I am reading a proof and having a hard time understanding on some parts.
In math language, the proof is trying to show the existence of continuous preorder $\succsim$ (i.e. preserved under limits, complete, transitive) on $X\subset\mathbb{R^L_+}$ as a real-valued function.
Im economics language, want to show rational and continuous preference relation on $X\subset\mathbb{R^L_+}$ can be represented as real-valued utility function. (Reference: Micro Theory MWG Ch.3)
Here's the part I have a question on:
Show for every bundle $x\in X\subset\mathbb{R^L_+}$, $\exists$ nonnegative real scalar $k$ such that $k(x)e\sim x$ where $e=(1,...,1)$.
Strategy for Proof:
Know $\succsim$ is continuous on X. So the upper contour and lower contour (i.e. $U(x)=\{y\in X:y\succsim x\}$, $L(x)=\{y\in X:x\succsim y\}$) are closed. Then, $K^+=\{k\in\mathbb{R}:ke\succsim x\}$, $K^-=\{k\in\mathbb{R}:x\succsim ke\}$ are also nonempty and closed. By completeness of $\succsim$, $\mathbb{R_+}\subset(K^+\cup K^-)$. With all the above with the fact that $\mathbb{R_+}$ is connected, there is some nonnegative real number $k$ such that $k(x)e\sim x$.
My questions are:
- Why does completeness of $\succsim$ imply the set of $K^+,K^-$ is the cover of the nonnegative real line?
- Is $\mathbb{R_+}=(K^+\cup K^-)$ also true?
- Why is it not obvious that $K^+,K^-$ from the beginning? Suppose $X\subset\mathbb{R_+}$ so $x$ is just a real number. $e$ will be 1. then, isn't it obvious that $K^+=\{k\in\mathbb{R}:ke\geq x\}$, $K^-=\{k\in\mathbb{R}:x\geq ke\}$ are nonempty and closed AND their intersection is the point $x$? I don't understand why the authors of the proof goes over these hoops of cover, connectedness of $\mathbb{R_+} and so forth to prove this.
If anything is unclear, please comment. I will try my best to clarify, and appreciate your 2 cent and help.
The pre-order being complete has no topological meaning, but purely set-theoretic. It means that for any two points $x,y$ in the domain of the pre-order, we must have $x \succsim y$ or $y \succsim x$ (or possibly both here, because we have a pre-order, so we can have both at the same time (invariant goods (?), or some such thing, economics is not my field, though I did do some of it).
Look at the definitions, where $x$ is now fixed for the rest of the proof (and $x \in X$, the domain of $\succsim$): $K^+ = \{k \in \mathbb{R}: ke \succsim x\}$ and $K^- = \{k \in \mathbb{R}: x \succsim ke \}$. So we look at all scalars $k$, compute $ke$ where $e = (1,1,\ldots)$ and compare it to the fixed point $x$, using the preference relation $\succsim$. Now as $\succsim$ is complete we must have that at least one of $x \succsim ke$ or $ke \succsim x$ holds, and we put $k$ into $K^+$ when the one thing happens, and in $K^-$ when the other thing happens, and in both if both happen. If we know that $ke$ is in $X$ for every $k$, we would know that $\mathbb{R} = K^+ \cup K^-$. But I think that in these kinds of economic settings, $X$ is a subset of the positive vectors in $\mathbb{R}^\mathbb{L}$, and $X$ is the domain of the ordering $\succsim$, so in this case I think that te assumption on $X$ is that we do know that $ke \in X$ for every positive $k$, but not necessarily for every $k$. So we can, in that case, only say that $\mathbb{R}^+ \subseteq K^+ \cup K^-$, and cannot say about the other $k$, as we do not know that $ke$ is even in $X$, so that comparison with $x$ is defined.
A special case of the reals does not show anything. The whole point is that $X$ can be any subset (often closed and convex though, and not quite arbitrary) and $\succsim$ any relation satisfying some conditions. The theory is to find the most general, useful, conditions that will guarantee some interesting theorems on $X$. Taking a trivial special case defeats the whole purpose. Here we can, from just saying it's complete (which is part of being a rational preference, IIRC) etc. we can prove things about the relation $\succsim$. So we need to use general facts about connectedness etc. to do that. The connectedness of $\mathbb{R}^+$ and the closedness of $K^+$ and $K^-$ (which follows from $\succsim$ being "continuous") is all that's needed to find a point in $K^+ \cap K^-$ which is the $k(x)$ that is calimed to exist.
I suggest trying other relations (like the lexicographic order) in the plane etc. to get a feel for what more general $\succsim$ relations can look like. If you use a good text (I don't know yours) this would have been done at the beginning, to motivate the whole theory in the first place. That's what I would do, anyway.