Cfrtjryk

Let $(H,mathcal T)$ is a Hausdorff space then is there any total order relation on $H$ such that the topological space induced by the order relation be the same $(H,mathcal T)$?

Thanks a lots beforehand.

A total order relation induces a normal topology, which means that if there is a Hausdorff space which is not normal as well, then it is not induced by an order relation.

Though I am also interested to see what counter-examples people will give you.

One example is the Sorgenfrey line (i.e., $mathbb{R}$ with the lower-limit topology). See the following quetion and its answer for an outline showing why it is not orderable:

• Sorgenfrey line is not orderable

An interesting thing about this space is that it is "close" to being orderable. A topological space $X$ is called suborderable if it is homeomorphic to a subspace of an ordered space. We can show rather quickly that the Sorgenfrey line is suborderable.

Consider $mathbb{R} times { 0 , 1 }$ with the lexicographic order: $$(x,i) prec ( y , j ) Leftrightarrow begin{cases}
x < y, &text{or} \
x = y, i=0, j=1.
end{cases}$$
Then the Sorgenfrey line is homeomorphic to the subspace $mathbb{R} times { 1 }$ of the above ordered space.