1. ## Boundedness and Convergence

Suppose that {x(n)} 1=n=infinity, is a sequence of real numbers with limit x, and suppose that a≤ x(n) ≤b, all n. Prove that a≤ x ≤b.

2. Originally Posted by Snooks02
Suppose that {x(n)} 1=n=infinity, is a sequence of real numbers with limit x, and suppose that a≤ x(n) ≤b, all n. Prove that a≤ x ≤b.
start with $a \le x_n \le b$. now take the limit as $n \to \infty$ of that system

3. Originally Posted by Snooks02
Suppose that {x(n)} 1=n=infinity, is a sequence of real numbers with limit x, and suppose that a≤ x(n) ≤b, all n. Prove that a≤ x ≤b.
If $x_n$ is convergent and $x_n \leq b$ then $x\leq b$ - where $x=\lim x_n$.
This is what we will prove. Say we prove that if $x_n \leq 0$ then $x\leq 0$. This is sufficient to complete the proof. Why? Because if $x_n \leq b$ with $x = \lim x_n$ then define $y_n = x_n - b$ so $y_n \leq 0$ and by above $y \leq 0 \implies x \leq b$ (here $y = \lim y_n$).

Now let $x_n\leq 0$. And assume, by contradiction, $x > 0$. Then there is $N$ so that $|x_N - x| < \frac{x}{2}$ and so $x_N > \frac{x}{2} > 0$ a contradiction.

Similary do it for $x_n \geq a \implies x \geq a$ by repeating the argument.