I have some problems with the visualization of this proof.
(I will present the problems I have with it in the end, with some intuitive thoughts related to them in italics.)

Theorem: Every real sequence has a monotonic subsequence.

Proof (Thurston):
Take any $(x_m) \in R^\infty$ and define $S_m := \{x_m, x_{m+1},\dots \}$ for each $m \in \mathbb{N}$. If there is no maximum element in $S_1$, then it is easy to see that $(x_m)$ has a monotonic subsequence. (Let $x_{m_1} := x_1$, let $x_{m_2}$ be the first term in the sequence $(x_2,x_3,\dots)$ greater than $x_1$, let $x_{m_3}$ be the first term in the sequence $(x_{m_{2}+1}, x_{m_{2}+2}, \dots)$ greater than $x_{m_2}$ , and so on.) By the same logic, if, for any $m \in \mathbb{N}$, there is no maximum element in $S_m$, then we are done. Assume then max $S_m$ exists for each $m \in \mathbb{N}$. Now define the subsequence $(x_{m_k})$ recursively as follows: $$ x_{m_1} := \text{max} S_1,\hspace{0.7cm} x_{m_2} := \text{max} S_{m_{1}+1}, \hspace{0.7cm}x_{m_3} := \text{max} S_{m_{2}+1}, \hspace{0.7cm} \dots $$ Clearly, $(x_{m_k} )$ is decreasing. QED


1) Why do we have to build $S_m$ for any $m > 1$?
I mean, to me it looks enough to do the construction in brackets for $S_1$, also because $S_m$ should be - by construction - a subset of $S_1$, right?
Basically, I have a lot of problems with the sentence "By the same logic, if, for any $m \in \mathbb{N}$, there is no maximum element in $S_m$, then we are done.". To me, we are done way before.

2) What about the fact that $x_m = x_{m+1}$?
Shouldn't it be covered explicitly? And isn't it actually explicitly ruled out (i.e. the use of "greater" without mentioning equality) by the proof?
Indeed, nowhere it is assumed that we are talking about strictly monotonic subsequences (this looks really as an hidden assumption to me).

3) Do we build up the $S_m$ set for any $m >1$ because we have to use it in the second part of the proof?

Many thanks for any hint or feedback.


1) If you take the sequence $10,20,30,1 - \frac{1}{4}, 1 - \frac{1}{5}, \ldots 1 - \frac{1}{n}, \ldots$, $S_1$, $S_2$ and $S_3$ all have a maximum element, whereas $S_4$ does not have one. In other words, if $S_m$ has a maximum element, so does $S_1$; but the converse is not true.

2) I believe that the word "greater" here should read "greater than or equal to". Of course, the sequence $1,1,1,1, \ldots $ has no strictly monotonic subsequence.

3) The $S_m$ are useful constructions for both parts of the proofs. Can you clarify the problems you had with it?

  • $\begingroup$ 1) So the sentence that bothers me (i.e. "By the same logic, if, for any m∈ℕ, there is no maximum element in Sm, then we are done") is just the result of a process that we apply in order to be sure that, even if all the $S_k$ have a maximum for any $k < m$, at a certain point there is no maximum anymore, and those are the $S_n$ sets (for any $n > k$ we are interested in with that part of the proof. Am I right? $\endgroup$ – Kolmin Feb 8 '14 at 17:28
  • 1
    $\begingroup$ Not exactly (if I understood correctly). I think the logic is as follows: there are two possibilities: (1) there exists some $m \in \mathbb{Z}$ such that $S_m$ does not have a maximum (2) for all $m \in \mathbb{Z}$, $S_m$ has a maximum. After we take care of (1), all we need to do is take care of (2); since in (2) we know that every $S_m$ has a maximum, we are free to construct the sequence $x_m = \max S_m$, which is clearly non-increasing. $\endgroup$ – Pedro M. Feb 8 '14 at 18:06

1) You need to be able to keep picking members of the sequence that are further on than the ones you have already used in the subsequence you're constructing - hence needing to keep considering $S_m$ (but yes, they will all be subsets of $S_n$ for all $n < m$ - and you need this fact to know that the last sequence will actually be decreasing).

2) The theorem is not true for all sequences if you insist on a strictly monotonic subsequence - consider the sequence 1,1,1,1,... for example.


1) $S_1$ may have a maximum.

2) Constant sequences seem are considered monotonic (though they are not strictly monotonic).

3) If we find a single $S_m$ without maximal element we are done. As it is possible that no $S_m$ has a maximal element we automatically have them all available for the fail case.

  • $\begingroup$ 1) So the idea is that we iterate the procedure until we find a $S_m$ without a maximum? $\endgroup$ – Kolmin Feb 8 '14 at 17:26

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