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?


Question: Drupal redirects to old url after moving to new server
Answer: Maybe you can use this module: 

Path Redirect (Drupal 6)

Redirect (Drupal 7)