<p>I solved this: </p>

<p>note - I use $R$ to denote the binary relation as defined above.</p>

<p><strong>Completeness</strong>: </p>

<p>I claim that this $R$ is not complete for $\alpha \neq \beta$. To illustrate, I use an extreme case. </p>

<p>Let $\alpha \to 1$ and $\beta \to 0$ s.t. $f_{\alpha} \approx x_1$ and $f_{\beta} \approx x_2$  </p>

<p>Now, let [x,y] $\subset \mathbb{R}^2$ s.t. $x_1 &gt;&gt; y_1$ and $x_2&lt;&lt;y_2$</p>

<p>Now:  $f_{\alpha}(x) \approx x_1 &gt;&gt; y_1 \approx f_{\alpha}(y) $ </p>

<p>and:  $f_{\beta}(x) \approx x_2 &lt;&lt; y_2 \approx f_{\beta}(y)$</p>

<p>so $f_{\alpha}(x) &gt;&gt; f_{\alpha}(y)$ but $f_{\beta}(x)&lt;&lt;f_\beta(y) $</p>

<p>I have used extreme case reasoning to demonstrate, WLOG, that anytime $\alpha \neq \beta \space \exists [x,y] \subset \mathbb{R}_+^2 s.t.$ neither $xRy$ or $yRx$ </p>

<p><em>note_1</em> - The intuition here is very similar to a delta/epsilon proof.  For any combination of alpha and beta (where the two are not equal) I can pick x,y so that the above proof always hold.</p>

<p><em>note_2</em> - That the case of $\alpha = \beta$ is complete is trivial. I assume its inclusion here is not necessary. </p>

<p><strong>Transitivity</strong>: </p>

<p>Let $[x,y,z] \subset \mathbb{R}^2$ s.t. $xRy, yRx$ </p>

<p>I claim $xRz$ </p>

<p>since $Xry$, $yRx$ and by the properties of $\mathbb{R}_+^2$ </p>

<p>$$f_{\alpha}(x) \geq f_{\alpha}(y) \geq f_{\alpha}(z)$$
$$f_{\beta}(x) \geq f_{\beta}(y) \geq f_{\beta}(z)$$ </p>

<p>$$\implies xRz$$ </p>

<p>and $R$ is transitive. </p>

<p><strong>Continuity</strong></p>

<p>proven above so I will not repeat this part of the proof. </p>

CC BY-SA 3.0
2016-05-29T15:58:56.563Z
2016-05-29T15:58:56.563Z
123
<p>I will be using the following definition of continuity for binary relations.</p>

<p><strong>Definition 1:</strong> A binary relation $\mathcal{R}$ on $\mathbb{R}^2_+$ is continuous if for every $x$, the following sets are closed.
$$
USC_x = \{y\in \mathbb{R}^2_+|y\mathcal{R}x\}
$$
$$
LSC_x = \{y\in \mathbb{R}^2_+|x\mathcal{R}y\}
$$</p>

<p><strong>Definition 2:</strong> A set $Z$ is closed if $z_n$ is a sequence in $Z$ and $z_n\rightarrow z$, then $z\in Z$.</p>

<p>Now, let $\mathcal{R}$ be defined as in your question, i.e.
$$
x\mathcal{R} y\iff f_\alpha(x) \geq f_\alpha(y)\text{ and }f_\beta(x) \geq f_\beta(y)
$$
Let $x\in \mathbb{R}^2_+$. Let $x_n$ be a sequence in $USC_x$ converging to $x^\ast$. Then, for each $n$, 
$$
f_\alpha(x_n) \geq f_\alpha(x)\text{ and }f_\beta(x_n) \geq f_\beta(x).
$$
Since $f_\alpha$ and $f_\beta$ are continuous, we have 
$$
f_\alpha(x^\ast) \geq f_\alpha(x)\text{ and }f_\beta(x^\ast) \geq f_\beta(x).
$$
implying that $x^\ast\in USC_x$; hence, $USC_x$ is closed. We can show also that $LCS_x$ is closed by using identical arguments. This concludes that $\mathcal{R}$ is continuous.</p>

