How many non-increasing sequences are there over the natural numbers?

Question

How many non-increasing sequences are there over the natural numbers? By splitting it to categories, I sort of got it has to be $\aleph_0$. Nevertheless, I haven't seen such a question and therefore I don't know if I am even correct in my result. I would appreciate your help.

The horrible term "non-increasing" could either mean "not increasing", as logic would suggest, or (more likely) weakly decreasing. The answer to the question depends on this distinction, so it might be useful to be unambiguous here. — Marc van Leeuwen, Feb 09 '15 at 14:58
@MarcvanLeeuwen: good comment, to me a 'non-increasing sequence' is a sequence that is not increasing. — Alexandre Halm, Feb 09 '15 at 15:00
@MarcvanLeeuwen Why do some authors use terms “non-ascending” and “non-descending” instead of ascending and descending?. — Git Gud, Feb 09 '15 at 17:05
@AlexH. Unfortunately most of the time you'll be wrong by interpreting properly, (i.e., the way you do). — Git Gud, Feb 09 '15 at 17:06
@GitGud See also why do we use 'non-increasing' instead of decreasing? (in fact I've now flagged the other one as a duplicate) — Marc van Leeuwen, Feb 09 '15 at 17:28
@MarcvanLeeuwen I love your first sentence. I had already read (and up voted) this answer, but didn't remember it. — Git Gud, Feb 09 '15 at 17:51

score 15 · Accepted Answer · edited Apr 13 '17 at 12:19

15

You are correct.

Since the sequences are not increasing, they contain a minimal element. By adding $1$ to every entry, you have a bijection between the sequences with minimal element $n$ and those with minimal element $n+1$. It is therefore sufficient to show that the sequences with minimal element $0$ are countable, because the countable union of countable sets is countable. These sequences are sequences of finite length (i.e. the index of the last nonzero entry), and by the same argument as above it is sufficient to show that the sequences of a fixed length $\ell$ form a countable set. But in fact, the set of sequences of length $\ell$ is just $\mathbb N^\ell$, which is countable.

edited Apr 13 '17 at 12:19

Community

1

answered Feb 09 '15 at 13:16

Jesko Hüttenhain

14,792

7

Any sequence of natural numbers contains a minimal element, it doesn't have to be non-increasing. – dtldarek Feb 09 '15 at 14:02
@dtldarek: Very true - the mental image was, of course, that this is so-to-speak the "limit point" of the sequence. – Jesko Hüttenhain Feb 09 '15 at 18:56
I didn't use functions but I did use combinatorics when looking at different kinds of non-increasing sequences so as to show that even when the aforementioned sequence is infinite, it might have countable number of elements, but once you determine the first, you only have a finite number of elements to work with. i.e, you will make $\aleph_0$ choices but every choice is finite. Thus applying the rule of product it cannot exceed $\aleph_0$... – Donna Feb 09 '15 at 20:17

score 11 · Answer 2 · edited Feb 09 '15 at 18:19

11

Let $(a_k)_k$ be such a sequence and let $m=\inf\{\,a_k:k\in\mathbb N\,\}$. Then almost all $a_k$ equal $m$, and we can "encode" the infinite sequence by writing down only the initial terms until the first occurrence of $m$. This way, the set of nonincreasing sequences can be viewed as subset of the set of finite sequences, which is countable.

edited Feb 09 '15 at 18:19

John Bentin

18,454

answered Feb 09 '15 at 13:18

Hagen von Eitzen

374,180

This is how I would do it, though I might use words like injection or one-to-one. – Henry Feb 09 '15 at 13:20

score 3 · Answer 3 · answered Feb 09 '15 at 13:59

Call a sequence $(u_1,u_2,\ldots)$ $n$-bounded if

$u_1 \le n$, and
for all $m > n, u_m = u_n$.

Then all non-increasing sequences are $n$-bounded for some $n$, and for any $n$, the number of $n$-bounded sequences is finite. So the number of non-increasing sequences is countable.

Marc van Leeuwen · Answer 4 · 2015-02-09T17:32:33.813

To a weakly decreasing sequence $a=(a_n)_{n\in\Bbb N}$ one can associate a function $f(a):\Bbb N\to\Bbb N\cup\{\omega\}$ mapping $m\mapsto\#\{\, n\in\Bbb N\mid a_n>m\,\}$. One can recover* $a$ from $f(a)$, and $f(a)$ takes only finitely many nonzero values, so the number of possibilities for $f(a)$ are countable. (In fact $f$ establishes a bijection to the set of weakly decreasing functions $\Bbb N\to\Bbb N\cup\{\omega\}$ of finite support.)

*One has $a_n=\#\{ \,m\in\Bbb N \mid f(m)>n\,\}$ for all $n$, considering that $0\in\Bbb N$.

score 1 · Answer 5 · answered Feb 09 '15 at 16:45

In order to fix ideas denote such a sequence by ${\bf x}=(x_0,x_1,x_2,\ldots)$ with $x_0\geq x_1\geq \ldots\geq0$. Note that only finitely many differences $x_{k-1}-x_k$ are $\geq1$. Let $S$ be the set of all these sequences.

Let $(p_k)_{k\geq1}$ be the sequence of prime numbers. Then $$\phi:\quad S\to{\mathbb N}_{\geq1}, \qquad{\bf x}\mapsto 2^{x_0} \>\prod_{k=1}^\infty p_k^{\>x_{k-1}-x_k}$$ is injective. This proves $\#(S)=\aleph_0$.

How many non-increasing sequences are there over the natural numbers?

5 Answers5