Partial ordering: is increasing pair equivalent with increasing subsequence

Question

An answer on this site states that

For partial ordering $\leq$, the claims

1. every infinite sequence $x_1, x_2, \dots$ has an increasing pair, i.e., there exist $i < j$, such that $x_i \leq x_j$
2. every infinite sequence $x_1, x_2, \dots$ has increasing subsequence, i.e., there exist $i_1 < i_2 < \cdots$, such that $x_{i_1}\leq x_{i_2}\leq \cdots$.

are equivalent

I wonder why is partial ordering needed here.

My thoughts:

Proving $2\Rightarrow 1$ is trivial: suppose 2 holds. Then, we have infinitely many such pairs. Proving $1\Rightarrow 2$ is a bit harder:

Suppose 1 holds. Let $A_1$ be the set $A_1 = \mathbb{N}$. Consider the sequence $(x_i)_{i\in A_1}$ (the initial sequence). Then, we can find a pair of indices $i_1 < j_1$, such that $i_1 < j_1$ and $x_{i_1} \leq x_{j_1}$. We repeat the same for the sequence $(x_i)_{i\in A_0, i > j_1}$ and obtain $(i_2, j_2)$. By repeating this, we obtain infinitely many increasing pairs of indices $(i_k, j_k)$, $k\in\mathbb{N}$, such that $$i_1 < j_1 < i_2 < j_2 < i_3 < j_3 < \cdots$$ Let's store them in the set $B_1$.

Then, we consider the set of indices $A_2 = \{ j_k \mid (i_k, j_k)\in B_1 \}$ and the subsequence $(x_i)_{i\in A_2}$. Again, we can construct infinitely many increasing pairs that are stored in $B_2$. However, every such pair now corresponds to an increasing triplet. For example, if $x_{j_2} \leq x_{j_{7}}$, then we also have $x_{i_2}\leq x_{j_2} \leq x_{j_7}$.

Then, we consider the set of indices $A_3 = \{ j_k \mid (i_k, j_k)\in B_2 \}$, and the subsequence $(x_i)_{i\in A_3}$. Again, every pair we find, extends some increasing triplet to an increasing quadruple.

We continue with the procedure and gradually construct (many) increasing subsequences.

Answer 1

I don't see how you are generating an infinite subsequence. Of course, you are generating finite subsequences of arbitrary length, but this is not the same as generating an infinite one.

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I guess for $1\implies 2$ you can try as follows. Let us assume $2$ does not hold for the sequence $x_1, x_2, x_3,\ldots$. Then this sequence reaches a maximal element, i.e. the element $x_{i_1}$ such that it is not $x_{i_1}\le x_j$ for any $j>i_1$. The reason why this is true is: otherwise you could just recursively create the increasing subsequence starting from $x_1$, and then keep finding new bigger elements one by one, further down the sequence.

Now take the sequence $x_{i_1+1}, x_{i_1+2}, x_{i_1+3}\ldots$ and apply the same logic. Again, as the original sequence was a counterexample for 2, this new sequence must also reach a maximum element $x_{i_2}$ such that $x_{i_2}\le x_j$ is false for all $j>i_2$.

And so on, we find indices $i_3, i_4,\ldots$ such that $x_{i_k}\le x_j$ is false for all $k$ and all $j>x_k$.

Finally, look at the subsequence $x_{i_1}, x_{i_2}, \ldots$: that subsequence does not have any pairs of elements in the right order: say if $u<v$, then $x_{i_u}\le x_{i_v}$ is false simply because $i_v>i_u$. This contradicts 1.

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The reason they mention "partial" ordering in the linked answer is, probably, that the word "partial" is there just to remind you that it is not necessarily a total ordering (i.e. if $a\le b$ is false, $b\le a$ doesn't need to be true; $a$ and $b$ may be incomparable). It does not imply that the ordering must not be total, it just means the statement is valid even without that assumption. After all, the linked question was about curious properties of partial orderings!