<p><strong>Remark 1:</strong> The arguments above provide a general strategy of analyzing statements regarding $\mathcal{R}$. Transitivity of $\mathcal{R}$ can be proven in a couple of simple steps. Your statement regarding this property is a bit vague, and should be more rigorous imho.</p>

<p><strong>Remark 2:</strong> Your explanation about completeness needs a clearly written argument as well. While it is true that $\alpha=\beta$ implies completeness of $\mathcal{R}$, the reasoning that makes it true is not exactly what you are proposing. It is also necessary to show that $\alpha\neq \beta$ implies that $\mathcal{R}$ wouldn't be complete (I believe that this should be the case). This would require a bit more work on your part. Let me know if you run into troubles in this.</p>

<p><strong>Bottomline:</strong> In my sincerely humble opinion, you should exercise a lot with proving things. Maybe grab a book on introduction to abstract mathematics and such. </p>

CC BY-SA 3.0
2015-10-11T18:56:23.513Z
I think I need to edit my question to make some things clear. For example, I do not intend for anything in my question to represent an attempt at a formal proof.
123
CC BY-SA 3.0
2015-10-11T19:04:33.917Z
I've edited the question. Hopefully I've cleared up a few things. I am looking for arguments about combinations of $\alpha , \beta$. Thanks for the initial answer.
123
CC BY-SA 3.0
2015-10-10T23:44:20.580Z
2015-10-11T02:07:56.687Z
2015-10-11T02:07:56.687Z
ramazan
<p>A few thoughts:</p>

<p>For completeness, I think you can make a slightly stronger case that for any $x_1  \geq y_1, x_2 \geq y_2$, this satisfies completeness $\forall \alpha, \beta$.</p>

<p>$$x_1^\alpha x_2^{1-\alpha} \geq y_1^\alpha y_2^{1-\alpha}$$
$$x_1^\beta x_2^{1-\beta} \geq y_1^\beta y_2^{1-\beta}$$</p>

<p>For continuity, there are two definitions you can use, one with sequences and one with epsilon-balls around a point in your binary relation.</p>

CC BY-SA 3.0
2015-10-11T19:07:04.097Z
It is alpha and beta that we are considering and not combinations of x,y. I included that comment originally just as a thought. However, we are looking at completeness based on combinations of alpha,beta and not the vectors x,y with fixed alpha,beta. My question was unclear. So, we need xRy or yRx for all x,y (which alpha,beta make this possible?). Hope I've cleared it up and sorry for the confusion.
123
CC BY-SA 3.0
2015-10-11T19:53:38.620Z
Ah I see you've edited the question. That makes a little more sense now, cool beans and best of luck.
Kitsune Cavalry
CC BY-SA 3.0
2015-10-10T23:46:37.947Z
2015-10-10T23:46:37.947Z
Kitsune Cavalry
<p>I am reviewing old midterms to prepare for my upcoming midterm and ran across this question: </p>

<p>Let $\alpha , \beta \in (0,1)$. Now, let $f_{\alpha}$ and $f_{\beta}$ on $\mathbb{R^2}$ be defined as $f_{\alpha}(x)=x_1^{\alpha} x_2^{1 - \alpha}$ and $f_{\beta}(x)=x_1^{\beta} x_2^{1 - \beta}$</p>

<p>Now, let R be a binary relation on $\mathbb{R_x^2}$. Let ${x,y} \subset R_+^2$ We have that: $$xRy \leftrightarrow f_{\alpha}(x) \geq f_{\alpha}(y) \land  f_{\beta}(x) \geq f_{\beta}(y)$$</p>

<p>For which combinations of $\alpha$ and $\beta$ is this binary relation complete, for which combinations is it transitive and for which combinations is it continuous. </p>

