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


Thanks in advance.

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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 "∃"

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

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