TITLE: Can unlimited hypernaturals be represented by increasing sequences?
QUESTION [6 upvotes]: Consider the usual ultra-power construction of the hyperreals $^*\mathbb R$ with ultrafilter $\mathcal F$. Let $K = [k_n] \in ^*\mathbb N$ be an unlimited hypernatural. My question is does there always exists a (strictly?) monotonically increasing sequence $(a_n)$ such that also $[a_n] = K$. And if this is the case, can $(a_n)$ be chosen as a sub-sequence of $(k_n)$?
I came up with this question after making the following observation: Consider the strictly monotonically increasing sub-sequence $(k_{n_m})$ constructed by setting $n_1 = 1$ and $n_{m+1} = \min\{n\in\mathbb N \mid k_n >k_{n_i} \text{ for all } i=1\ldots m\}$. 
E.g. for the sequence
$$ 1,3,2,4,3,6,5,7,6,\ldots$$
generated by alternatingly adding $2$ and substracting $1$, the subsequence would be $1,3,4,6,7,\ldots$
Now let $A = \{n_m \mid m\in\mathbb N\}$. Then there are 2 possibilities: either $A\in\mathcal F$ or  $A^c = \mathbb N \setminus A \in\mathcal F$. In the first case we can simply set $a_n = k_n$ if $n\in A$, and fill up the gaps such that $(a_n)$ is monotonially increasing, e.g. in the example sequence we could choose
$$ 1,3,3,4,4,6,6,7,7,\ldots$$
In the latter case, we would repeat the construction, ignoring any terms in $A$ giving us again a strictly monotonically increasing sub-sequence with indices in $B\subseteq A^c$ such that $A^c = B \uplus B^c$ (with $B^c = A^c \setminus B$). Then either $B$ or $B^c$ must be in $\mathcal F$ (standard property of ultra-filters), at which point we can apply the same arguments as above again. It seems plausible that this iterative process should stop at some point since after every iteration we ignore another infinite number of terms (maybe Zorn's Lemma could help?).

REPLY [5 votes]: A slightly more elementary answer: the answer is negative in general because of the following nice characterisation of P-point ultrafilters: an ultrafilter is a P-point if and only if every infinitesimal in $\mathbb{R}^{\mathbb N}/\mathcal F$ is representable by a sequence tending to zero.  Thus if the ultrafilter happens not to be a P-point (a property indepedent of ZFC), then there is an "inaccessible" infinitesimal $\epsilon$, i.e., not representable by a sequence tending to zero. In particular, the reciprocal $\frac{1}{\epsilon}$ cannot be represented by a monotone sequence. More details on inaccessible infinitesimals can be found in this article.