I have only the uniform convergence case for (c), (d), and (e) to consider:
Assume fn converges uniformly to f on a set A, then decide which are true.
(c) If each fn has a finite number of discontinuities, then f has a finite number of discontinuities.
(d) If each fn has fewer than M discontinuities (where M is a natural number fixed), then f has fewer than M discontinuities.
(e) If each fn has at most a countable number of discontinuities, then f has at most a countable number of discontinuities.
Thanks for the comments. Yes I should include my attempt. And accordingly 6.2.10. was just removed.
For (c) and (e), I am trying to use Cantor set. Take [0, 1] as the domain. I want the limit function f to be 1 on every point of Cantor set and 0 otherwise so that I create a function with uncountably many discontinuities. Let f1 take the value of 1 on [0, 1/3] & [2/3, 1] and the value of 0 otherwise; Let f2 take the value of 1-1/2 on [0, 1/9] & [2/9, 1/3] & [2/3, 7/9] & [8/9, 1] and value 0 otherwise..., then I get a sequence (fn) converging uniformly to f.
Then for any given ε, there is an N such that for any n greater than N, fn(x) is either 0 (which is the same as f(x)) or sufficiently close to f(x), for any x in [0, 1].
So (c) and (e) are both wrong.
Am I right here?