Using the following definition of continuity: $\succsim$ is continuous if for any bundles $x,y,z$ such that x$\succ$y$\succ$z, there exists $\alpha \in (0,1)$ such that $\alpha x + (1-\alpha)z \sim y$.

I am unable to show continuity/not continuity using this definition.


1 Answer 1


Are you sure that the definition you want?

With your definition then yes its true.

You should show that $x\succ y\succ z\implies x>y>z$. Hence, you can always find a $\alpha$ such that $\alpha x+(1-\alpha)y=z$, so $u(\alpha x+(1-\alpha)y)=u(z)$ which then implies $\alpha x+(1-\alpha)y\sim z$.

Perhaps the definition you want to use is for all $x^n$ and $y^n$ are two sequences with $x^n\to x$ and $y^n\to y$ and $x^n\succeq y^n$ for all $n$ then $x\succeq y$? But maybe not, I dont want to presume.

  • $\begingroup$ That is exactly my concern. I used the $x^n$ and $y^n$ definition and got that it is not continuous. Then I used my initial definition and got continuity. My lecture notes claim that the two definitions are equivalent, and that is why I posted the question. $\endgroup$
    – econ86
    Sep 28, 2019 at 21:18
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    $\begingroup$ You might want to recheck your proof. Yes, $u(x)$ is not continuous but the underlying preferences are $x^n\succeq y^n$ implies $x^n\geq y^n$... $\endgroup$
    – guest12382
    Sep 28, 2019 at 21:28
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    $\begingroup$ Real analysis is a bit rusty, but I think $\lim_{n\to\infty} \lfloor 2-2/n \rfloor = 1$ so $y \sim x$. $\endgroup$
    – Art
    Oct 2, 2019 at 17:03
  • $\begingroup$ I guess you wanted to write $$\alpha x+(1-\alpha)z=y \text{, so}\hspace{0.2em} u(\alpha x+(1-\alpha)z)=u(y)$$? $\endgroup$
    – user17900
    Oct 29, 2019 at 14:48
  • $\begingroup$ It is certainly not continuous under the usual sequence definition. For $n>1$, we get $\lfloor 1-1/n\rfloor=0\leq 0=\lfloor 0\rfloor$. If the relation would be continuous, we would get $1=\lfloor 1\rfloor=\lfloor \lim_n 1-1/n\rfloor\leq 0=\lfloor 0\rfloor$. $\endgroup$ Jun 24, 2020 at 23:37

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