<p>Are you sure that the definition you want? </p>

<p>With your definition then yes its true.</p>

<p>You should show that <span class="math-container">$x\succ y\succ z\implies x&gt;y&gt;z$</span>. Hence, you can always find a <span class="math-container">$\alpha$</span> such that <span class="math-container">$\alpha x+(1-\alpha)y=z$</span>, so <span class="math-container">$u(\alpha x+(1-\alpha)y)=u(z)$</span> which then implies <span class="math-container">$\alpha x+(1-\alpha)y\sim z$</span>.</p>

<p>Perhaps the definition you want to use is for all <span class="math-container">$x^n$</span> and <span class="math-container">$y^n$</span> are two sequences with <span class="math-container">$x^n\to x$</span> and <span class="math-container">$y^n\to y$</span> and <span class="math-container">$x^n\succeq y^n$</span> for all <span class="math-container">$n$</span> then <span class="math-container">$x\succeq y$</span>? But maybe not, I dont want to presume.</p>

CC BY-SA 4.0
2019-09-28T21:18:23.420Z
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
CC BY-SA 4.0
2019-09-28T21:28:11.547Z
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
CC BY-SA 4.0
2019-10-02T17:03:00.937Z
Real analysis is a bit rusty, but I think $\lim_{n\to\infty} \lfloor 2-2/n \rfloor = 1$ so $y \sim x$.
Art
CC BY-SA 4.0
2020-06-24T23:37:56.903Z
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
CC BY-SA 4.0
2019-09-28T21:04:54.320Z
2019-09-28T21:04:54.320Z
guest12382
<p>Using the following definition of continuity: <span class="math-container">$\succsim$</span> is continuous if for any bundles <span class="math-container">$x,y,z$</span> such that x<span class="math-container">$\succ$</span>y<span class="math-container">$\succ$</span>z, there exists <span class="math-container">$\alpha \in (0,1)$</span> such that <span class="math-container">$\alpha x + (1-\alpha)z \sim y$</span>.</p>

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

<microeconomics><utility><preferences>
Study whether $\succsim$ represented by $u(x)=\lfloor x \rfloor$ is continuous
econ86
# Study whether $\succsim$ represented by $u(x)=\lfloor x \rfloor$ is continuous

(Asked by: username_0 on 2019-09-28)

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.

**Tags:** microeconomics, utility, preferences


---

## 1 answers

### Answer by username_1 on 2019-09-28. Score: 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.

> **Comments:**
> **username_0**: 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.  
> **username_1**: You might want to recheck your proof. Yes, $u(x)$ is not continuous but the underlying preferences are $x^n\succeq y^n$ implies $x^n\geq y^n$...  
> **username_2**: Real analysis is a bit rusty, but I think $\lim_{n\to\infty} \lfloor 2-2/n \rfloor = 1$ so $y \sim x$.  
> **username_3**: 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 $1=\lfloor 1\rfloor=\lfloor \lim_n 1-1/n\rfloor\leq 0=\lfloor 0\rfloor$.  

---

