Let $(P,\leq)$ be a partially ordered set (poset). We define the ordering dimension $\textrm{dim}_\textrm{ord}(P)$ of $(P,\leq)$ to be the smallest cardinal $\kappa$ such that there exist a set of linear orders $\mathcal{L}$ on $P$ with $|\mathcal{L}|=\kappa$ such that the ordering relation $\leq$ equals $\bigcap \mathcal{L}$. (It follows from Szpilrajn's theorem that the intersection of the collection of all linear orders extending $\leq$ equals $\leq$, so $\kappa$ is well-defined.)

Now the poset $(P,\leq)$ can be endowed with the interval topology $\tau_i(P)$, which is generated by the subbasis $$\{P\setminus \downarrow x: x\in P\} \cup \{P\setminus \uparrow x: x\in P\},$$ where $\downarrow x := \{p\in P: p\leq x\}$ and similarly for $\uparrow x$.

We denote the Lebesgue covering dimension $(P,\tau_i(P))$ by $\textrm{dim}_L(P)$.

Is there a poset $(P,\leq)$ such that $\textrm{dim}_\textrm{ord}(P) < \textrm{dim}_L(P)$?

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    $\begingroup$ I think that you need sharp inequality in your definition of the sub-base of the interval topology (or else your interval topology would be discrete). $\endgroup$ – Włodzimierz Holsztyński Nov 4 '14 at 9:12
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    $\begingroup$ For a finite poset the topology is discrete, isn't it? So the topological dimension is zero, but the order dimension can be any natural number. So your question is whether the topological dimension can ever exceed the order dimension. $\endgroup$ – bof Nov 4 '14 at 9:48
  • $\begingroup$ @bof that's correct - will edit my question accordingly. $\endgroup$ – Dominic van der Zypen Nov 4 '14 at 11:21
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    $\begingroup$ Do you have any reason to think they are related? They seem wildly different. Note that a totally ordered set (such as $\omega_1$) can have uncountable covering dimension. $\endgroup$ – Eric Wofsey Nov 4 '14 at 11:27
  • $\begingroup$ @EricWofsey Since the interval topology $\tau_i(P)$, the covering dimension, and the order dimension all come from the same source (the ordering $\leq$), I was hoping that the two notions of dimension are somehow connected. Your example seems to destroy my hope -- but it's of course a useful answer - thanks! $\endgroup$ – Dominic van der Zypen Nov 4 '14 at 14:37

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