<p>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 <span class="math-container">$X$</span> be a topological space and <span class="math-container">$\succeq$</span> be a complete preorder on <span class="math-container">$X$</span> such that for every <span class="math-container">$x\in X$</span>, the sets <span class="math-container">$\{y\in X\mid y\succeq x\}$</span> and <span class="math-container">$\{y\in X\mid x\succeq y\}$</span> are closed. Define <span class="math-container">$\succ$</span>, <span class="math-container">$\sim$</span> in the usual way and endow <span class="math-container">$X/\sim$</span> with the <a href="https://en.wikipedia.org/wiki/Quotient_space_(topology)" rel="noreferrer">quotient topology</a>. Then <span class="math-container">$X/\sim$</span> is a Hausdorff space. Write <span class="math-container">$[x]$</span> for the equivalence class corresponding to <span class="math-container">$x$</span> and view <span class="math-container">$\succeq$</span> by abuse of notation as a linear order on <span class="math-container">$X/\sim$</span>.
Now, let <span class="math-container">$[x]\neq [y]$</span>. Without loss of generality, let <span class="math-container">$x\prec y$</span>.</p>
<p>There are two cases: Either there exists <span class="math-container">$z\in X$</span> such that <span class="math-container">$x\prec z\prec y$</span> or there does not. If there exists such a <span class="math-container">$z$</span>, then the sets <span class="math-container">$\{v\mid v\prec z\}$</span> and <span class="math-container">$\{v\mid v\succ z\}$</span> are open neighborhoods of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, respectively. By the definition of the quotient topology, the sets <span class="math-container">$\{[v]\mid v\prec z\}$</span> and <span class="math-container">$\{[v]\mid v\succ z\}$</span> are open neighborhoods of <span class="math-container">$[x]$</span> and <span class="math-container">$[y]$</span>, respectively. If there exists no such <span class="math-container">$z$</span>, then the sets <span class="math-container">$\{v\mid v\prec y\}$</span> and <span class="math-container">$\{v\mid v\succ x\}$</span> are open neighborhoods of <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, respectively. Again by the definition of the quotient topology, the sets <span class="math-container">$\{[v]\mid v\prec y\}$</span> and <span class="math-container">$\{[v]\mid v\succ x\}$</span> are open neighborhoods of <span class="math-container">$[x]$</span> and <span class="math-container">$[y]$</span>, respectively.</p>
<p>It follows that <span class="math-container">$X/\sim$</span> is a Hausdorff space and the problem can be reduced to representing a preference ordering on a Hausdorff space.</p>

CC BY-SA 4.0
2022-05-14T16:35:02.783Z
Wow where did you learn all those things? So impressive.
High GPA
CC BY-SA 4.0
2022-05-14T17:21:20.763Z
Thank you; one can learn a lot over time.
Michael Greinecker
CC BY-SA 4.0
2022-05-26T20:57:19.060Z
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?
High GPA
CC BY-SA 4.0
2022-05-27T04:48:48.627Z
Then the sets would not be disjoint; they would both contain $z$.
Michael Greinecker
CC BY-SA 4.0
2022-05-14T16:10:36.153Z
2022-05-14T16:10:36.153Z
Michael Greinecker
<p>By Debreu's theorem of ordinal utility, any continuous weak order on <span class="math-container">$X$</span> is represented with a continuous utility function, if <span class="math-container">$X$</span> is a second countable or connected separable topological space.</p>
<p>My question is, does the theorem require <span class="math-container">$X$</span> to be endowed with T0 or T1 or T2 (Hausdorff) topology?</p>

<microeconomics><utility><preferences><decision-theory><theory>
Does Debreu's representation theorem of ordinal utility require Hausdorff topology?
High GPA
# Does Debreu's representation theorem of ordinal utility require Hausdorff topology?

(Asked by: username_0 on 2022-05-14)

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?

**Tags:** microeconomics, utility, preferences, decision-theory, theory

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## 1 answers

### Answer by username_1 on 2022-05-14. Score: 7 | Accepted answer

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](https://en.wikipedia.org/wiki/Quotient_space_(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.

> **Comments:**
> **username_0**: Wow where did you learn all those things? So impressive.  
> **username_1**: Thank you; one can learn a lot over time.  
> **username_0**: 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?  
> **username_1**: Then the sets would not be disjoint; they would both contain $z$.  

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