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I tried to assume otherwise and derive a contradiction, but since $x_n - x_{n+1}$ is an arbitrary small distance, I don't see that convergence can get the job done, nor whether my conjecture is true. I am guessing not, but the truth of this statement would help me with another proof which I already checked as answered. Now I am not satisfied with my understanding.

NS248
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Try $$ x_{2n} = b - \frac 1n\\ x_{2n+1} = b- 2^{-n} $$ for a counterexample.

mookid
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  • I had a sinking feeling that a counterexample could be constructed. I played around with points on paper and thought there could be one. Perhaps you can look art my question about left handed limits and give me some advice there. I don't see how to do tot without a monotonic sequence. In any case, thanks. I will check this question as answered as soon as the time limit is up. – NS248 Mar 18 '14 at 20:42
  • Can you give a link to that question? – MPW Mar 18 '14 at 20:45
  • @MPW http://math.stackexchange.com/questions/716380/prove-that-if-f-is-bounded-and-nondecreasing-on-a-b-then-lim-fx-as-x – NS248 Mar 18 '14 at 20:51
  • the first answer is perfect imo/ – mookid Mar 18 '14 at 20:52
  • Beyond hints, what am I missing there, if you don't mind helping me? I can see that the function will converge for the subsequence, but how does that establish the limit? – NS248 Mar 18 '14 at 20:53
  • you show that every converging subsequence converge to the same limit. When you consider a given converging subsequence, consider a monotonic sub-subsequence. The limit is always the same (this is the limit of the monotonic sequences). – mookid Mar 18 '14 at 20:56