Lower and Upper Bounds for Sequences

Theorem
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

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

Then:


 * $(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$;
 * $(2): \quad \forall n \in \N: x_n \le b \implies l \le b$.

Proof
$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$:

Let $\epsilon > 0$.

Then:
 * $\exists N \in \N: n > N \implies \left|{x_n - l}\right| < \epsilon$

So from Negative of Absolute Value:
 * $l - \epsilon < x_n < l + \epsilon$

But $x_n \ge a$, so:
 * $a \le x_n < l + \epsilon$

Thus, for any $\epsilon > 0$:
 * $a < l + \epsilon$

From Real Plus Epsilon it follows that $a \le l$.

$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$:

If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used.

Warning
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

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

Then it is not the case that:


 * $(1): \quad \forall n \in \N: x_n > a \implies l > a$;
 * $(2): \quad \forall n \in \N: x_n < b \implies l < b$.

Take the examples:


 * $(1): \quad \left \langle {x_n} \right \rangle = \dfrac 1 n$
 * $(2): \quad \left \langle {y_n} \right \rangle = -\dfrac 1 n$

Then :
 * $\forall n \in \N_{>0}: \dfrac 1 n > 0, -\dfrac 1 n < 0$

From Power of Reciprocal: Corollary, we have
 * $x_n \to 0$
 * $y_n \to 0$

as $n \to \infty$.

However, it is clearly false that $0 > 0$ and $0 < 0$.