<p><strong>My thoughts</strong> </p>

<ul>
<li><p>It seems this can only be complete if $\alpha=\beta$ but I can't quite finish the proof whenever I proceed WLOG with $\alpha &lt; \beta$ Can anyone here offer an attempt at formally proving that this is complete iff $\alpha = \beta$ ?  </p></li>
<li><p>I think that anytime we have xRy, yRz we will necessarily have xRz. That is. And so, I think this is transitive for all combinations of $\alpha,\beta$. My proof involves using the well-ordering of the reals and the definition given for this particular relation. If anyone thinks that this is not true for all $\alpha,\beta$ please let me know why/how. </p></li>
<li><p>I know what continuity is and how to prove it. However, I am not sure for which combinations of $\alpha,\beta$ this relation is continuous. I suspect it is continuous for all combinations of $\alpha,\beta$. Is this true? If so, can you prove it?   </p></li>
</ul>

<preferences>
Binary Relations for Cobb-Douglas
123
# Binary Relations for Cobb-Douglas

(Asked by: username_0 on 2015-10-10)

I am reviewing old midterms to prepare for my upcoming midterm and ran across this question:

Let $\alpha , \beta \in (0,1)$. Now, let $f\_{\alpha}$ and $f\_{\beta}$ on $\mathbb{R^2}$ be defined as $f\_{\alpha}(x)=x\_1^{\alpha} x\_2^{1 - \alpha}$ and $f\_{\beta}(x)=x\_1^{\beta} x\_2^{1 - \beta}$

Now, let R be a binary relation on $\mathbb{R\_x^2}$. Let ${x,y} \subset R\_+^2$ We have that: $$xRy \leftrightarrow f\_{\alpha}(x) \geq f\_{\alpha}(y) \land f\_{\beta}(x) \geq f\_{\beta}(y)$$

For which combinations of $\alpha$ and $\beta$ is this binary relation complete, for which combinations is it transitive and for which combinations is it continuous.

**My thoughts**

* It seems this can only be complete if $\alpha=\beta$ but I can't quite finish the proof whenever I proceed WLOG with $\alpha < \beta$ Can anyone here offer an attempt at formally proving that this is complete iff $\alpha = \beta$ ?
* I think that anytime we have xRy, yRz we will necessarily have xRz. That is. And so, I think this is transitive for all combinations of $\alpha,\beta$. My proof involves using the well-ordering of the reals and the definition given for this particular relation. If anyone thinks that this is not true for all $\alpha,\beta$ please let me know why/how.
* I know what continuity is and how to prove it. However, I am not sure for which combinations of $\alpha,\beta$ this relation is continuous. I suspect it is continuous for all combinations of $\alpha,\beta$. Is this true? If so, can you prove it?

**Tags:** preferences

---

## 3 answers

### Answer by username_0 on 2016-05-29. Score: 0 | Accepted answer

I solved this:

note - I use $R$ to denote the binary relation as defined above.

**Completeness**:

I claim that this $R$ is not complete for $\alpha \neq \beta$. To illustrate, I use an extreme case.

Let $\alpha \to 1$ and $\beta \to 0$ s.t. $f\_{\alpha} \approx x\_1$ and $f\_{\beta} \approx x\_2$

Now, let [x,y] $\subset \mathbb{R}^2$ s.t. $x\_1 >> y\_1$ and $x\_2<

Now: $f\_{\alpha}(x) \approx x\_1 >> y\_1 \approx f\_{\alpha}(y) $

and: $f\_{\beta}(x) \approx x\_2 << y\_2 \approx f\_{\beta}(y)$

so $f\_{\alpha}(x) >> f\_{\alpha}(y)$ but $f\_{\beta}(x)<

I have used extreme case reasoning to demonstrate, WLOG, that anytime $\alpha \neq \beta \space \exists [x,y] \subset \mathbb{R}\_+^2 s.t.$ neither $xRy$ or $yRx$

*note\_1* - The intuition here is very similar to a delta/epsilon proof. For any combination of alpha and beta (where the two are not equal) I can pick x,y so that the above proof always hold.

*note\_2* - That the case of $\alpha = \beta$ is complete is trivial. I assume its inclusion here is not necessary.

**Transitivity**:

Let $[x,y,z] \subset \mathbb{R}^2$ s.t. $xRy, yRx$

I claim $xRz$

since $Xry$, $yRx$ and by the properties of $\mathbb{R}\_+^2$

$$f\_{\alpha}(x) \geq f\_{\alpha}(y) \geq f\_{\alpha}(z)$$
$$f\_{\beta}(x) \geq f\_{\beta}(y) \geq f\_{\beta}(z)$$

$$\implies xRz$$

and $R$ is transitive.

