# Study whether $\succsim$ represented by $u(x)=\lfloor x \rfloor$ is continuous [closed]

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.

• 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. – econ86 Sep 28 '19 at 21:18
• You might want to recheck your proof. Yes, $u(x)$ is not continuous but the underlying preferences are $x^n\succeq y^n$ implies $x^n\geq y^n$... – guest12382 Sep 28 '19 at 21:28
• Real analysis is a bit rusty, but I think $\lim_{n\to\infty} \lfloor 2-2/n \rfloor = 1$ so $y \sim x$. – Art Oct 2 '19 at 17:03
• I guess you wanted to write $$\alpha x+(1-\alpha)z=y \text{, so}\hspace{0.2em} u(\alpha x+(1-\alpha)z)=u(y)$$? – user17900 Oct 29 '19 at 14:48
• 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 $1=\lfloor 1\rfloor=\lfloor \lim_n 1-1/n\rfloor\leq 0=\lfloor 0\rfloor$. – Michael Greinecker Jun 24 at 23:37