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By Debreu's theorem of ordinal utility, any continuous weak order on $X$ is represented with a continuous utility function, if $X$ is a second countable or connected separable topological space.

My question is, does the theorem require $X$ to be endowed with T0 or T1 or T2 (Hausdorff) topology?

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No. However, the problem can be reduced to representing preferences on a Hausdorff space. Instead of trying to represent a complete preorder on a set, one can try to represent linear orders on the space of indifference classes. On the latter space, Hausdorffness is automatic if the preference relation is continuous. To see this, let $X$ be a topological space and $\succeq$ be a complete preorder on $X$ such that for every $x\in X$, the sets $\{y\in X\mid y\succeq x\}$ and $\{y\in X\mid x\succeq y\}$ are closed. Define $\succ$, $\sim$ in the usual way and endow $X/\sim$ with the quotient topology. Then $X/\sim$ is a Hausdorff space. Write $[x]$ for the equivalence class corresponding to $x$ and view $\succeq$ by abuse of notation as a linear order on $X/\sim$. Now, let $[x]\neq [y]$. Without loss of generality, let $x\prec y$.

There are two cases: Either there exists $z\in X$ such that $x\prec z\prec y$ or there does not. If there exists such a $z$, then the sets $\{v\mid v\prec z\}$ and $\{v\mid v\succ z\}$ are open neighborhoods of $x$ and $y$, respectively. By the definition of the quotient topology, the sets $\{[v]\mid v\prec z\}$ and $\{[v]\mid v\succ z\}$ are open neighborhoods of $[x]$ and $[y]$, respectively. If there exists no such $z$, then the sets $\{v\mid v\prec y\}$ and $\{v\mid v\succ x\}$ are open neighborhoods of $x$ and $y$, respectively. Again by the definition of the quotient topology, the sets $\{[v]\mid v\prec y\}$ and $\{[v]\mid v\succ x\}$ are open neighborhoods of $[x]$ and $[y]$, respectively.

It follows that $X/\sim$ is a Hausdorff space and the problem can be reduced to representing a preference ordering on a Hausdorff space.

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    $\begingroup$ Wow where did you learn all those things? So impressive. $\endgroup$
    – High GPA
    May 14 at 16:35
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    $\begingroup$ Thank you; one can learn a lot over time. $\endgroup$ May 14 at 17:21
  • $\begingroup$ In the second part of the second paragraph, if we assume that $z$ does not exist and $x\prec y$, then you constructed an open neighborhood for $x$: $\{v|v\prec y\}$. I am guessing that such construction might also be used in the first case when $z$ does exist. Is my understanding correct? $\endgroup$
    – High GPA
    May 26 at 20:57
  • $\begingroup$ Then the sets would not be disjoint; they would both contain $z$. $\endgroup$ May 27 at 4:48

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