When does order topology coincide with the norm topology?

Stream: maths

[Yongxi Lin (Aaron)]

In #34770, I proved that the optional stopping theorem is true for a submartingale taking values in a type E such that

[NormedAddCommGroup E] [NormedSpace ℝ E] [CompleteSpace E]
[LinearOrder E] [OrderTopology E] [HasSolidNorm E] [SecondCountableTopology E] [IsOrderedAddMonoid E] [IsStrictOrderedModule ℝ E]

I start to worry whether this is really meaningful, as I can't think of a nontrivial example (i.e. E ≠ ℝ) such that E satisfies all these properties. Are there any references that can help me 1. find out related (counter)examples 2.better understand the interplay between orders and norms?

[Sébastien Gouëzel]

There is just $\R$ and ${0}$. If your space is a linear order with the order topology, then all points (but at most two) should be cut-points, i.e., E \ {x} should not be connected. But on a real vector space with dimension at least 2, then E \ {x} is connected. So your space should have dimension 0 or 1.

[Yongxi Lin (Aaron)]

You are right, so indeed I didn't prove anything new...I'll modify my PR