# Checking Proof for Monotone Convergence

I've been given a proof for convergence of a monotone increasing sequence bounded from above, and am attempting to replicate it for a decreasing sequence bounded from below (we're working in the real space). I'm hoping this is a valid proof, any improvements/criticisms would be helpful:

Let $$\left\{x_n|n \in \mathbb{N}\right\}$$ be a monotone decreasing sequence of positive real numbers bounded below.

By completeness of $$\mathbb{R}$$ the sequence has a greatest lower bound which we will denote as $$x^*$$. Thus $$x_n \geq x^* \enspace\forall\enspace n \in \mathbb{N}$$.

Let $$\epsilon > 0$$. Then $$x^* + \epsilon$$ is not a lower bound of $$\left\{x_n\right\}$$. $$\exists$$ $$k \in \mathbb{N} : x^* + \epsilon >x_k$$. Since $$\left\{x_n\right\}$$ is decreasing we can say: $$x^* + \epsilon > x_n \enspace\forall\enspace n \geq k$$.

So we have: $$x^* + \epsilon > x_k \geq x_n \geq x^* > x^*-\epsilon \enspace \forall \enspace n \geq k$$

$$\therefore \epsilon> x_n - x^* > -\epsilon \enspace \forall \enspace n \geq k$$

Which gives: $$\left|x_n - x^*\right| < \epsilon \enspace\forall\enspace n \geq k$$

Showing that $$\left\{x_n\right\} \to x^* \blacksquare$$

This is correct! Just that you might be required to show proof for the existence of the largest lower bound as this is a consequence of the completeness axiom.

Also, "So ∃ at least one" is rather redundant. You can say "So there exists ..." or just "∃"

• Oops meant to write greatest lower bound as opposed to least upper bound. Will edit. Thanks! – rory_c Aug 9 at 18:31