# Two Views of the Monotone Sequence Theorem

Originally published: 2012-11-05
Canonical URL: /2012/Nov/two-views-of-the-monotone-sequence-theorem/

If a sequence of real numbers $(a_n)$ is _bounded_ and _monotone_ (and I'm actually going to say _nondecreasing_, without loss of generality), then it _converges_. I'm going to tell you _why_ and I'm going to tell you _twice_.

If our sequence is bounded, the completeness of the reals ensures that it has a _least_ upper bound, which we'll call, I don't know, _B_, but there have to be sequence elements arbitrarily close to (but not greater than) _B_, because if there weren't, then _B_ couldn't be a _least_ upper bound. So for whatever arbitrarily small ε, there's an _N_ such that $a_N > B - \varepsilon$, which implies that $|a_N - B| < \varepsilon$, but if the sequence is nondecreasing, we also have $|a_n - B| < \varepsilon$ for $n \ge N$, which is what I've been trying to tell you—

—_twice_; suppose by way of contraposition that our sequence is _not_ convergent. Then there _exists_ an ε such that for all _N_, there exist _m_ and _n_ greater or equal to _N_, such that $|a_m - a_n|$ is greater or equal to ε. Suppose it's monotone, without loss of generality _nondecreasing_; that implies that for all _N_, we can find $n > m \ge N$ such that $a_n - a_m \ge \varepsilon$. Now suppose our sequence is bounded above by some bound _B_. However, we can actually describe an algorithm to find sequence points greater than _B_, thus showing that this alleged bound is really not a bound at all. Start at $a_1$. We can find points later in the sequence that are separated from each other by at least ε, but if we do this $\lceil(B - a_1)/\varepsilon\rceil$ times, then we'll have found a sequence point greater than the alleged bound.