Bah, my computer ate my post! Well, here it is again I guess.
1) Find a function which is discontinuous at infinitely many points
lol, that would be the stair-step equation
Yep, that's exactly what I was thinking of.
f(x) = FloorOf(x)
Damn, the lack of math symbols that have plagued computers for so long!
2) Find a function which is discontinuous at an uncountably infinite number of points.
same as #1, just stated in different terms that make it redundant
Nope, the step function is only discontinuous at countably infintly many points. (At all the integers).
3) Find a function which is continuous at only a countably infinte number of points.
define countably infinate...
Ahh, now we get to the good question.
Formally, a countable set can have a bijection formed between it and a subset of the natural numbers. (A bijection is a one-to-one and onto function, basically it creates a pairing between elements of one set with elements of another set. Basically, a countable set can be "listed" in such a way that no element is missed. Also, sets are considered to be subsets of themselves. So a subset of the natural numbers would include the natural numbers themselves.)
For instance, you can list the integers so that none are missed:
0, 1, -1, 2, -2, 3, -3, 4, -4, ...
(Hence they have the pairing (1, 0), (2, 1), (3, -1), (4, 2), (5, -2), (6, 3), (7, -3), ...)
You can also list the positive rational numbers in a special way, where the rational numbers are laid out in a 2D array with numerator and denomintors increasing along corresponding axis. You can then traverse diagonallay through the array. Note: You can't traverse along each row or column since they are infinte and you would never get to the next row or column, hence missing all those numbers. Also, as a technicality, you should skip over repeated numbers such as 2/2 = 1/1. Anyways, once you have a list of positive rational nubmers, you can form a list of all rational numbers in the same was as the integers were listed above.
Non countable sets would include the irrational numbers and the real numbers. There is no way to pair them with the natural numbers and not miss any of them.
Also, just because a set is strictly larger or smaller than another set, doesn't mean they have different cardinality. That holds only in the finite case. For instance, the natural numbers are a subset of the integers which are a subset of the rational numbers, but they all have the same cardinality (countably infinite). Thus, even though there are "more" rational numbers than natural numbers, these sets have the same "size". (At least if you're defining size in terms of cardinality, which it usually is).
impossible, as for it to be continuous, it must have other poins of reference that would be continuous as well
Entirely possible. Continuity is defined at a point, the properties at that point need not hold for surrounding points. Granted, the definition requires considering "nearby" points, but there is no specification as to how near those points are. Essentially what you need to do is find a way to break up "nearby" points everywhere except at the one point you want to be continuous. At this point, "nearby" points bear similarity to the continuous point, but not so much to each other.
6) Find a continuous function which if not differentiable anywhere.
lol, if I had time, I would answer this, but I would need at least half an hour to write and check it
This one is kinda tough. I don't think I could write a proof of it in half an hour without my textbook handy. I'm tempted to give a hint to this one. Essentially what you want is sharp corners, so the function is not differentiable at the point, but the function is still continuous. But to be non differentiable everywhere, you need these sharp corners essentially everywhere (well, "almost everywhere"). There are a few hints I'd like to give about how to go about doing this but I figure I should let the question stand for a little while longer.
