Binary Relations for Cobb-Douglas 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 $\alpha &lt; \beta$ 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?
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.