<p>Yes it is: <br>
If direction
$$
x \succ y \Rightarrow x \not \precsim y \Rightarrow u(x) &gt; u(y).
$$
Only if direction: <br>
For all $x, y  \in X$, 
$$
x \succsim y \iff u(x) \geq u(y)
$$
implies
$$
x \sim y \iff u(x) = u(y).
$$
Also
$$
u(x) &gt; u(y) \Rightarrow u(x) \geq u(y) \Rightarrow  x \succsim y ,
$$
$$
u(x) &gt; u(y) \Rightarrow u(x) \not = u(y) \Rightarrow  x \not\sim y.
$$
and
$$
x \succsim y \mbox{ AND } x \not\sim y \Rightarrow x \succ y.
$$</p>

CC BY-SA 3.0
2015-10-13T21:42:05.113Z
2015-10-13T21:54:33.150Z
2015-10-13T21:54:33.150Z
Giskard
<p>Suppose I have a rational preference relation $\succsim$ on some consumption set $X$. </p>

<p>Suppose also that there is a utility function $u:X \to \mathbb{R}$ representing $\succsim$. </p>

<p>Definition: A function $u: X \to \mathbb{R}$ is a utility function representing preference relation $\succsim$ if, for all $x, y  \in X$, $$x \succsim y \iff u(x) \geq u(y)$$</p>

<p>Is it possible to prove that $x \succ y \iff u(x) &gt; u(y)$ without a continuity condition on $\succsim$?</p>

<p>My intuition says no, but am having difficult finding a suitable counter example. Any help is appreciated.</p>

<consumer-theory><utility><preferences>
Strict preference relations and utility representations
möbius
# Strict preference relations and utility representations

(Asked by: username_0 on 2015-10-13)

Suppose I have a rational preference relation $\succsim$ on some consumption set $X$.

Suppose also that there is a utility function $u:X \to \mathbb{R}$ representing $\succsim$.

Definition: A function $u: X \to \mathbb{R}$ is a utility function representing preference relation $\succsim$ if, for all $x, y \in X$, $$x \succsim y \iff u(x) \geq u(y)$$

Is it possible to prove that $x \succ y \iff u(x) > u(y)$ without a continuity condition on $\succsim$?

My intuition says no, but am having difficult finding a suitable counter example. Any help is appreciated.

**Tags:** consumer-theory, utility, preferences

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## 1 answers

### Answer by username_1 on 2015-10-13. Score: 7 | Accepted answer

Yes it is:   
If direction
$$
x \succ y \Rightarrow x \not \precsim y \Rightarrow u(x) > u(y).
$$
Only if direction:   
For all $x, y \in X$,
$$
x \succsim y \iff u(x) \geq u(y)
$$
implies
$$
x \sim y \iff u(x) = u(y).
$$
Also
$$
u(x) > u(y) \Rightarrow u(x) \geq u(y) \Rightarrow x \succsim y ,
$$
$$
u(x) > u(y) \Rightarrow u(x) \not = u(y) \Rightarrow x \not\sim y.
$$
and
$$
x \succsim y \mbox{ AND } x \not\sim y \Rightarrow x \succ y.
$$


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