Let $P$ be a finite poset and for $n\in\mathbb N$, let $\bf n$ denote the $n$-element totally ordered set.
If $m,n\in\mathbb N$ and $1<m<n$, is the dimension of $P\times \bf m$ equal to the dimension of $P\times\bf n$?
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2$\begingroup$ Please explain what you mean by "dimension". $\endgroup$– F. C.Commented Aug 11, 2021 at 13:58
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$\begingroup$ The smallest number of totally ordered sets such that a poset can be order-embedded into a product of those totally ordered sets. en.wikipedia.org/wiki/Order_dimension $\endgroup$– TriCommented Aug 11, 2021 at 14:20
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$\begingroup$ Tri, "product" or "intersection"? $\endgroup$– Wlod AACommented Oct 12, 2023 at 1:44
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Let $k$ be the dimension of $P$. This means that $P$ can be embedded into a direct product of $k$ chains, but it cannot be embedded into a direct product of less than $k$ chains.
Now if $$ P \hookrightarrow C_1 \times \dots \times C_k $$ then of course $$ P \times \mathbf{n} \hookrightarrow C_1 \times \dots \times C_k \times \mathbf{n} $$ which shows that the dimension of $ P \times \mathbf{n} $ is at most $k+1$. On the other hand, it is also clear the dimension of $P$ is less than or equal than the dimension of $P \times \mathbf{n}$.
So the only possible values for the dimension of a product of $P$ and a finite chain are either $k$ or $k+1$. We can infer that for infinitely many values of $m$ and $n$ indeed the dimensions of $P \times \mathbf{n}$ and $P \times \mathbf{m}$ will be the same. I conjecture that they will all be equal to $k+1$ and thus equal to each other; but I cannot as yet prove or disprove it.
