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Let $\{X_i\}_{i\in I}$ be a family of totally ordered set. Assign each $X_i$ with the order topology. Can we define a total order on $\prod\limits_{i\in I} X_i$ so that the induced order topology coincides with the product topology? What if $I$ is countable or finite? I came up with this question when I read chapter 2 in Munkres' Topology.

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The usual topology on $\mathbb R$ is an order topology, but I don't think the product topology on $\mathbb R\times\mathbb R$ is an order topology. For one thing, a nontrivial connected linearly ordered topological space can be disconnected by removing one point. The plane can't be disconnected by removing one point.

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  • $\begingroup$ In fact, if $X$ is orderable, $\dim(X) \le1$, while $\mathbb{R}^n=n$. $\endgroup$ – Henno Brandsma May 22 '18 at 4:26
  • $\begingroup$ @bof My bad. This is a satisfying answer, thanks. $\endgroup$ – William Sun May 22 '18 at 4:35

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