Bounded monotonic non converging sequences I remember reading a couple of weeks ago about some examples of sequences which are bounded, monotonic, but not convergent. 
As far as I can remember, the proof was carried out without assuming the axiom of choice or in constructive mathematics. However, as I m new to this field, I just don't really know how to go back to that results. 
Any clue about this story?
Thanks a lot!
 A: Consider the space $\mathbb R$, as you may know it is a complete metric space. This means that every bounded monotonic sequence is convergent.
For example, $a_n=\dfrac1n$ converges to $0$.
If, however, we consider only a subspace of the real numbers, for example $(0,1)$ - all the numbers strictly between $0$ and $1$ then the above sequence does not converge in that space. Why? For a sequence to be convergent the space has to know the possible limit point. The limit is $0$ which $(0,1)$ does not know about.
If you take the number $\sqrt 2$, as you may know it is not a rational number. Take the rational sequence defined by longer and longer decimal expansions of $\sqrt 2$: $1,1.4,1.41,\ldots$ 
In the real numbers this is a monotonic and bounded sequence (all those are below $\sqrt 2$) and thus converge (to $\sqrt 2$ as luck would have it). Now you can see that all those elements in the sequence are rationals - the decimal expansion is finite - therefore this sequence is also a sequence in $\mathbb Q$.
In the rational numbers this is a monotonic and bounded sequence but the limit point is $\sqrt 2$ and this is a number that $\mathbb Q$ does not know about and therefore cannot tell that the sequence is converging.
A: If the sequence lives in a subspace of the reals, then since it is bounded and monotonic it is necessarily a Cauchy sequence: If it were not, there would be an $\epsilon\gt0$ such that infinitely many pairs of consecutive members differ by more than $\epsilon$, and since the sequence is monotonic these steps would all go in the same direction, contradicting the fact that the sequence is bounded. (That proof doesn't require the axiom of choice.)
Thus, if the space is a subspace of the reals, such a sequence can exist only if the space is not complete, i.e. if there are Cauchy sequences that don't converge. One example of this would be the rationals, in which the sequence of successive decimal approximations of $\sqrt2$ doesn't converge.
On the other hand, if the sequence lives in some other space, all bets are off, since we know nothing about the relationship between the order defining "monotonic" and the metric defining "bounded" and "convergent". For instance, consider the interval $[0,1]$ with the usual metric, but with an order such that all rational numbers are ordered lexicographically according to their numerators and denominators in reduced form. Then the sequence of all rational numbers in the interval in order is bounded and monotonic but does not converge even though the space is complete.
A: According to https://mathoverflow.net/questions/87214/axiom-of-choice-and-convergence
In constructive mathematics, it is possible to define a bounded increasing sequence called Specker sequence (https://en.wikipedia.org/wiki/Specker_sequence), whose limit is not a computable number, so its supremum does not exist constructively. I think here constructive mathematics means intuitionistic logic or some type theory.
A: It is essentially dependent on the field you are looking at- if it is reals, then such a result cannot be true. Let us consider a subspace of reals, say I, the set of irrationals. Consider the sequence a(n) = 0.5 + sqrt(2)/n, a sequence of irrationals, converging to 0.5 in reals, but not convergent on the set of irrational numbers, I, since 0.5 does not belong to I. However, 0.5 + sqrt(2)/n is monotonically decreasing, and is bounded below by, just to give a specific value, -sqrt(2) in I. Basically, this example has been generated, exploiting the fact that the LUB Property does not hold in I.
