Here's a partial answer (how to deal with the bottom bracket, not what bike it is):

The easiest solution (and likely best) is to buy the universal bottom bracket from Velo Orange (or similar). These basically expand to fit in, and don't affect or use the threads so they'll work on most standard BB road bike frames. It will be JIS (square taper). There is some more information at this link. 

I wouldn't bother trying non-square taper -- you're likely not to find a way to fit anything else in. If you want to match the original threading, you'll likely have to buy from Phil Wood or Velo Orange and end up with a square taper anyway. Phil Wood will cost a lot of money, but Velo Orange will be around 40-50 bucks for the right threading (and around 70 for the universal). 

Then, just buy your favorite square taper crankset (there are still a lot of good ones on the market) and go to town.

Identify my Motobécane racing bike
This is problem 17 in baby Rudin's chapter on continuity. He has a hint to use triplets of rationals that bound each simple discontinuity on the left, right, and in between the values of the limits from the left and right. It seems like this can be weakened to just rationals to the left and right. 

Simple discontinuities are those in which the limit from the left and right exist, so there must be intervals to the left and right of a simple continuity on which no other simple discontinuity can exist. More precisely, let $c$ and $c'$ be simple discontinuities for $f$ on $(a,b)$ and consider the limit, $l$, of $f$ approaching $c$ from the left:

$\forall \epsilon &gt;0, \exists \delta&gt;0 : c-x&lt;\delta \Rightarrow |l-f(x)|&lt;\epsilon$

but $ \exists \epsilon '&gt;0 \forall \delta ' &gt;0 : |c'-x|&lt;\delta ' \Rightarrow |f(c')-f(x)|&gt;\epsilon '
\\ \therefore \epsilon = \epsilon ' , \delta ' = \delta \rightarrow \leftarrow
$

Therefore, you can make an injection from the set of simple discontinuities to a subset of rationals by associating each simple discontinuity with one rational in the aforementioned "free" interval to its left; from there compose with the map from rationals to integers to show countable. Is this argument correct?
