such that:

$\forall n \in \N: n \ge N \implies \size {a_n - l} < \epsilon$

Equivalently:

$\forall n \in \N: n \ge N \implies \size {l - a_n} < l - c$

For each $a_n$, either $a_n \ge l$ or $a_n < l$.

If $a_n < l$, then $0 < l - a_n$, so $\size {l - a_n} = l - a_n$.

Then:

\(\ds \forall n \in \N: \, \)

\(\ds n\)

\(\ge\)

\(\ds N\)

\(\ds \leadsto \ \ \)

\(\ds \size {l - a_n}\)

\(\lt\)

\(\ds l - c\)

\(\ds \leadsto \ \ \)

\(\ds l - a_n\)

\(\lt\)

\(\ds l - c\)

\(\ds \leadsto \ \ \)

\(\ds a_n\)

\(\gt\)

\(\ds c\)

If $a_n \ge l$, and $l > c$, then $a_n > c$.

So:

$\forall n \in \N: n \ge N \implies a_n > c$

$\blacksquare$

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Note

Not to be confused with the Principle of Inertia.