**Continuity**

proven above so I will not repeat this part of the proof.


---

### Answer by username_1 on 2015-10-10. Score: 2

I will be using the following definition of continuity for binary relations.

**Definition 1:** A binary relation $\mathcal{R}$ on $\mathbb{R}^2\_+$ is continuous if for every $x$, the following sets are closed.
$$
USC\_x = \{y\in \mathbb{R}^2\_+|y\mathcal{R}x\}
$$
$$
LSC\_x = \{y\in \mathbb{R}^2\_+|x\mathcal{R}y\}
$$

**Definition 2:** A set $Z$ is closed if $z\_n$ is a sequence in $Z$ and $z\_n\rightarrow z$, then $z\in Z$.

Now, let $\mathcal{R}$ be defined as in your question, i.e.
$$
x\mathcal{R} y\iff f\_\alpha(x) \geq f\_\alpha(y)\text{ and }f\_\beta(x) \geq f\_\beta(y)
$$
Let $x\in \mathbb{R}^2\_+$. Let $x\_n$ be a sequence in $USC\_x$ converging to $x^\ast$. Then, for each $n$,
$$
f\_\alpha(x\_n) \geq f\_\alpha(x)\text{ and }f\_\beta(x\_n) \geq f\_\beta(x).
$$
Since $f\_\alpha$ and $f\_\beta$ are continuous, we have
$$
f\_\alpha(x^\ast) \geq f\_\alpha(x)\text{ and }f\_\beta(x^\ast) \geq f\_\beta(x).
$$
implying that $x^\ast\in USC\_x$; hence, $USC\_x$ is closed. We can show also that $LCS\_x$ is closed by using identical arguments. This concludes that $\mathcal{R}$ is continuous.

**Remark 1:** The arguments above provide a general strategy of analyzing statements regarding $\mathcal{R}$. Transitivity of $\mathcal{R}$ can be proven in a couple of simple steps. Your statement regarding this property is a bit vague, and should be more rigorous imho.

**Remark 2:** Your explanation about completeness needs a clearly written argument as well. While it is true that $\alpha=\beta$ implies completeness of $\mathcal{R}$, the reasoning that makes it true is not exactly what you are proposing. It is also necessary to show that $\alpha\neq \beta$ implies that $\mathcal{R}$ wouldn't be complete (I believe that this should be the case). This would require a bit more work on your part. Let me know if you run into troubles in this.

**Bottomline:** In my sincerely humble opinion, you should exercise a lot with proving things. Maybe grab a book on introduction to abstract mathematics and such.

> **Comments:**
> **username_0**: I think I need to edit my question to make some things clear. For example, I do not intend for anything in my question to represent an attempt at a formal proof.  
> **username_0**: I've edited the question. Hopefully I've cleared up a few things. I am looking for arguments about combinations of $\alpha , \beta$. Thanks for the initial answer.  

---

### Answer by username_2 on 2015-10-10. Score: 0

A few thoughts:

For completeness, I think you can make a slightly stronger case that for any $x\_1 \geq y\_1, x\_2 \geq y\_2$, this satisfies completeness $\forall \alpha, \beta$.

$$x\_1^\alpha x\_2^{1-\alpha} \geq y\_1^\alpha y\_2^{1-\alpha}$$
$$x\_1^\beta x\_2^{1-\beta} \geq y\_1^\beta y\_2^{1-\beta}$$

For continuity, there are two definitions you can use, one with sequences and one with epsilon-balls around a point in your binary relation.

> **Comments:**
> **username_0**: It is alpha and beta that we are considering and not combinations of x,y. I included that comment originally just as a thought. However, we are looking at completeness based on combinations of alpha,beta and not the vectors x,y with fixed alpha,beta. My question was unclear. So, we need xRy or yRx for all x,y (which alpha,beta make this possible?). Hope I've cleared it up and sorry for the confusion.  
> **username_2**: Ah I see you've edited the question. That makes a little more sense now, cool beans and best of luck.  

---

