Inequality Rule for Real Sequences

Theorem
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.

Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:


 * $\displaystyle \lim_{n \mathop \to \infty} x_n = l$
 * $\displaystyle \lim_{n \mathop \to \infty} y_n = m$

Let there exist $N \in \N$ such that:
 * $\forall n \ge N: x_n \le y_n$

Then:
 * $l \le m$

Proof
Suppose $l > m$.

Then:
 * $m = \dfrac m 2 + \dfrac m 2 < \dfrac {l + m} 2 < \dfrac l 2 + \dfrac l 2 = l$

Let $\epsilon = \dfrac {l - m} 2$.

Then:
 * $\epsilon > 0$

We are given that:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = l$

By definition of the limit of a real sequence, we can find $N_1$ such that:
 * $\forall n \ge N_1: \size {x_n - l} < \epsilon$

where $\size {x_n - l}$ denotes the absolute value of $x_n - l$

Suppose $n \ge N_1$

If $x_n \ge l$ then $x_n > \dfrac {l + m} 2$

If $x_n < l$ then:

In either case $x_n > \dfrac {l+m} 2$

We are also given that:
 * $\displaystyle \lim_{n \mathop \to \infty} y_n = m$

Similarly we can find $N_2$ such that:
 * $\forall n > N_2: \size {y_n - m} < \epsilon$

Suppose $n \ge N_2$

If $y_n \le m$ then $y_n < \dfrac {l + m} 2$

If $y_n > m$ then:

In either case $y_n < \dfrac {l + m} 2$

Let $N = \max \set {N_1, N_2}$.

Then if $n > N$, both the above inequalities will be true:
 * $n > N_1$
 * $n > N_2$

Thus $\forall n > N$:
 * $y_n < \dfrac {l + m} 2 < x_n$

It has been shown that:
 * $l > m \implies \forall n \in \N: \exists m \ge n: x_n > y_n$

Taking the contrapositive:
 * $\exists N \in \N: \forall n \ge N: x_n \le y_n \implies l \le m$

Hence the result.