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.


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.

| improve this answer | |
  • $\begingroup$ 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 '19 at 21:18
  • 1
    $\begingroup$ 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 '19 at 21:28
  • 3
    $\begingroup$ 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 '19 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 '19 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 $1=\lfloor 1\rfloor=\lfloor \lim_n 1-1/n\rfloor\leq 0=\lfloor 0\rfloor$. $\endgroup$ – Michael Greinecker Jun 24 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.