CLASS THUS FAR

After today I will have completed 1 week of classes. I still don't think I have a good indication of what the semester is going to be like. 1 of my classes has yet to assign a problem set, which when it does come, I'm sure it will be hefty. I still haven't decided what to do about the never-ending story that is my dilemma between microeconomics in the Econ Department (super-hefty) or microeconomics in the Woodrow Wilson School (not so hefty). Thus far, I've been attending both. The WWS version makes sense. The Econ version makes sense while I am in class, but the moment I leave class and look at the problem sets my brain melts. Most of the problems are proofs of some sort or another. I've not done a lot of proving in the past, so here's my assessment of proofs thus far.

Step 1: Write down something that makes absolutely no sense or is completely obvious.

Example: Show that {#<<|*(c<=3)} is [(0)-$3-->R++] OR Show that 1 is 1.

Step 2: Stare blankly for several hours and wonder either:

a) Did the person fall asleep on their keyboard to create this proof?
or
b) Why is this not already completely obvious?

Step 3: Use lots of symbols and definitely more symbols than words.

I recommend < > # $ * ~ and {}. These seem to be popular choices.

Step 4: The few words you use must come from the following set:

assume, suppose, implies, exist, therefore, we have, & as desired

Step 5: Mix symbols with words. Never use actual numbers. Never use the word "example".

Step 6: Write the same thing at the end that appears at the beginning.


Done! You now have proven something. Granted, you are still confused and most likely frustrated, but take a deep, satisfying breath and know you've filled lots of white space.

In case you don't belive me, here is a sample from the first day's material.

Let B` = A ∪ B. Then, WARP yields c(A) ∩ B` != ∅ implies c(B`) ∩ A ⊂ c(A).
But since c is a choice function, we have c(A) ∩ B` = c(A) ∩ (A ∪ B) != ∅ and therefore
c(A∪B) ∩A = c(B`) ∩A ⊂ c(A) as desired. Next, assume that c(A∗) is a singleton for all
A∗ and c satisfies α. Also assume that c(A)∩B != ∅. Then, c(A) = {x} for some x ∈ A∩B.
Let {y} = c(A ∪ B). If y ∈ A, then α and single-valuedness implies {y} = c(A) = {x}.
Hence, by α, we have {y} = c(B) implying c(B) ∩ A ⊂ c(A) as desired. If y /∈ A, then
{y} = c(B) and c(B) ∩ A = ∅ ⊂ c(A), again, as desired.