I am required to show that if: $f:R^L \rightarrow R$ is a strictly monotonic function and $u:R^L \rightarrow R$ is a utility function representing a preference relation $\succsim$, then the function $v:R^L \rightarrow R$ defined by $v(x) = u(f(x))$ does not always represent the preference $\succsim$.

For such a proposition, can I assume that if $x>y$, then $u(x) > u(y)$. The context is that I am trying to prove this by contradiction and my contradiction hinges on $x >y \implies u(x) > u(y)$, where x,y are elements of $R^L$.

  • $\begingroup$ If the domain of $u$ is $R^L$, and the codomain of $f$ is $R$, then $u(f(x))$ doesn't make sense. Also, if $x$ and $y$ are in $R^L$, then $x>y$ doesn't make sense unless you've defined an ordering on $R^L$. $\endgroup$ Oct 17, 2021 at 4:55

1 Answer 1


You have to show that something does not hold universally true. To show this, you just have to show that there is at least one exception- a counterexample. For this counterexample, you can make any assumption that does not contradict the assumptions of your problem.

  • $\begingroup$ This makes sense. Thanks! $\endgroup$
    – Kinno
    Oct 16, 2021 at 22:59
  • 1
    $\begingroup$ In general can you assume that more is better, as in more quantity brings you more utility? $\endgroup$
    – Rumi
    Oct 16, 2021 at 23:00
  • $\begingroup$ Yes, and you can even find one explicit utility function and one explicit increasing function $f$ that gives you a counterexample. $\endgroup$ Oct 16, 2021 at 23:55

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