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:


 * $$\forall n \in \N: x_n \ge a \implies l \ge a$$;
 * $$\forall n \in \N: x_n \le b \implies l \le b$$.

Proof

 * $$\forall n \in \N: x_n \ge a \implies l \ge a$$:

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 $$a \le l$$.


 * $$\forall n \in \N: x_n \le b \implies l \le b$$:

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:


 * $$\forall n \in \N: x_n > a \implies l > a$$;
 * $$\forall n \in \N: x_n < b \implies l < b$$.

Take the examples:


 * $$\left \langle {x_n} \right \rangle = \frac 1 n$$
 * $$\left \langle {y_n} \right \rangle = -\frac 1 n$$

Then :
 * $$\forall n \in \N^*: \frac 1 n > 0, -\frac 1 n < 0$$.

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 $$0 < 0$$.