For all $m$ and $n$ there is a list of $mn$ pairwise distinct reals with no strictly increasing sublist of length $m+1$ and no strictly decreasing sublist of length $n+1$

If $m=0$ or $n=0$, then $mn=0$ and the empty list has no sublist of the positive lengths $m+1$ and $n+1$, proving the assertion in these boundary cases. Hence for the construction below assume $m,n\ge 1$.

The index bijection. Define $\beta :m\times n\to \mathbb{N}$ by $\beta (b,j):=bn+j$. Its values lie in $mn$: from $j<n$ and $b<m$ we get $j+1\le n$ and $b+1\le m$, so $\beta (b,j)+1=bn+(j+1)\le bn+n=\sigma (b)\cdot n\le mn$ by [L1], whence $\beta (b,j)<mn$.

$\beta$ is injective. Suppose $\beta (b,j)=\beta (b^{\mathrm{\prime }},j^{\mathrm{\prime }})$ with $(b,j),(b^{\mathrm{\prime }},j^{\mathrm{\prime }})\in m\times n$. If $b<b^{\mathrm{\prime }}$ then $\sigma (b)\le b^{\mathrm{\prime }}$, so $\beta (b,j)<bn+n=\sigma (b)n\le b^{\mathrm{\prime }}n\le \beta (b^{\mathrm{\prime }},j^{\mathrm{\prime }})$ by [L1], a contradiction; symmetrically $b^{\mathrm{\prime }}<b$ is impossible, so $b=b^{\mathrm{\prime }}$ by [L1], and then $j=j^{\mathrm{\prime }}$ by cancellation.

The list. By [L2] and [L3], the injection $\beta$ of step 1.3 is a bijection of $m\times n$ onto $mn$, so every index $i<mn$ is $\beta (b,j)$ for exactly one pair, and $a_{\beta (b,j)}:=\iota (bn+n-j)(b<m,j<n)$ defines a list $a:(mn)\to \mathbb{R}$. Write $\mathrm{b}\mathrm{l}\mathrm{k}(\beta (b,j)):=b$ for the block of an index.

Inside a block the values decrease. Let $b<m$ and $j<j^{\mathrm{\prime }}<n$. Then $\beta (b,j)<\beta (b,j^{\mathrm{\prime }})$ by [L1], while $n-j^{\mathrm{\prime }}\ge 1$ and $n-j=(n-j^{\mathrm{\prime }})+(j^{\mathrm{\prime }}-j)$ with $j^{\mathrm{\prime }}-j\ge 1$, so $n-j^{\mathrm{\prime }}<n-j$ and hence $a_{\beta (b,j^{\mathrm{\prime }})}<a_{\beta (b,j)}$ by [L1] and [L5].

Across blocks the values increase, and the block is monotone in the index. Let $b<b^{\mathrm{\prime }}<m$ and $j,j^{\mathrm{\prime }}<n$. Then $bn+n-j\le bn+n=\sigma (b)n\le b^{\mathrm{\prime }}n<b^{\mathrm{\prime }}n+(n-j^{\mathrm{\prime }})$, the last step because $n-j^{\mathrm{\prime }}\ge 1$; so $a_{\beta (b,j)}<a_{\beta (b^{\mathrm{\prime }},j^{\mathrm{\prime }})}$ by [L5]. Moreover $\beta (b,j)<\beta (b^{\mathrm{\prime }},j^{\mathrm{\prime }})$, by the computation of step 1.3; equivalently, $\mathrm{b}\mathrm{l}\mathrm{k}$ is nondecreasing along the index order.

The list is pairwise distinct. Two indices with different blocks carry different values by step 3.2, and two indices in the same block carry different values by step 3.1; since every index has exactly one block by step 2.1, distinct indices carry distinct values.

No strictly increasing sublist of length $m+1$. Let $s:L\to mn$ be a strictly increasing sublist. The map $\mathrm{b}\mathrm{l}\mathrm{k}\circ s:L\to m$ is injective: if $p<q<L$ had $\mathrm{b}\mathrm{l}\mathrm{k}(s(p))=\mathrm{b}\mathrm{l}\mathrm{k}(s(q))$, then $s(p)<s(q)$ lie in one block, so $a_{s(p)}>a_{s(q)}$ by step 3.1, contradicting that the sublist increases. Hence $L\le m$ by [L4] and natural-order trichotomy, so $L\mathrm{\ne }m+1$.

No strictly decreasing sublist of length $n+1$. Let $s:L\to mn$ be a strictly decreasing sublist and let $p<q<L$. Then $s(p)<s(q)$, so $\mathrm{b}\mathrm{l}\mathrm{k}(s(p))\le \mathrm{b}\mathrm{l}\mathrm{k}(s(q))$ by step 3.2; and $\mathrm{b}\mathrm{l}\mathrm{k}(s(p))<\mathrm{b}\mathrm{l}\mathrm{k}(s(q))$ would give $a_{s(p)}<a_{s(q)}$ by step 3.2, contradicting that the sublist decreases. Thus, if $L>0$, all the $s(p)$ lie in the block of $s(0)$, say $s(p)=\beta (b,j_p)$, and $p\mapsto j_p$ is injective into $n$ because $s$ is injective and $\beta$ is a bijection. If $L=0$, the empty map is already an injection $L\to n$. In either case [L4] and natural-order trichotomy give $L\le n$, so $L\mathrm{\ne }n+1$.

Together with the boundary cases in step 1.1, the list of step 2.1 is therefore a pairwise distinct list of $mn$ reals with no strictly increasing sublist of length $m+1$ and no strictly decreasing sublist of length $n+1$, which is the assertion.