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I asked on Math SE a bit about level sets here.

Based on what I learned, it seems like we usually assume a level set $L(f)$ and its function $f$ defined on $\Bbb{R}^n$ have the following properties:

  1. $f$ is continuous
  2. Thus, the curves in the level set are closed curves. (not sure why)
  3. All closed curves are "converging" to a point. (not sure why)
  4. This point will be a maximum for the function: the derivative in the point will be zero, since if it was not then the point would be a 1-manifold, which is absurd.
  5. $\nabla f \neq 0$ on the rest of the points

I don't understand how we know a preference relation on a consumption bundle has a utility function representation such that these properties hold. According to my professor, we use level sets to describe indifference curves.

Therefore, if indifference curves are described by a level set, that level set must have the properties listed above. Note, for economics, we are considering a function $U(\mathbf{x})$, where $\mathbf{x}=(x_1,...,x_n)$, and the level set $L(U)$. Also, note $U: \Bbb{R}^n\rightarrow \Bbb{R}$.

My Question:

In economic terms, what assumption do we have to make to ensure, $U$, $L(U)$ have these properties to create a level set used for indifference curves?

Ideally, it would be great to have an enumerated list of the assumptions (ie 1,2,3).

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Look take a rational preference $\succeq$ defined on X, endow X with some metric (thus a assume it is a metric space). Assume also X is separable (e.g. $\mathbb{R}^n$ satisfies this conditions but is more general). Now we let $\succeq$ be (i) rational (complete, transitive), (ii) continuous (it means that if $x^n\rightarrow x$ , $y^n\rightarrow y$ and $x^n \succeq y^n$ $\forall n$, then $x \succeq y$. Under this assumptions, then it can be represented by a continuous utility function. 1. u is continuous by the above conditions. 2. Curves in the level set are closed. It follows from the fact that the indifference relation is a closed set and the continuous utility representation: take $x^n \sim y\quad \forall n$ by continuity of the preferences $x \sim y$, now this implies that $u(x^n)$ is a sequence on the level curve $u(x^n)=u(y)$ for fixed y and by above $u(x)\rightarrow u(y)$ thus making it closed. The other properties are deeper in the sense that one needs more assumptions, for instance to have the gradient of u it has to be differentiable, uniqueness of the extremum needs some sort of convexity of the level curves and so on.

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Level sets are always well defined. No property of utility functions has to be assumed. For any utility level $\bar u$ just define the level set to be $\{x\in \mathbb{R}^n|U(x)=\bar u\}$. No property of $U$ must be assumed. This is well-defined regardless of the properties of $U$. Properties of $U$ are useful if one wants to derive specific properties of the level sets, but are not needed to define them.

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  • $\begingroup$ What if we aren't using $\Bbb{R}^n$? Or is that always assumed? $\endgroup$ – Stan Shunpike Apr 6 '15 at 17:15
  • $\begingroup$ Even then, the above definition should is fine. Just replace in the above definition $\mathbb{R}^n$ by whatever set $X$ the agent chooses from. $\endgroup$ – TMB Apr 6 '15 at 17:19
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As stated by TMB, levels sets of points at which the same utility is achieved ($\{x\in \mathbb{R}^n|U(x)=\bar u\}$). These depend on pretty much nothing, except that the ensemble has an equality relation - not much to ask really.

1 and 2. The curves are defined as $f^{-1} (\bar u)$ and $\{\bar u\}$ is closed, so the antecedent by a continuous function of $\{\bar u\}$ is closed too.

3,4 and 5 sounds like bullshit to me, maybe you can give us the reference of your course or a course refereeing to the same assumption? How do you define a maximum in $\Bbb{R}^n$?

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