Do you mean by this that theonlysubsequential is 0? Any sequence converging to 0 will do.

Find a sequence, , that converges to 0, a sequence, , that converges to 1, and "interleave" them: .b) limit point of subsequence = {0,1}

It is a property of all real numbers, that, given any , there exist a rational number within distance of the real number. Since the rational numbers are countable, we can order them: . That is, the set of all rational numbers form a sequence having all real numbers as subsequential limits, and there are, of course, an infinite number of them.c) limit point of subsequence = {infinitely many points}

Since there are an infinite number of rational numbers in 0, 1, any example for d is also an example for c. But, unfortunately, the example I gave for c does not work since its set of sequential points is too big- it includes much more than the rational numbers. I don't think you will be able to give a specific, numerical, example. You might do something like this:d) limit point of subsequence = Q ∩ [0,1]

Since the set of rational numbers in [0, 1] is countable, it is possible to put them into a sequence: . It is a property of all real numbers, and in particular of rational numbers, that given any rational number, , there exist a sequence of rational numbers, [tex]\{a_{ij}\}[tex] that converges to that rational . Now "interleave" those: .