[Math] How to improve the proof on monotone sequences converging if bounded

convergence-divergenceproof-writing

My professor gave me 3/5 points for this proof and just wanted to know how I could make it better.

Prove a monotone sequence converges if and only if it is bounded.
If {$x_n$} is monotone then it converges which means it is bounded above if it is strictly increasing or bounded below if it is strictly decreasing. If it is an increasing sequence then $x_k$->sup{$x_n$} and if it is a decreasing sequence then $x_k$->inf{$x_n$}. If {$x_n$} is not a bounded, monotone increasing sequence, then it is not bound from above, so {$x_n$} goes to infinity. If {$x_n$} is not a bounded, monotone decreasing sequence, then it is not bounded from below, so {$x_n$} goes to negative infinity.

Any help is appreciated!

Best Answer

Probably your professor wanted you to prove your assertions (which are correct) that:

  1. If $(x_k)$ is increasing and bounded above, then $x_k \to \sup\{x_k\}$
  2. If $(x_k)$ is decreasing and bounded below, then $x_k \to \inf\{x_k\}$
  3. If $(x_k)$ is increasing and unbounded above, then $x_k \to \infty$
  4. If $(x_k)$ is decreasing and unbounded below, then $x_k \to -\infty$

I'll provide a proof for (1) and (3). The others are very similar.

(1) Suppose that $(x_k)$ is increasing and bounded above. Since the set $\{x_k\}$ is bounded above, it has a supremum, call it $M$. We claim that $x_k \to M$. Let $\epsilon > 0$. Then there is some $k$ for which $M - \epsilon \leq x_k \leq M$. (If this were not true, then $M - \epsilon$ would be an upper bound for $\{x_k\}$.) Since $(x_k)$ is increasing, this means that $M - \epsilon \leq x_n \leq M$ for all $n \geq k$. Since $\epsilon$ can be chosen arbitrarily small, this proves that $x_k \to M$.

(3) Suppose that $(x_k)$ is increasing and unbounded above. Given any integer $M$, there is some $k$ for which $x_k > M$. (Otherwise, $M$ is an upper bound for $\{x_k\}$.) Since $(x_k)$ is increasing this means that $x_n > M$ for all $n \geq k$. Since $M$ can be chosen arbitrarily large, this shows that $x_k \to \infty$.

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